UNIT 6 ( PAGE 2)

ENTHALPY (H) AND ENTHALPY CHANGE(H)
Energy change occurring during the reaction at constant temperature and volume is given by internal energy change. However, most of the reactions in the laboratory are carried out in open beakers or test tubes etc. In such cases, the reacting system is open to atmosphere . Since atmospheric pressure is almost constant therefore, such reactions may involve the changes in volume. The energy change occurring during such reactions may not be equal to internal energy change. In order to understand this let us assume that a chemical reaction involving gaseous substances which proceeds with the evolution of heat. When the reaction is carried at constant pressure, two possibilities arise.
i) If the reaction proceeds with increase in volume, the system has to expand against atmospheric pressure and energy is required for this purpose. The heat evolved in this case would be little less than the heat evolved at constant volume.
ii) If the reaction proceeds with decrease in volume at constant pressure, the work is done on the system and heat evolved will be greater than the heat evolved at constant volume.
Thus, it can be concluded that heat changes occurring at constant pressure and constant temperature are not simply due to the changes in internal energy alone , but also include energy changes due to expansion or contraction against the atmospheric pressure. In order to study the changes of chemical reactions at constant temperature and pressure, a new function enthalpy is introduced. Enthalpy is the total energy associated with any system which includes its internal energy and also energy due to environmental factors such as pressure-volume conditions. This can be understood as follows :
A substance has to occupy a space in its surroundings depending up on its volume (V). It does so against the compressing influence of the pressure(P). Due to this, the substance possesses an additional energy called PV energy.
The sum of energy and P-V energy of any system , under a particular set of conditions is called Enthalpy. It is denoted by H.
Mathematically, it may be put as :
H = U + P V
Enthalpy is a state function and also known as heat content of the substance.
It is not possible to determine the absolute value of enthalpy of a system because absolute value of internal energy (U) is not known. However, change in enthalpy (H) taking place during the process can be determined. Change in enthalpy is equal to difference between the enthalpies of products and reactants. The change in enthalpy may be expressed as :
H = H(products)  H(reactants)
= HP  HR
The enthalpy change of a reaction is equal to the heat absorbed or evolved during a reaction at constant temperature and pressure.
Thus,
H = Heat evolved or absorbed in a reaction at constant
temperature and pressure.
The change in enthalpy (H) is experimentally determined by carrying out the reaction at constant pressure in a well-insulated calorimeter. The heat evolved or absorbed changes the temperature of the system. From the change in temperature, the heat evolved or absorbed during the change can be calculated. This heat change is equal to the enthalpy change (H ).
Origin of Enthalpy change in a reaction
During chemical reactions certain bonds are cleaved and certain bonds are formed. Energy is required to break bonds and is released when new bonds are formed. Enthalpy changes taking place during reactions can be related to these energy changes. As an illustration, let us consider a reaction which involves only gaseous species, such as :

For such reactions, taking place at constant pressure, enthalpy change is given as under :

In the case of above mentioned reaction, the change should be equal to the difference between the energy required to break HH and ClCl bonds and the energy released when two HCl bonds are formed.

Energy required to break HH bond = 437 kJ mol1
Energy required to break ClCl bond = 244 kJ mol1

Energy released in the formation
of HCl bond = 433 kJ mol1
Therefore for the reaction :
H2(g) + Cl2(g)  2 HCl (g)
Enthalpy change H = (437+ 244)  (2 x 433)
= 185 kJ mol1
If the reaction involves reactants in dissolved state then the interaction between the reactants and solvents have to accounted for. Similarly, for liquid or solid reactants the interactions among the neighboring molecules have to be considered.
Problems
15. Calculate the enthalpy change for the reaction :
C(s) + O2(g)  CO2(g)
Given the energy required for sublimation of one mole of carbon is 716.7 kJ mol1. Also the energy required to break one mole of O=O and C=O bonds is 494 kJ mol1 and 707 kJ mol1 respectively.
Relationship between H and U
Consider a reaction involving gases. Let the process be isothermal and carried out at constant pressure (P). If VA be the total volume of the gaseous reactants and VB be the total volume of the gaseous products, also nA be the number of moles gaseous reactants and nB be the number of moles of gaseous products, then :
P VA = nA R T
P VB = nB RT
P VB  P VA = (nB  nA) R T
P( VB  VA ) = (nB  nA) R T
P V = n R T …….. (9)
n , here refers to the change in number of moles of gaseous species during the process.
H = U+ PV ………(10)
Equation (9) can also be written as :
or H = U + ng R T
qP = qv + PV
or qP = qv + ng R T
The difference between H and U will depend up on the magnitude as well as the sign of n. For example :
If n = 0 ; H = U
n =  ve ; H < U and n = +ve ; H > U



Note
i) For chemical reactions, involving no change in volume, or those involving same number of gas moles in reactants and products :
qP = qv or H = U
ii) While solving numerical problems, R must be taken equal to 0.002 k cal mol1 if H , U values are in k cal mol1and it is taken to be 8.314 J K1mol1, if H and U values are in joules.
Problems
16. The enthalpies of all elements in their standard states are :
(i) unity (ii) zero
(iii) <0 (iv) different for each element 17. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100C is 14 kJ mol1. Calculate the internal energy change , when : (i) 1 mol of water is vapourised at 1 bar pressure and 100C. (ii) 1 mol of water is converted into ice. 18. The enthalpy change for the reaction : N2(g) + 3 H2(g)  2 NH3(g) is 92.3 kJ at 298 K. What is U at 298 K ? 19. Calculate the difference between the heats of reaction at constant pressure and constant volume for the reaction at 298 K in Joules ? 2 C6H6(ℓ) + 15 O2(g)  12 CO2(g) + 6 H2O(ℓ) 20. The heat of combustion of methane is measured in a bomb calorimeter at 298.2 K and is found to be  885.50 kJ / mol. Find the value of enthalpy change H. HEAT CAPACITY ( C) The heat capacity of a system is defined as the amount of heat required to raise the temperature of the system through 1C. If q is the amount of heat supplied to a system and as a result , the temperature of the system rises from T1 to T2 , then the heat capacity (C) of the system is given by However, since the heat capacity varies with temperature, therefore the value of C has to be considered over a very narrow temperature range. Thus if dq is the small amount of heat absorbed by a system which raises the temperature of the system by a small amount dT ( say from T to T + dT) , then the heat capacity of the system will be given by The specific heat of a substance is defined as the amount of heat required to raise the temperature of 1 gram of the substance by 1C. If instead of 1 g , one mole of the substance is taken , the term used is called ‘molar heat capacity’. Molar heat capacity of a substance is defined as the amount of heat required raise the temperature of one mole of the substance through 1C. The specific heat capacity ,c , is the quantity of heat required to raise the temperature of unit mass of a substance by one degree celsius (or one kelvin). For finding out the heat , q required to raise the temperature of a sample , we multiply the specific heat capacity of the substance, c, by the mass m and temperature change , T. q = c m T The specific heat capacity of water is 4.18 J/g K and in terms of calories , it will be 1.00 cal/g K Problem 21. It has been found that 60.8 J are required to change the temperature of 25.0 g of ethylene glycol by 1 K. Calculate the specific heat capacity of ethylene glycol ? Types of heat capacities or molar heat capacities Since ‘q’ is not a state function and depend upon the path followed, therefore C is also not a state function. Hence to know the value of C , the conditions such as constant volume or constant pressure have to be specified which define the path. Thus there are two types of heat capacities, which are: (i) Heat capacity at constant volume (Cv) (ii) Heat capacity at constant pressure (Cp) The heat supplied to a system to raise its temperature through 1C.keeping the volume of the system constant is called heat capacity at constant volume. Similarly , the heat supplied to a system to raise its temperature through 1C keeping the external pressure constant is called heat capacity at constant pressure. According to the first law of thermodynamics , we know that When the volume is kept constant, dV = 0 and therefore , equation (4) becomes For an ideal gas , this equation may be simply be written as Thus heat capacity at constant volume may be defined as the rate of change of internal energy with temperature at constant volume. When the pressure is kept constant during the absorption of heat, equation (4) becomes Also we know that the heat content or ethalpy of a system is given by : H = U + P V Differentiating w.r.t T at constant P , we get Combining equations (7) and (8), we get For an ideal gas this equation may simply be put in the form Thus heat capacity at constant pressure may be defined as the rate of change of enthalpy with temperature at constant pressure. Relationship between Cp and Cv If the volume of the system is kept constant and heat is added to a system, then no work is done by the system. Thus the heat absorbed by the system is used up completely to increase the internal energy of the system. Again if pressure of the system is kept constant and heat is supplied to the system, then some work of expansion is also done by the system in addition to the increase in internal energy. Thus if at constant pressure , the temperature of the system to be raised through the same value as constant volume, then some extra heat is required for doing the work of expansion. Hence Cp > Cv.
The difference between the heat capacities of an ideal gas can be obtained by subtracting equation (6) from equation(10). So we have

But H = U + P V (by definition)
and P V = R T ( for 1 mol of an ideal gas)
H = U + RT
Differentiating w . r. T, we get

Combining equations (11) and (13) we get

for 1 mole of an ideal gas. Thus Cp is greater than Cv by the gas constant R i.e., approximately 8.314 Joules.
MEASUREMENT OF INTERNAL
ENERGY CHANGE (U)
The heat changes in reactions can be measured with the help of calorimeters. There are different types of calorimeters depending upon the requirements of a particular experiment. In general, the reactions taking place at constant volume and involving gases are carried out in a closed container with rigid walls which could withstand high pressure developed. One such vessel is bomb calorimeter. It is made of heavy steel. The steel vessel is coated inside with gold or platinum to avoid oxidation of steel during the chemical reactions. The vessel is fitted with tight screw cap. There are two electrodes R1 and R2 which are connected to each other through a platinum wire S shown in Fig 3. The wire S remains dipping in a platinum cup below it. A small amount of substance under investigation is taken in a platinum cup. The vessel is then filled with excess of oxygen at about 20 to 25 atmospheres and is subsequently sealed. It is now dipped in an insulated water bath which is also provided with a mechanical stirrer and a thermometer, sensitive enough to read up to 0.010C (Beckmann thermometer).

The Bomb calorimeter
The initial temperature of water is noted and the combustion is initiated by passing electric current through the platinum wire. The heat evolved during the chemical reaction rises the temperature of water which is recorded in the thermometer. By knowing the heat capacity of the calorimeter and also the rise in temperature, heat of combustion at constant volume can be calculated by using the expression.
U = Z T ( M / m)
where,
Z = Heat capacity of calorimeter system.
T = Rise in temperature
M = Molecular mass of the substance
m = mass of substance taken
MEASUREMENT OF ENTHALPY CHANGE (H)
The enthalpy change of a reaction is measured in a calorimeter. The calorimeter used is kept open to atmosphere.

Calorimeter for measuring heat changes at constant pressure (atmospheric pressure)
The calorimeter is immersed in an insulated water bath fitted with stirrer and thermometer. The temperature of the bath is recorded in the beginning and after the end of the reaction and change in temperature is calculated. Knowing the heat capacity of water bath and calorimeter and also change in temperature, the heat absorbed or evolved can be calculated. This gives the enthalpy change H of the reaction.
Since H = qp (at constant pressure P) and therefore , heat absorbed or evolved , qp at constant pressure is called the heat of reaction or enthalpy of reaction, rH.
In an exothermic reaction, heat is evolved and the system loses heat to surroundings. Therefore , qp will be negative and rH will also be negative . Similarly in an endothermic reaction, heat is absorbed , qp is positive and rH will be positive.
EXOTHERMIC AND ENDOTHERMIC REACTIONS
The chemical reactions which proceed with evolution of heat energy to the surroundings are called exothermic reactions. Therefore qp will be negative and H will also be negative. The chemical reactions which proceed with absorption of heat energy are called endothermic reactions. Since heat is absorbed qp is positive and H will also be positive. In brief, H will be positive for endothermic reactions and negative for exothermic reactions. In exothermic reactions , loss of energy shows that reactants were at higher level of enthalpy than the products. Similarly in endothermic reactions, the reactants are at lower level of enthalpy than the products. This is represented in Fig.

Exothermic reactions
H2(g) + ½ O2(g)  H2O(ℓ) : H =  286 kJ
CH4(g) + O2(g)  CO2(g) + 2 H2O(ℓ) ; H = 890 kJ
N2(g) + 3H2 (g)  2 NH3(g) ; H = 890 kJ
Endothermic reactions
C(s) + 2 S(ℓ)  CS2(ℓ) ; H = + 92.00 kJ
H2(g) + I2(g)  2 HI(g) ; H = + 53.6 kJ
C(s) + H2O(g)  CO(g) + H2(g) ; H = +131.2 kJ
Problems
Problem
22. 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmosphere pressure according to the equation :
C(graphite) + O2(g)  CO2(g)
During the reaction , temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ? (N 164 P. 6.6)
23. (a) The energy needed to raise the temperature of 10.0 g
of iron from 25C to 500C if specific heat capacity of
iron is 0.45 J(C)1g1.
(b) What mass of gold (specific heat capacity
0.13J(C)1g1 can be heated through the same
temperature difference when supplied with the same
amount of energy as in (a)
24. The reaction of cyanamide , NH2CN(s) , with dioxygen was carried out in a bomb calorimeter and U was found to be 742.7 kJ mol1 at 298K. Calculate enthalpy change for the reaction at 298 K. (N 182 Ex 6.8)
NH2CN(g) + (3/2) O2(g)  N2(g) + CO2(g) + H2O(ℓ)
25. A 1.25 g sample of octane (C8H18) is burned in excess of oxygen in a bomb calorimeter. The temperature of calorimeter rises from 294.5 K to 300.78 K. If heat capacity of the calometer is 8.93 kJ/K , find the heat transferred to calorimeter.
26. 20.0 g of ammonium nitrate is dissolved in 125 g of water in a coffee-cup calorimeter, the temperature falls from 296.5 K to 286.4 K. Find the value of q for the calorimeter(Hint : Treat heat capacity of water as heat capacity of the calorimeter and its content)
27. A chemist while studying the properties of gaseous C2F2Cl2, a chlorofluoro carbon refrigerant , cooled a 1.25 g sample at constant atmosphere pressure of 1.0 atm from 320K to 293 K. During cooling , the sample volume decreased from 274 to 248 mL. Calculate H and U for the fluorochloro carbon for this process. For C2F2Cl2, Cp = 80.7 J/(mol K)
28. 0.562 g of graphite kept in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure was burnt according to the equation.
C(graphite) + O2(g)  CO2(g)
During the reaction, the temperature rises from 298 K to 298.9 K. If the heat capacity of calorimeter and its content is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ?
ENTHALPY CHANGE , rH OF A REACTION – REACTION ENTHALPY
In a chemical reaction , reactants are converted into products and is represented by,
Reactants  Products
The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction , is given by rH.
rH = [ sum of enthalpies of products] 
[ sum of enthalpies of reactants ]
=  ai Hproducts   bi Hreactsnts
i i
Here symbol  is used for summation and ai and bi are stoichiometric coefficients of products and reactants respectively in the balanced chemical reaction. For example, for the reaction
CH4(g) +2 O2(g)  CO2(g) + 2 H2O(ℓ)
rH =  ai Hproducts   bi Hreactsnts
i i
= [ Hm (CO2, g) + 2 Hm (H2O, ℓ )] 
[ Hm ( CH4 , g) + 2 Hm ( O2 , g )
Here Hm is the molar enthalpy.
Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
STANDARD ENTHALPY OF REACTIONS
Reaction enthalpies depend upon the conditions under which a reaction is carried out. It is therefore necessary that we must specify some standard condition. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances (elements and compounds) are in their standard states. The standard state of a substance at specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 298 K and 1 bar. ; the standard state of solid iron at 500 K is pure iron at 500 K and 1 bar. Usually the data chosen are at 298 K . The standard enthalpy change for a reaction or for physical process is the difference between enthalpy of products in their standard states and enthalpy of the reactants in their standard states all at the same specified temperature.
Standard conditions are denoted by adding the superscript Ѳ to the symbol H. For the combustion of one mol of methane under standard conditions , we write the equation
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(ℓ)
: cHѲ=  890.4 kJ/mol
Here the subscript ‘c’ is used for combustion in standard states. If the enthalpy change is measured at 298 K , we add 298 K and denote as cHѲ (298 K). The equation for combustion of methane indicates that in the combustion of 1 mol of CH4 in sufficient oxygen or air , 1 mol of CO2 and 2 mol of H2O are formed and 890.4 kJ of heat is released at a pressure of 1 bar and at 298 K. It might seem surprising that the combustion can apparently be carried out at 298 K. In fact, when methane is ignited , its temperature rises rapidly as it burns and produces heat, so that combustion is in fact not occurring at 298 K. But if reactant(s) are at 298 K at the start of the reaction and products are at 298 K at the completion of the reaction, all eventually transferred to the calorimeter during measurement. If the reactants are initially at 298 K and products formed also finally come to a temperature of 298 K, it is immaterial that during the reaction temperature is higher than 298 K. Similarly in endothermic reaction, which is started with the reactants at 298 K and proceeds with an absorption of heat, the products are finally to be at 298 K and as such heat has to be supplied to the system from some external source(surroundings). This heat taken from the external source at one bar will be the measure of standard enthalpy of the reaction. For a reaction, standard enthalpy change is denoted by rHѲ
Thermochemical equations
A balanced chemical equation together with a designation of its value of H is called a thermochemical equation. For example,
C2H5OH(ℓ) + 3 O2(g)  2 CO2(g) + 3 H2O(ℓ) ;
rHѲ =  1367 kJ/mol
The above equation describes the combustion of liquid ethanol , C2H5OH(ℓ) at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction.
The reverse reaction would require the absorption of 1367 k J under the same conditions, i.e., endothermic.
2 CO2(g) + 3 H2O(ℓ) C2H5OH(ℓ) + 3 O2(g) ;
rHѲ = +1367 kJ/mol
Thus reactions which are endothermic in the forward direction are exothermic in the reverse direction. This rule applies to both chemical and physical processes. This is according to the law of conservation of energy.
Conventions for writing Thermochemical Equations
1. The coefficients in a balanced thermochemical equation refer to the number of moles of reactants and products involved in the reaction. In the thermodynamic interpretation of an equation , we never interpret the coefficients as number of molecules. Thus , it is acceptable to write as fractions rather than as integers , whenever necessary.
2. The numerical value of rHѲ refers to the number of moles of substances specified by an equation. Standard enthalpy change rHѲ is , therefore expressed as enthalpy change per moles shown in equation. In brief , the unit of rHѲ will remain kJ /mol even if more than one mole of the reactant or product are involved but its magnitude will change.
3. When a chemical equation is reversed, the value of H is reversed in sign. For example,
N2(g) + 3 H2(g)  2 NH3(g) : rHѲ =  91.8 kJ/mol
2 NH3(g)  N2(g) + 3 H2(g) : rHѲ = + 91.8 kJ/mol
4. Physical states of all species is important and must be specified in a thermochemical equation as H depends on the phase of the substances. For two states of water in the following two equations , value of rHѲ is different.
2 H2(g) + O2(g)  2 H2O(g) : rHѲ =  483.7 kJ/mol
2 H2(g) + O2(g)  2 H2O(ℓ) : rHѲ =  571.6 kJ/mol
5. The value of rH usually shows small change with increase of temperature (provided there are no phase changes when the temperature is increased).
6. In case the coefficients in the chemical equation are multiplied or divided by some number, the H value must also be multiplied or divided by the same number. For example,
H2(g) + ½ O2(g)  H2O(ℓ) : H = 286 kJ
If the whole equation is multiplied by 2 , the H for the
new thermochemical equation is given as :
2 H2(g) + O2(g)  2 H2O(ℓ) : H = 2 x ( 286) kJ
Problems
29. Red phosphorus reacts with liquid bromine in a exothermic reaction
2 P(s) + 3 Br2 (ℓ)  2 PBr3 (g) : rHѲ = 243kJ/mol
Calculate the enthalpy change when 2.63 g of phosphorus reacts with an excess of bromine in this way.
Factors on which Enthalpy of a reaction(H) depend
The enthalpy of a reaction depends on the following factors :
1. Physical state of reactants and products: The change of physical state also involve heat changes. Thus, the value of H does depend upon the physical state of the reactants and products.
2. Quantities of reactants : The amount of heat evolved or absorbed depends upon the amounts of reactants. For example, combustion of 12 g of carbon produces 393.5 kJ , while combustion of 1.2 g of carbon produces 39.35 kJ of heat.
3. Allotropic modifications : For different allotropic forms of the substances different amount of heat is evolved or absorbed.
C(graphite) + O2(g)  CO2(g) ; H = 393.5 kJ
C(diamond) + O2 (g)  CO2(g) ; H = 395.4 kJ
4. Temperature : The value of H also depends upon the temperature at which reactants and products are considered. For example for the reaction :
H2(g) + Cl2(g)  2 HCl (g)
The value of H at 298 K and 348 K are 184.6 kJand
184.8 kJ respectively.
5. Conditions of constant pressure and volume The heat of reaction also depends upon the conditions of constant volume or constant pressure at a particular temperature
VARIOUS TYPES OF ENTHALPIES OF REACTIONS
1. Enthalpy of formation f H : It is the enthalpy change accompanying the formation of one mole of a compound from its constituent elements. It is generally denoted by f H . For example, enthalpy of formation of carbon dioxide and ethyl alcohol are 393.5 and 277.0 kJ/mol respectively. These are expressed as :
C(s) + O2(g)  CO2(g) ;  f H = 393.5 kJ
2 C(s) + ½ O2(g) +3 H2(g)  C2H5OH(ℓ); f H = 277.0 kJ
When all the species of the chemical reactions are in their standard states, the enthalpy of formation is called Standard Heat of Formation. The standard enthalpy of formation is denoted by fHѲ . The standard heat of formation of a compound is defined as the heat change accompanying the formation of one mole of a compound from its constituent elements , all substances being in their standard states( 1 atm pressure and 298 K).
The standard heats of formation of compounds are also called their standard enthalpies and they are denoted as HѲ. The values of standard enthalpies of formation of some compounds are given below :
Standard Heats of formation of some compounds.
Substance fHѲ Substance fHѲ
H2O(ℓ) 28.96 CO(g) 111.3
HCl(g) 92.4 CO2 393.5
HBr 36.12 CH4(g) 7.18
SO2 298.2 C2H4 +52.5
H2S 20.16 C2H5OH 278.9
NO2 +33.5 H2SO4 814.8
It may be noted that by convention standard enthalpies of elementary substances are taken to be zero.
Importance of enthalpies of formation
The knowledge of standard enthalpies of various substances can help us to calculate the standard enthalpy change of any reaction. Standard enthalpy change of a reaction is equal to the difference of the standard enthalpies of all the products and standard enthalpies of all the reactants.

or rHѲ =  f HѲ (products)   f HѲ (reactants)
For a hypothetical reaction :
a A + b B  ℓ L + m M
rH0 =  f HѲ (products)    f HѲ (reactants)
= [ ℓ  f HѲ (L)+m  f HѲ (M)]  [ a  f HѲ (A)+ b  f HѲ(B)]
Problems
29. Calculate the enthalpy change for the reaction :
CO2(g) + H2(g)  CO (g) + H2O(g)
Given that  f HѲ for CO2(g), CO, H2O(g) as 393.5 ,
 111.3,  241.8 kJ/mol respectively.
30. Calculate the standard enthalpy of formation of C2H4(g) from
the following thermochemical equation:
C2H4(g) + 3 O2 (g)  2 CO2(g) + 2 H2O(g)
: HѲ = 1323 kJ.
Given that  f HѲ of CO2(g) , H2O(g) as 393.5 and 249 kJ/mol respectively.



2. Heat of combustion or enthalpy of combustion
It is the heat change accompanying the complete combustion of one mole of a substance in excess of oxygen or air. Some examples are given below:
i) Combustion of carbon.
C(s) + O2(g)  CO2(g) ; H = 393.5 kJ
ii) Combustion of methane.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(ℓ) ;cH =  891 kJ
iii) Combustion of butane.
C4H10(g) +(13/2)O2(g)  4 CO2(g) + 5 H2O(ℓ)
; cH = 2658 kJ
Combustion reactions are always accompanied by the evolution of heat, therefore for such reactions the H is always negative.
Problems
31. Uө of combustion of methane is  X kJ mol1. The value of Hө is : (N 182 Ex 6.4)
(i) = Uө (ii) > Uө (iii) < Uө (iv) = 0
32 i) A cylinder of cooking gas is assumed to contain 11.2 kg of butane. The thermochemical equation for combustion of butane is :
C4H10(g) +(13/2)O2(g)  4 CO2(g) + 5 H2O(ℓ)
; cH = 2658 kJ
If the family needs 15,000 kJ of energy per day for cooking, how long would cylinder last ?
ii) Assuming that 30% of gas is wasted due to
incomplete combustion, how long would the cylinder last ?
Combustion of food in our body
Our body requires energy for performing various functions. This energy mainly comes from carbohydrates and fats that we take as food. Carbohydrates are decomposed into glucose or its derivatives in the stomach. Glucose is soluble in blood. It is transported by blood cells to various parts of the body where it reacts with oxygen in a series of steps producing carbon dioxide, water and energy.
C6H12O6 (s) + 6 O2(g)  6 CO2(g) + 6 H2O
; cH = 2900 kJ
This combustion of glucose in the human body take place at the body temperature because of the catalytic action of enzymes.
Efficiencies of Foods and Fuels
Just as the major source of energy in our bodies is combustion of foods , the major sources of energy for industries is combustion of fuels. The efficiency of foods and fuels is compared in terms of Calorific Values. The calorific value of food or fuel is defined as the amount of energy (heat) produced when one gram of the substance (food or fuel) is completely burnt. The calorific values of some common foods and fuels are given below :
Food Calorific value
( kJ / g) Fuel Calorific value
( kJ / g)
Curd 2.5 wood 17
Milk 3.2 Charcoal 33
Egg 7.3 kerosene 48
Meat 12.0 Fuel oil 45
Honey 13.3 Butane(LPG) 55
Ghee 37.6 Hydrogen 150


3. Heat of Neutralisation or Enthalpy of Neutralisation
It is the enthalpy change accompanying the complete neutralisation of one gram equivalent of an acid by a base or vice versa in dilute aqueous solutions. For example, enthalpy change accompanying the neutralisation of NaOH and HCl is represented as :
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(ℓ)
; nH = 57.1 kJ
The heat of neutralisation of all strong acids and strong bases is always constant i.e., 57.1 kJ. The explanation to this generalisation can be provided on the basis of Arrhenius theory of ionisation. The strong acids and bases are almost completely ionised in dilute aqueous solutions. The neutralisation of strong acid and strong bases simply involve the combination of H+ ions(from acid) and OHions (from base) to form water molecules. It has been found experimentally that when one mole of water is formed from H+ ions and one mole of OH ions 57.1kJ of energy is released.
H+(aq) + OH(aq)  H2O(ℓ) ; H = 57.1 kJ
Now the net process in the neutralisation of HCl(aq) and NaOH(aq) can be represented as :
H+(aq) + Cl (aq) + Na+(aq) + OH(aq) 
Na+(aq)+Cl(aq)+H2O(ℓ)
Cancelling common ions on both sides, we get:
H+(aq) + OH(aq)  H2O(ℓ)
In a similar way, the neutralisation of HNO3(aq) and KOH(aq) can be represented as :
H+(aq) + NO3(aq) + K+(aq) + OH(aq) 
K+(aq)+ NO3(aq) + H2O(ℓ)
H+(aq) + OH (aq)  H2O(ℓ)
Thus during the neutralisation of all strong acids and bases the same reaction takes place. Hence the value of enthalpy of neutralisation of strong acid by a strong base is constant.
However, if either acid or base is weak, then its ionisation is not complete in solution. Therefore part of the energy liberated during combination of H+ and OH ions is utilised for the ionisation of weak acid(base). Consequently, the value of enthalpy of neutralisation of weak acid /strong base or strong acid/weak base is numerically less than 57.1 kJ. For example, enthalpy of neutralisation of acetic acid and sodium hydroxide is 56.1 kJ. The neutralisation of acetic acid and sodium hydroxide can be represented as:
CH3COOH ⇌ CH3COO + H ; Hionisation = +1.0 kJ
(weak acid)
NaOH  Na+ + OH
H+(aq) + OH(aq)  H2O(l) : H = 57.1 kJ
Net reaction:
CH3COOH + Na+ + OH ⇌ CH3COO + Na+ + H2O
; H = 56.1 kJ
or CH3COOH + OH ⇌ CH3COO + H2O : H = 56.1 kJ
Thus, enthalpy of neutralisation of acetic acid and sodium hydroxide is 56.1 kJ. Similarly, enthalpy of neutralisation of ammonium hydroxide (weak base) and hydrochloric acid (strong acid ) is 51.5 kJ.
Experimental Determination of Heat of neutralisation
The enthalpy of neutralisation of an acid with a base or vice versa can be determined with the help of a simple calorimeter consisting of a polythene bottle.
The bottle is fitted with a cork having two holes through which a thermometer and stirrer are passed as shown in Fig.

Experimental determination of enthalpy of neutralisation.
Let us discuss the method to determine the enthalpy of neutralisation of HCl and NaOH.
A known volume (say 100 cm3) of HCl solution of known concentration(say 0. 5M ) is taken in a polythene bottle. An equal volume of 0.5 M NaOH is taken in another polythene bottle. The temperature of the two solutions are recorded. If their temperature is not same, then the solution having lower temperature is stirred with a test tube containing boiling water till temperatures of the two solutions become equal. The solutions are then mixed and throughly stirred. The final temperature is recorded. The heat of neutralisation is calculated as follows:
The initial temperature of the base = T1
The final temperature of the solution after mixing = T2
Rise in temperature = (T2  T1 )
For the purpose of simplicity, the heat capacity of the polythene bottle may be neglected because it is very small as compared to that of solution. The specific heat capacity and specific gravity of the solution are assumed to be same as that of water.

= 200 x 4.2 J/K

= 840 (T2  T1 ) J
Heat of neutralisation is the amount of heat released during neutralisation of one equivalent of acid. In the case of HCl, one mole of HCl can give one mole of H+ ions, therefore,
Gram equivalent of HCl = Gram molecular mass of HCl

Problems
33. What would be the heat evolved when :
a) 0.5 mole of HCl is neutralised by 0.5 mole of NaOH in aqueous solutions.
b) An aqueous solution containing 0.5 mole of HNO3 is mixed with an aqueous solution containing 0.3 mole of NaOH.
c) 200 cm3 of 0.1 M H2SO4 is mixed with 150 cm3 of 0.2 M KOH.
34. Acetic acid and hydrochloric acid react with KOH solution. The enthalpy of neutralisation of acetic acid is 55.8 kJ/mol while that of hydrochloric acid is  57.3 kJ/mol. Can you think of how are these different ?
35. The heat evolved in the combustion of glucose is shown in
the following equation:
C6H12O6(s) + 6 O2(g) 6 CO2g) + 6 H2O(ℓ) ;
H = 2,840 kJ.
What is the requirement of energy for the production of
0.36 g of glucose ?
4. Enthalpy of solution
It is the enthalpy change accompanying the dissolution of one mole of a substance in a large excess of solvent so that further addition of solvent does not produce any more heat change.
If water is the solvent, then the symbol aq(aqueous) is used to represent large dilution. For example, the thermochemical equations for the dissolution of KCl and CuSO4 are :
KCl(s) + aq  KCl(aq) ; H = +18.6 kJ
CuSO4(s) + aq  CuSO4(aq) ; H = 66.5 kJ
It may be noted that :
i) Hsol of hydrated salts like CuSO4.5H2O, CaCl2.6H2O etc. or salts which do not form hydrates like NaCl, KCl, NH4Cl etc. is generally positive.
ii) Hsol of anhydrous salts which form hydrates like CuSO4 etc. are negative.
While doing thermochemical calculations, if any equation is multiplied by any integer, the word aq which represents large dilutions is not multiplied by it and is retained as such.
When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But salvation of these ions (hydration in the case of water) also occurs at the same time. This is shown diagrammatically , for an ionic compound , AB (s).

The enthalpy of solution of AB(s) , sol HѲ , in water is determined by the selective values of lattice enthalpy , lattice HѲ and enthalpy of hydration of ions, hydHѲ as
sol HѲ = latticeHѲ + hydHѲ
For most of ionic compounds , sol HѲ is positive and dissociation process is endothermic. Therefore solubility of most salts in water increases with rise of temperature. If the lattice enthalpy is very high, the dissolution of the compound may not take place at all.
Lattice Enthalpy
The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
Na+Cl  Na+(g) + Cl(g) : latticeHѲ = +788 kJ mol1
Since it is impossible to determine lattice enthalpies directly by experiment , we use an indirect method where we construct an enthalpy diagram called Born – Haber Cycle (Fig).
Let us calculate the enthalpy of Na+Cl (s) by the following steps given below :
i) Na(s)  Na(g) , sublimation of sodium metal, subHѲ = 108.4 kJ mol1
ii) Na(g)  Na+(g) + e , the ionization of sodium atoms , ionization enthalpy , iHѲ = 496 kJ mol1
iii) ½ Cl2(g)  Cl(g) , the dissociation of chlorine, the
reaction enthalpy is half the bond dissociation enthalpy.
½ bondHѲ = 121 kJ mol1
iv) Cl(g) + e  Cl (g) ; electron gained by chlorine atoms. The electron gain enthalpy , eg HѲ =  348.6 kJ mol1.
v) Na+ + Cl  Na+Cl(s)
These sequence of steps is shown in Fig and is known as a Born Haber Cycle. The importance of the cycle is that , the sum of the enthalpy changes round a cycle is zero.
Applying Hess’s law we get.
latticeHѲ = 411.2 + 108.4 + 121 + 496  348.6
latticeHѲ = +788 kJ
For NaCl(s)  Na+(g) + Cl (g)
Internal energy is smaller by 2 RT ( because ng = 2) and is equal to +783 kJ mol1 Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression :
sol HѲ = latticeHѲ + hyd HѲ
For one mole of NaCl(s) ,
Lattice enthalpy = +788 kJ mol1
and hydHѲ =  784 kJ mol1 (from literature)
solHѲ = +788 kJ mol1  784 kJ mol1
= +4 kJ mol1
The dissolution of NaCl(s) is accompanied by very little heat change.


Enthalpy diagram for lattice enthalpy of NaCl

5. Heat of hydration
It is the enthalpy change accompanying the hydration of one mole of an anhydrous salt by combining with specific number of moles of water. For example, heat of hydration of anhydrous copper sulphate can be represented as:
CuSO4(s) + 5 H2O(ℓ)  CuSO4.5H2O(s) ; H = 78.2 kJ
ENERGETICS OF PHASE TRANSITIONS
Energy is required to convert a solid into liquid or a liquid into gas. The conversion of solid into liquid is called melting or fusion and the process of conversion of liquid into gas is termed as vaporisation. The processes are collectively called Phase transitions or phase changes.
a) Enthalpy of fusion or Heat of fusion
It is the enthalpy change accompanying the conversion of 1 mole of a solid substance into the liquid state at its melting point. For example, when one mole of ice changes into water at its melting point(273K) , 6.0 kJ of heat is absorbed. It is represented as :
H2O(s)  H2O(ℓ) ; fusH = +6.0 kJ
The value of enthalpy of fusion give an idea about the magnitude of inter-particle forces in solids. For example, ionic solids have very strong interparticle forces. Consequently, they have high values of enthalpy of fusion. On the other hand, molecular solids have weak inter-particle forces which reflect that their values of enthalpy are low.
b) Enthalpy of vaporisation or Heat of vaporisation
It is the enthalpy change accompanying the conversion of one mole of a liquid into its vapours at its boiling point. For example, when one mole of water is converted into steam at 373 K, enthalpy change accompanying the process is 40.6 kJ. It can be expressed as :
H2O(ℓ)  H2O(g) ; vapH = + 40.6 kJ
The value of enthalpy of vaporisation give some idea about the magnitude of inter-particle forces in liquids.
c) Enthalpy of Sublimation or Heat of Sublimation
It is the enthalpy change accompanying the conversion of one mole of a solid directly into its gaseous state at a temperature below its melting point. For example, when one mole of solid iodine sublimes, the enthalpy change is 62.4 kJ and is represented as:
I2 (s)  I2 (g) ; subH = +62.4 kJ
The enthalpy of sublimation can be calculated with the help of Hess's law . The process of sublimation consists of the conversion of solid state directly into vapour state in one step. The same change may be brought about in two steps.
i) The melting of solid into its liquid and the enthalpy change is fusH.
ii) The vaporisation of the liquid into gaseous state and the enthalpy change is vapH.
subH =  fus H + vapH
Problems
36. A swimmer coming out from a pool is covered with a film of water weighing about 80 g. How much heat must be supplied to evaporate this water. (vapHѲ for water = 40.79 kJ/mol).
37. The enthalpy of vaporisation of CO is 6.04 kJ/mol. Calculate the enthalpy change when 2.38 g of carbon monoxide vaporize at its normal boiling point.
38. Standard vaporization enthalpy of benzene at its boiling point is 30.8 kJ/mol ; for how long would a 100 W electric heater have to operate in order to vaporize a 100 g sample of benzene at its boiling point ? ( power = energy/ time; 1 W = 1 J/s)
HESS'S LAW OF CONSTANT HEAT SUMMATION
Statement : The law states that the enthalpy change in a chemical or physical process is the same whether the process is carried out in one step or in several steps.
If the reaction takes place in more than one steps, the enthalpy of the reaction is equal to the algebraic sum of the enthalpies of various reactions.

Consider a process involving the conversion of reactant A into product B in one step(path I) . The enthalpy change of the process is represented by H. Now suppose the process is carried out in two stages involving a change from A to C and C to B according to the path II . Let H1 and H2 be the enthalpy changes from A to C and C to B respectively.
H = H1 + H2
Hess's law is simply a corollary of the law of conservation of energy. It implies that the enthalpy change of the reaction depends only on the states of reactants and products and not on the manner by which the change is brought about.
Illustration of Hess's law
Consider the formation of carbon dioxide from carbon and oxygen. There are two ways by which the change can be brought about.
a) Conversion of carbon to carbon dioxide.

b) Conversion of carbon to carbon monoxide and subsequent oxidation of carbon monoxide to carbon dioxide.


Now according to Hess's law:
H = H1 + H2
393.5 kJ = 110.5 + (283.0) = 393.5 kJ
Applications of Hess's law
The practical utility of Hess's law lies in the fact that, it allows us to carry out thermochemical calculations to predict the enthalpies different reactions, whose direct measurement is not possible. The thermo-chemical equations like algebraic equations, can be added, subtracted, multiplied or divided by any numerical factor. Some of the important applications of Hess's law are discussed below.
1. Determination of Heat of formation
There are a large number of compounds such as methane, benzene, carbon monoxide , ethane etc. which cannot be directly synthesised from their elements. Hence their heats of formation cannot be determined by the calorimetric method. In the case of such substance, the heat of formation can be determined by indirect method based on Hess's law. The unknown reaction is made one segment of the closed cycle of reactions. The enthalpy changes of the reactions which represent other segments of the cycle can be determined experimentally. Then by applying Hess's law the desired value of H can be calculated .
For example, consider the calculation of enthalpy of formation of SO3. The enthalpy of formation(H1) and enthalpy of combustion(H2) of SO2 have been found to be
297.4 kJ/mol and 97.9 kJ/mol respectively. These changes have been shown in Fig.

According to Hess's law:
H = H1 + H2
= 297.4 + (97.9) = 395.3kJ
The same result can be arrived at by means of thermochemical calculations.
i) Conversion of sulphur to sulphur dioxide.

ii) Combustion of sulphur dioxide to sulphur trioxide.

Adding equation (i) and (ii) :
S(s) + (3/2) O2(g)  SO3(g) : H = H1 + H2
= 297.4 + (97.9)
= 395.3 kJ
Problem
39. The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O(ℓ) are produced and 3267.0 kJ of heat is produced. Calculate the standard enthalpy of formation fHѲ of benzene.
2. Determination of Enthalpy of transition
Transition implies the conversion of one allotropic form of a substance to another. For example, change of graphite to diamond, red phosphorus to yellow phosphorus and rhombic sulphur to monoclinic sulphur etc. Such reactions are very slow and enthalpy change accompanying them cannot be measured directly. However, Hess's law is quite helpful in determining the enthalpy of transition.
For example, the enthalpies of combustion of monoclinic sulphur and rhombic sulphur have been found to be 296.4 and 295.1 kJ/mol respectively. These changes have been shown in Fig.

H1 = H + H2
H = H1  H2
= 295.1 (296.4) = 1.3 kJ
3. Determination of enthalpy of hydration
Hydration means the conversion of an anhydrous salt into its hydrate by combining with specific number of moles of water. For example, one mole of anhydrous copper sulphate combines with five moles of water to form one mole of hydrated copper sulphate.
CuSO4(s) + 5 H2O(ℓ)  CuSO4. 5 H2O(s)
anhydrous hydrated
Direct measurement of enthalpy of hydration is not possible. However, use of Hees's law makes it possible as described below:
The enthalpies of dissolution of hydrated copper sulphate and that of anhydrous copper sulphate have been experimentally found to be +11.70 kJ/mol and  66.5 kJ/mol respectively. These changes are shown in Fig.

According to Hess's law:
H1 = H+ H2
H = H1 H2
= 66.5 kJ  11.70 kJ = 78.20 kJ
4. Determination of standard enthalpy of reaction
Hess's law cycle can also be used for the determination of standard heat of reaction from the knowledge of standard heats of formation of various reactants and products. The Hess's law cycle for elements in their standard states, reactants and products can be written as shown in Fig.


H1 = Sum of standard enthalpy of formation of reactants
= HѲf (Reactants)
H2 = Sum of the standard enthalpy of formation of
products.
= HѲf (Products)
HѲ = Standard enthalpy change of reaction
Now according to Hess's law:
H2 = HѲ + H1
HѲ = H2  H1
=  HѲf (Products)  HѲf (Reactants).
Problem
40. With the help of thermochemical equations given below , determine rHѲ at 298 K for the following reaction .
C(graphite) + 2 H2 (g)  CH4(g) : rHѲ = ?
C(graphite) + O2 (g)  CO2 (g)
: rHѲ=  393.5 kJ/mol
H2 (g) + ½ O2 (g)  H2O(ℓ)
; rHѲ =  285.8 kJ/mol
CO2 (g) + 2 H2O(ℓ)  CH4(g) + 2 O2 (g)
; rHѲ = +890.3 kJ/mol


Calculation of Bond Energies (bond HѲ)
When a bond is formed between the atoms, energy is released. The same amount of energy is required to break the bond. The energy required to break a particular bond in a gaseous molecule is referred to as Bond dissociation energy. It is a definite quantity and is expressed in kJ/mol.
In diatomic molecule , the bond dissociation energy is same as bond energy. For example, the energy required to break one mole of H H bonds in gaseous state is 435.4 kJ. Therefore, bond dissociation energy or bond energy of H H bond is 435.4 kJ/mol. This is expressed as:
H H H = 435.4 kJ/mol
However, in a polyatomic molecule like H2O , the bond dissociation energy is not the same as that of for successive bonds although they may be of the same type. Therefore, in such a case the bond energy is not equal to the bond dissociation energies of the two OH bonds and differ from one another as described below:
HOH(g)  H(g) + OH(g) ; H = 498 kJ
OH(g)  O(g) + H(g) ; H = 430 kJ
In such a case, therefore the bond energy is expressed as the average of the bond dissociation energies of various similar bonds. For example, the bond energy of OH bond is expressed as :

= 464 kJ mol1
Thus, bond energy may be defined as the average amount of energy required to break one mole of bonds of that type in gaseous molecules. The bond energies of various bonds have been given in TABLE.
TABLE
Bond Bond energy kJ mol1 Bond length
pm
HH 435 74
HC 414 110
HN 389 100
HO 464 97
HS 368 132
HF 569 101
HCl 431 136
HBr 368 151
HI 297 170
NN 163 145
N=N 418 123
NN 949 109
NO 230 136
N=O 590 115
CC 347 154
C=C 611 134
CC 837 120
CN 305 147
C=N 615 128
CN 891 116
CO 360 143
C=O 728 123
CCl 326 177
OO 142 145
O=O 498 121
FF 159 128
ClCl 243 199
BrBr 192 288
II 151 266
The thermochemical data makes it possible to calculate bond energies of different bonds. For example, bond energy of CH bond in methane can be obtained if the heat changes (H) for the following reaction is known.
CH4(g)  C(g) + 4H(g) : H = ?
The CH bond energy is equal to one-fourth of the H for the above reaction. The H for the reaction can be calculated indirectly from the data given.
a) the values of Hf of H(g) and C(g) and CH4 are given.
The value of Hf of : H(g) = +218 kJ/mol
C(g) = +717 kJ/mol
CH4(g) = 75 kJ/mol
If , H =  fH (Products)  fH(Reactants)
= 4 fH H(g) + fH C(g) fH CH4(g)
= 4 (218) + 717 (75) = 1664 kJ/mol
Bond energy of C H bond = (1664) / 4
= 416 kJ/mol
b) If the data has values other than Hf :
In such cases, the thermochemical calculations have to be carried out as described below:
Let the data be :
subH of carbon = +717 kJ/mol
Hdissociation of H2(g) = +436 kJ/mol
Hcombustion of Carbon, Hydrogen and methane are 394 kJ, 286 kJ and 891 kJ respectively.
The thermochemical equation can be written as follows:
i) Heat of combustion of methane.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(ℓ) : H1 = 891 kJ
ii) Heat of combustion of carbon.
C(s) + O2(g)  CO2(g) ; H2 = 394 kJ
iii) Heat of combustion of hydrogen.
H2(g) + ½ O2(g)  H2O(ℓ) ; H3 =286 kJ
iv) Heat of sublimation of carbon.
C(s)  C(g) ; H4 = +717 kJ
v) Heat of dissociation of hydrogen molecule.
H2(g)  2 H(g) : H5 = +436 kJ
From the above equations, heat change for the reaction :
CH4(g)  C(g) + 4 H(g)
can be calculated as follows:
Multiplying equation (v) and also equation (iii) by 2. Now add equation (i) , equation (iv) and twice of equation (v). From the resulting expression subtract equation (ii) and twice of equation (iii). The resulting value of enthalpy change comes out to be :
H = H1 H2 (2 x H3) + H4 + (2 x H5)
= 891  (394) ( 2 x 286) + (717) + (2 x 436)
= +1664 kJ
This represents the energy required for the cleavage of four C H bonds. Therefore the bond energy of C H bond in methane is 1664/4 = 416 kJ/mol.
Enthalpy of atomization ( aHѲ )
It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
H2(g)  2 H(g) : aHѲ = 435.0 kJ mol1
In the case of diatomic molecules, like dihydrogen, the enthalpy of atomization is also called the bond dissociation enthalpy. The other examples of enthalpy of atomization can be
CH4(g)  C(g) + 4 H(g) ; aHѲ = 1665 kJ mol1
Problems
41. Propane has the structure H3CCH2CH3. Use the average bond enthalpies to estimate the change in enthalpy, H, for the following reaction :
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
42. The bond enthalpies of HH , BrBr and HBr are 435 kJ/mol, 192 kJ/mol and 368 kJ/mol respectively. Estimate the enthalpy change for the reaction
H2(g) + Br2 (g)  2 HBr(g)
43. The standard enthalpy change rHө =  1.584 x 103 kJ/mol for the reaction :
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
and bond energies of CC , CH , C=O and OH are 347
kJ/mol, 414 kJ mol1, 728 kJ/mol and 464 kJ mol1
respectively. Calculate the energy of oxygen-oxygen bond in
O2 molecules.
44. Calculate the heat of formation of acetic acid, if its heat of combustion is 867 kJ mol1. The heats of formation of CO2(g) and H2O(ℓ) are393.5 kJ mol1and 285.9 kJ mol1respectively.
45. Calculate the heat of combustion of glucose from the
following data:
i) C(graphite) + O2(g)  CO2(g) ; H= 395 kJ
ii) H2(g) + ½ O2(g)  H2O(ℓ) ; H = 269.4 kJ
iii) 6C(graphite) + 6 H2(g) + 3 O2(g)  C6H12O6(s)
; H = 1,169.8 kJ
46. Calculate the heat of formation of benzene from the following
data :
i) C6H6(ℓ) + 7 ½ O2(g)  6 CO2(g) + 3 H2O(ℓ) ;
; H = 3280.1 kJ
ii) C(s) + O2(g)  CO2(g) ; H= 395 kJ
iii) H2(g) + ½ O2(g)  H2O(ℓ) ; H = 285.9 kJ
47. The heat of combustion of methane is890.65 kJ mol1and heats of formation of CO2 and H2O are 393.5 and 286.4 kJ mol1respectively. Calculate the enthalpy of formation of methane.
48. Use standard enthalpies of the formation to calculate the
value of rH for the reaction :
2 H2S(g) + 3 O2(g)  2 H2O(ℓ) + 2 SO2(g)
standard molar enthalpies of formation of H2S, H2O and SO2
are  20.60 kJ,  285.83 kJ and  296.83 kJ respectively.

49. Calculate rHѲ for the reaction :

The average bond enthalpies of C H and C Cl bonds are 416 and 325 kJ mol1 respectively.
50. Calculate the CC bond energy from the following data:
i) 2 C(graphite) + 3 H2(g)  C2H6(g)
; H =  84.67 kJ
ii) C(graphite)  C(g) ; H = 716.7 kJ
iii) H2(g)  2 H(g) ; H = 435.9 kJ
Assume 416 as CH bond energy.
SPONTANEOUS PROCESSES
Processes which takes place by itself under a given set of conditions without the intervention of any kind are known as spontaneous processes. All the processes that take place in nature are spontaneous in character, proceed only in one direction and are therefore thermodynamically irreversible. They can be reversed only by the help of an external agency. Let us consider some examples of spontaneous changes to understand their irreversible nature and the conditions which determine their feasibility.
i) We find that a beaker containing hot water kept in room cools spontaneously by losing heat to the surroundings. The first law of thermodynamics is obeyed because the energy lost by the system (i.e., beaker containing water) is gained by the surroundings. However, the first law will still be obeyed in the reverse process, namely, when the surrounding lose energy and the system gains it. We never observe such situation where the beaker of water at room temperature becomes hot on its own. In short , the cooling of hot body in a room proceeds spontaneously but not its heating even though both processes are allowed by the first law.
ii) If we add a drop of ink in a beaker containing water , the ink spreads throughout the beaker making the whole solution coloured. The reverse process where dispersed ink spontaneously accumulates to form a single drop does not take place.
iii) Two gases under the same pressure in two identical bulbs connected through a stopcock , mix and give a homogeneous mixture on opening the stopcock.

Spontaneous mixing of two gases
(a) before mixing (b) after mixing
Consider few examples involving chemical changes.
(i) If equal volumes of equimolar solutions of hydrochloric acid and sodium hydroxide are mixed, the neutralisation reaction proceeds spontaneously yielding sodium chloride and water. However, if sodium chloride is dissolved in water, the reverse reaction does not take place on its own.
In general, all transformations under given experimental conditions have a characteristic direction in which they proceed spontaneously ; i.e., on their own with out intervention of an outside agency. The reverse process i.e., the beaker of water can be made hot by heating, but this is not a spontaneous process because an outside agency (a gas burner) has to be used. Similarly an ink solution can be processed to obtain the ink drop ; and two gases in a mixture can be separated if allowed to diffuse through a porous plug, and a solution of sodium chloride can be made to yield hydrochloric acid and sodium hydroxide. In each case, the change has to be brought about using some external agency. It does not take place on its own.
There is another feature of all spontaneous changes that may be noted. A beaker of hot water cools until it attains the temperature of the surroundings after which no further change in temperature is observed. Now the beaker of water and the surroundings are in thermal equilibrium. The drop of ink similarly separates until it is distributed uniformly. Mixing of two gases continues until each is evenly distributed throughout both the containers. Thus all spontaneous changes proceed till equilibrium is achieved.