UNIT 6 ( PAGE 3)

Energy and Spontaneity

            Common experience shows that spontaneous processes involving macroscopic objects proceed with a decrease in potential energy.

(i) A rock spontaneously rolls down a slope if it is dislodged, but never rolls up to the top again (Fig)

The rolling of a rock down hill          The rolling of a rock uphill

is a spontaneous process.                  is a non-spontaneous process.

The rock eventually comes to

Equilibrium at the bottom of the

hill.

(ii)     If we drop a bottle , it falls spontaneously to the ground.

(iii)    When octane is ignited , it burns spontaneously in oxygen to give CO₂ and H₂O according to the reaction :

2 C₈H₁₈ + 25 O₂(g) ® 16 CO₂(g) + 18 H₂O(ℓ)

                                                   : ∆H^ө = - 10951 kJ/mol

Heat is evolved and the reaction is exothermic. Thus the system moves from a position of higher to lower energy.

But there are many reactions which are spontaneous for which ∆H^ө is +ve or zero.

(iv)    The evaporation of liquid water from an open vessel is spontaneous, but it is a process in which the system requires energy to evaporate water molecules.

(v)     Dissolution of a solute in a solvent is spontaneous eventhough the process may be endothermic, i.e., it may require energy.

(vi)    Dinitrogen pentoxide (N₂O₅) decomposes spontaneously at room temperature into NO₂ and O₂ although the reaction is endothermic.

    2 N₂O₅(g) ® 4 NO₂ (g) + O₂ (g) : ∆H^ө = 219.0  kJ/mol

(vii)   Expansion of an ideal gas through a pinhole is a spontaneous process. In this expansion there is no energy change of the system.

Experience tells us that highly exothermic processes are generally spontaneous but some endothermic process may also be spontaneous showing that lowering of energy is not  the only determining factor for spontaneity.

Disorder, Entropy and Second law of Thermodynamics

            During the spontaneous evaporation of water, there is an increase in energy of the molecules and the process is endothermic. However, this leads to greater disorder amongst water molecules in the vapour states compared to the liquid state. The disorder is the manifestation of another thermodynamic property called entropy and is represented by the symbol S. Entropy like any other thermodynamic property such as internal energy (U) and enthalpy(H) should be a state function. The entropy change ,  DS is given by the equation :

                       …………..(1)

Here q_rev is the heat absorbed when the process is carried out reversibly and isothermally (i.e., at constant T). If q is expressed in Joule(J) and temperature in Kelvin (K), then entropy change is given in unit of J/K or J K^-1. The DS defined by equation (1) is independent of path , as it is a state function.

            q_rev / T is is a measure of increase in disorder of the system. This can be justified on the grounds that the absorption of heat into a system leads to the vigorous movement of the molecules due to increased kinetic energy. In other words, the more heat the system absorbs, the more disordered it becomes.  If heat is absorbed at low temperature it becomes more disordered than when the same amount of heat is added at higher temperature. This is obvious as T is placed in denomenator in equation 1.

Physical meaning of Entropy

            The physical meaning of entropy is that entropy is a measure of disorder (or randomness) of a system. The relation between entropy and disorder provides a suitable explanation for entropy change in various processes. The greater the disorder in a system, the higher the entropy. For a given substance the solid state is the state of lowest entropy (most ordered state) , the gaseous state is the state of highest entropy and liquid state is intermediate between the two.

            Thus, when a sample changes from solid to liquid to gas, the particles become increasingly disordered and therefore its entropy increases. This shown in Fig (a)

                                    (a)

            

                                 (b)

Increase in disorder or entropy when a solid changes to liquid to gas.

Fig (a) shows the increase in disorder when solid changes to liquid to gas. When the solid is heated its entropy increases as its temperature is raised. The entropy increases sharply when solid melts to form more disordered liquid. It further gradually increases again upto boiling point. A second larger jump in entropy occurs when the liquid changes into vapour at the boiling point. The changes are shown in Fig(b).

            Similarly , in the case of mixing of two gases when the stopcock is opened, the gases mix to achieve more randomness or disorder. In this case, there is no exchange of matter or energy between the system and surroundings. The change occurs from ordered state (less entropy) to disordered state (higher entropy) . Thus the change in entropy is positive.

Driving force as the overall tendency for a process

The spontaneous processes occur because of two tendencies.

i)        Tendency of a system to acquire a state of minimum energy.

ii)      Tendency of a system to acquire a state of maximum randomness.

The overall tendency for a process to take place by itself is called driving force. Regarding these two tendencies , it is very  important  to note the following points.

i)        The two tendencies act independent of each other.

ii)      The two tendencies may work in the same or in opposite directions in a process.

            It is clear from the discussion that when there is no change in energy, the second tendency is the governing factor i.e., reaction will be spontaneous or non-spontaneous depending upon whether randomness factor favours or opposes. On the other hand, if there is no change in randomness, the energy factor is the controlling factor. However, this is not so simple in actual practice as both tendencies operate simultaneously. In these cases, the significance of magnitude of these tendencies determines whether the process is feasible or not.

Second Law of Thermodynamics

            The second law of thermodynamics introduces the concept of entropy and its relation with spontaneous process. In an isolated system, such as mixing of gases , there is no exchange of energy and matter between system and surroundings. But due to increase in randomness, there is an increase in entropy. Thus we can say that, for a spontaneous process in an isolated system, the change in entropy is positive , i.e., DS > 0 . However, if the system is not isolated, we have to take into account the entropy changes of the system and the surroundings. Then total entropy changes (DS_total) will be equal to the sum of the change in entropy  of the system and the surroundings (DS_surroundings). i.e.,

              DS_total = DS _system + DS_surroundings

For  a spontaneous process , DS_total must be positive i.e.,

             DS_total  =  DS_system  + DS_surroundings  > 0

But the system and surroundings constitute ‘universe’ for thermodynamic point of view so that for spontaneous change

                   DS_universe > 0  ……… (2)

The relation (2) (DS_universe > 0) forms the basis of most useful form of Second Law of Thermodynamics which states that : the entropy of the universe always increases in the course of every spontaneous (natural) change. Thus combining statements of First and Second Laws of thermodynamics, we conclude that :  In any natural or spontaneous process, the energy  of the universe remains constant but the entropy of the universe always increases.

Entropy change and equilibrium

For a spontaneous process DS_total > 0. The careful study of the spontaneous processes reveals that each of them gradually and ultimately  passes into a state of equilibrium. For example, in intermixing of gases, the process of intermixing continues till a uniform mixture is formed. Similarly in the evaporation of water in a closed vessel , soon a state of equilibrium is reached.  Although these processes are favoured by increase in the value of DS_total  , yet no further increase in entropy occurs once the equilibrium is established.  This leads to conclude that :

At equilibrium,

                          DS_total = 0

Following  the above discussion, the entropy criterion can be summed up as follows:

·        DS_total     > 0  : process is spontaneous

·        DS_total     = 0  :  equilibrium state

·        DS_total     < 0  : process is non-spontaneous

Let us apply , the entropy criterion to decide the feasibility of the process  of conversion of water to ice, at 1 atm pressure,

H₂O(ℓ) ®  H₂O(s)

The entropy changes for different temperatures 272, 273 and 274 K are given below.

Temperature (K)	DSsystem J/K/mol	DSsurroundings     J/K/mol	DStotal    J / K / mol
272	- 21.85	+ 21.93	+ 0.08
273	- 21.99	+ 21.99	0.00
274	- 22.13	+22.05	-  0.08

i)        DS_total is positive at 272 K, therefore freezing of liquid water  to ice is spontaneous.

            H₂O(ℓ) ® H₂O(s)  : spontaneous at 272 K

ii)      At 274 K, DS_total is negative, therefore the freezing of water is not spontaneous at 274 K.

iii)     At 273 K, DS_total is zero, therefore the process is at equilibrium, i.e., neither freezing nor melting is spontaneous. At this temperature, water and ice are in equilibrium and no net change is observed.

              H₂O(ℓ) ® H₂O(s)     : equilibrium at 273 K

Thus, we observe that DS_total is a criterion for spontaneity of a change.

Characteristics of Entropy

The impotant characteristics of entropy are summed up as below:

(i)      Entropy is an extensive property. Its value depends upon the amount of the substance present in the system.

(ii)     Entropy of the system is a state function. It depends upon the state variables (T, P, V and n) . Thus , the change in entropy is given as :

DS  = S_{final state} - S_{initial state}

(iii)    The change in entropy in going from one state to another is independent of the path .

(iv)    The change in entropy for a cyclic process is always zero.

(v)     The total entropy change of an non-isolated system is equal to the entropy change of the system and entropy change of the surroundings. The sum is called entropy change of the universe.

DS  = DS_system  +   DS_surr

(vi)    In a reversible process in equilibrium, DS_universe = 0 and therefore

DS_system  =   -DS_surr

(vii)   In a irreversible process,

DS_universe  >   0 . This means that there is an increase in entropy of universe in spontaneous changes.

Entropy changes in system and surroundings and total entropy change for Exothermic and Endothermic Reactions

Heat increases the thermal motion of the atoms or molecules and increases their disorder and hence their entropy. In the case of exothermic process, the heat escapes into the surroundings and therefore , entropy of the surroundings increases. On the other hand in the case of endothermic process , the heat enters the system from the surroundings and therefore , the entropy of the surroundings decreases. These changes are shown in Fig.

Entropy changes in (a) Exothermic and (b) Endothermic processes.

In general , there will be an overall increase of the total entropy (or disorder) whenever the disorder of the surroundings is greater than the decrease in disorder of the system. The process will be spontaneous only when the total entropy increases.

            Let us consider an exothermic process. In exothermic processes , heat released by the reaction increases the entropy of the surroundings. The overall entropy changes is certainly positive when the entropy of the system is positive (Fig a)

Entropy changes for an exothermic reaction

In some exothermic reactions, entropy of the system may decrease. This is possible in the case of reactions which involve the conversion of a gas into solid product. However, if the reaction is highly exothermic and increase in entropy of the surroundings is very high, the total entropy change will be positive and the reaction will be spontaneous (Fig b).

            For example, consider the reaction of oxidation of magnesium to magnesium oxide. The reaction is highly exothermic.

   2 Mg(s) + O₂(g) ® 2 MgO(s)  : DH^Θ = - 1202 kJ  mol^-1

But the reaction proceeds by decrease in entropy and DS^Θ for the reaction is -217 J K^-1mol^-1.

            The heat released to the surroundings will result into increase in entropy of the surroundings. The same formula applies to change in entropy of the surroundings that accompanies a chemical reaction.

Since DS_total is is positive , the reaction will be spontaneous.

            In case of endothermic reactions, reactants absorb energy and the products are at higher energy state and the temperature of the system falls. In other words, heat flows from surroundings into the system. As a result, the entropy of the surroundings decreases. The reactions will be spontaneous only when entropy of system increases enough to overcome the decrease in entropy of the surroundings so that overall entropy change is positive. This is shown in Fig.

Endothermic reaction is spontaneous only if the entropy of the system increases enough to overcome the decrease in entropy of the surroundings.

Thus,

·         A chemical reaction is spontaneous if it is accompanied by an increase in the total entropy of the system and the surroundings.

·         Spontaneous exothermic reactions are common because they release heat that increases the entropy of the surroundings.

·         Endothermic reactions are spontaneous only if the reaction mixture undergoes a large increase in entropy.

Entropy Changes of Phase transitions

            Phase transition refers to changes of one state (solid, liquid or gas) to another. Such change occurs at a definite temperature. For example, conversion of solid into liquid occurs at the melting point, while changes of liquid into vapour occurs at the boiling point. The entropy change taking place during such transitions  can be calculated as follows:

1. Entropy of fusion :  Entropy of fusion may be defined as the entropy  change taking place when one mole of substance changes from solid state into liquid state at its melting point.

                         Solid   ⇌   Liquid

During this phase transition, the system absorbs heat at constant temperature and pressure. The heat absorbed is equal to the enthalpy of fusion (DH_fusion). The entropy change taking place during this transition can be calculated as :

Here,       DS_fusion = Entropy of fusion

              DH_fusion = Enthalpy of fusion

                     S_ℓ= Entropy of one mole of liquid at its

                            melting point

                     S_s = Entropy of one mole of solid substance

                             at its melting point.

                     T_f = Melting point of the substance.

Since DH_fusion is positive , therefore DS_fusion is also positive. Hence S_ℓ > S_s i.e., entropy of liquid substance is greater than the entropy of the substance in the solid state.

2. Entropy of vaporisation : Entropy of vaporisation may be defined as the entropy change taking place when one mole of the substance changes from liquid state into vapour at its boiling point.

                        Liquid  ⇌   Gas

During this transition, the system absorbs heat at constant temperature and pressure. The heat absorbed is equal to the enthalpy of vaporisation(DH_vap) The entropy change for the transition may be calculated as :

                

Here, DSvap = Entropy of vaporisation

        DHvap = Enthalpy of vaporisation

              S_ℓ  = Entropy of one mole of substance in liquid

                      state at its boiling point.

              S_g = Entropy of one mole of substance in the gaseous

                     state.

              T_b = Boiling point of the substance

Since DH_vap is positive, therefore DS_vap is also positive.

Hence S_g > S_ℓ   i.e., entropy of substance in gaseous state is greater than the entropy of substance in liquid state.

3.       Entropy  of sublimation : Entropy of sublimation is

the entropy change when one mole of the solid changes into vapour at a particular temperature T.      Thus,

where DS_sub = DS_vap + DS_fusion

DS_sub  =  molar entropy of sublimation at temperature T in Kelvin.

Problems

51.   State whether each of the following processes will increase or decrease total energy content of the system :

(a)       Heat transferred to surroundings.

(b)       Work done by the system

(c)       Work done on the system

52.  Place the following systems in the order of increasing randomness :

(a)       1 mol of gas X

(b)       1 mol of solid X

(c)       1 mol of liquid X

53.  Which of the following processes are accompanied by an increase of entropy :

(a)       dissolution of iodine in a solvent

(b)       HCl is added to AgNO₃ solution and precipitate is obtained.

(c)       A partition is removed to allow two gases to mix.

54.   Predict the sign of entropy change for each of the following changes of state :

(a)       Hg (ℓ)        ® Hg(g)

(b)       AgNO₃(s)   ® AgNO₃(aq)

(c)       I₂(g)          ® I₂(s)       

(d)       C(graphite) ® C(diamond)

44.   Predict which of the following , entropy increases/decreases :

        (i)   A liquid crystallizes into a solid.

      (ii)   Temperature of a crystalline solid is raised from 0 K to

              115 K.

     (iii)    2 NaHCO₃(s) ®  Na₂CO₃ + CO₂(g) + H₂O(g)

     (iv)    H₂(g) ®  2 H(g)   (N 177 P 6.9)

55. The enthalpy of vaporisation of liquid diethyl ether is          26 kJ/mol at its boiling point(35°C). Calculate DS^Θ for the conversion of :

(a)       liquid to vapour, and

(b)       vapour to liquid at 35°C.

56.  The enthalpy change for the transition of liquid water to steam, DH_vap is 40.8 kJ/mol at 373 K . Calculate the entropy change for the process.

57.   At 0⁰C ice and water are in equilibrium and DH = 6 kJ mol^-1. For the process:

H₂O(s) ® H₂O(ℓ)

        Calculate DS for conversion of ice to liquid water.

58.   Calculate the entropy change of 1 g of ice to water at 273 K and one atmospheric pressure. The enthalpy of fusion of ice is 6.025 kJ/mol.

59.     Calculate the enthalpy of vaporisation per mol for ethanol.  Given DS = 109.8 J K^-1mol^-1  and boiling point of  ethanol = 78.5°C.

60.   The enthalpy of vapourisation of benzene is 30.8 kJ mol^-1 at  its boiling point(80.1°C). Calculate the entropy change in going from (i) liquid to vapour and (ii) vapour to liquid at 80.1°C.

61.    A reaction A + B ®  C + D + q is found to have a positive entropy change. The reaction will be : (N 182 Ex 6.6)

       (i)    possible at high temperature.

       (ii)   possible only at low temperature

       (iii)  not possible at any temperature

       (iv)  possible at any temperature

Predicting the sign of  DS

            The process in which entropy of product is greater than the entropy of the reactants have positive value of DS. Such a process is accompanied by increase in randomness.

            In many reactions, it becomes possible to predict the sign of  DS. Such processes are accompanied by increase in randomness. In many reactions, it becomes possible to predict the sign of DS by simply observing the reactants and products. If the products are more random than the reactants DS is positive and if the products are less random than the reactants, DS is negative.

Now, let us consider some reactions and predict the sign of DS.

i)       C(s) + O₂(g) ® CO₂(g)

In this reaction, the product is more random than the reactants because the product molecules are in gaseous state , whereas one of the reactants is in the solid state. DS would be positive for this reaction.

ii)                     2 SO₃(g) ® 2 SO₂(g) + O₂(g) 

In this reaction also products are more random than the reactants because more molecules are present in gaseous state in product than in reactants . So DS is +Ve.

iii)      NaCl(s) + H₂O(ℓ) ® Na⁺(aq)  + Cl^- (aq)

This process is also accompanied by increase in randomness because Na⁺ and  Cl^- ions are free to move more than  in solid NaCl. Hence DS is positive  for this type of  dissolution process.

Gibb’s Energy (G) and Gibb’s Energy Change (DG)

            Gibb’s energy of a system is defined as the amount of energy available to a system during a process that can be converted into useful work. In other words, it is a measure of capacity of a system to do useful work denoted by symbol G and is given by :

                     G =  H  -  T S

where H is the enthalpy of the system, S is the entropy and T is the absolute temperature.

Now,  H =  U +  P V

         G =  U + PV - T S

The change in Gibb’s energy may be expressed as :

DG = DU  +  D(PV) - D TS

If the process is carried out at constant temperature and pressure, the terms D(PV) and D(TS) becomes :

D(PV)  =  P DV  and D(TS) = TDS

\  DG = DE + PDV -  TDS

But , DU + PDV   = DH

or                  DG = DH - T DS   ……..  (11)

This equation is called Gibb’s energy equation and is very useful in predicting the spontaneity of a process.

Units of DG

DG has units of energy ie., kJ/mol or J mol^-1 because DH and TDS are are energy terms.

DG( J mol^-1)= DH( J mol^-1)   - TDS ( K  J K^-1 mol^-1)

Physical significance of Gibb’s energy

            In order to understand the physical significance of Gibb’s energy, let us consider the relationship between heat (q) and work (w). According to first law of thermodynamics :

DU =  q  + w  

Here w includes the pressure-volume work (- PDV ) as well as non-expansion work. Therefore, the above relation can be written as :

DU =  q  - PDV +  w_{non-expansion}

  q = DU + PDV -  w_{non-expansion}

     = DH -  w_{non-expansion}   ……… (15)

For a change taking place under isothermal and reversible conditions, the entropy change DS is given by :

DS = (q / T)

or             q =  T DS  ……..  (16)

Substituting this value of q in Equation (15) :

       TDS = DH - w_{non-expansion}

            DH - TDS  =  w_{non-expansion}

                    DG_T,P = w_{non-expansion}

                 - DG_T,P = - w_{non-expansion}

Decrease in Gibb’s energy = Non-expansion work done by the system

The non-expansion work may be regarded as useful work. Thus, the decrease in Gibb’s energy of a system during a process is a measure of maximum useful work done during the change.  Therefore Gibb’s energy  G of a system during a process is a measure of the maximum useful work done during the change.  Therefore Gibb’s  energy , G of a system is a measure of its capacity of doing useful work. Therefore greater the Gibb’s energy change , the greater is the amount of work that can be obtained from the process.

In our bodies , the food stuffs are oxidised to obtain energy to do work. For example, consider the metabolism of glucose in our body. When one mole of glucose (180.2 g) is oxidised

               C₆H₁₂O₆(s) + 6 O₂(g) ® 6 CO₂(g) + 6 H₂O(ℓ)

                   : DH^Θ  = -2808 kJ/mol    : DG^Θ  = -2870 kJ/mol

This means that 2870 kJ is the maximum amount of work that can be done by a person as a result of metabolizing 1 mol of glucose. Suppose a man (mass 60 kg) climbs a height of 100 m . The work a person must do in climbing a height h is given by ,   w = m g h where g is accleration due to gravity (9.8 m s^-2). Thus the man has to work

Work has to do = (60 kg) (9.8 m s^-2) x 100 m

                       = 58,800 kg m² s^-2 = 58,800 J = 58.8 kJ

In order to do this much work, he need to metabolize minimum of (58.8 / 2870) = 0.0205 mol or 0.0205 x 180 = 3.69 g of glucose.

            However, in practice, the conversion of energy to the work is not 100% efficient , therefore the person would need more than 3.7 g of glucose to climb the height.

Gibb’s Energy Change and Electrical work done

In the case of electrochemical cells, Gibb’s energy change  DG is related to the electrical work done in the cell. If E is the EMF of the cell and n mole of electrons are involved , the electrical work will be :

DG = - n F E

where F is Faraday constant = 96,500 C

If the products are in their standard states :

DG^Θ = - n F E^Θ

where E^Θ is the standard cell potential.

Gibb’s Energy  Changes for predicting feasibility of a reaction

Let us develop the criterion of spontaneity in terms of Gibb’s energy change.  The total entropy change during the process is given by :

DS_total =  DS_system +  DS_surrounding  …… (3)

Consider a process being carried out at constant temperature and pressure. Suppose heat equal to q is lost by the surroundings. Now heat gained by the system at constant temperature and pressure q_p represents its enthalpy change (DH). Therefore.

Negative sign   before q indicates that total heat has been lost by the surroundings. Now the heat gained by the system at constant temperature and pressure q_p represents the enthalpy change (DH). Therefore,

Substituting this value in equation (12) , we get :

Multiplying throughout by T :

The equation (5) forms the basis for the criterion of spontaneity in terms of Gibb’s energy change of the system only. For example, for a spontaneous process DS_total is positive. So that DG_system is negative.

 Thus  for a process, at a particular temperature and pressure to be spontaneous , the Gibb’s energy of the system must decrease, i.e., DG_T,P < 0.

Summary of Gibb’s energy criteria of spontaneity

i)         If  DS_total is positive : DG_.T,P  is negative. The process will be spontaneous.

ii)        If  DS_total = 0 ; DG_T,P = 0 ; the process will be at equilibrium.

iii)       If DS_total  is negative : DG_T,P is positive. The process will

        be non-spontaneous. 

Conditions for DG to be negative

            Let us formulate the conditions for DG to be negative. It can be done by considering Gibb’s Helmholtz equation.

DG = DH - T DS

i)         When both energy factor as well as entropy factor are favourable, i.e., DH is negative and DS is positive. Under these conditions DG would certainly be negative and the process would occur spontaneously.

ii)        Energy factor favours but entropy factor opposes, i.e., DH is negative and DS is negative. Under these conditions DG would be negative if  DH is greater than TDS in magnitude.

iii)       Energy factor opposes but entropy factor favours, i.e., DH is positive and DS is also positive. Under these conditions, DG would be negative if TDS is greater than DH in magnitude.

These conditions have been summarised in the following TABLE.

Sign of DH	Sign of DS	Sign of DG, occurrence of a reaction
-	+	- ve (spontaneous)
-	-	- ve ( if DH > TDS)
+	+	- ve ( if TDS > DH)

Effect of Temperature on Feasibility of a Reaction

Gibb’s Helmholtz equation helps us in predicting the feasibility of a process at a particular temperature. The spontaneity of a process depends on :

i)         the energy factor DH and

ii)        the entropy factor DS.

As is evident from the equation :   DG = DH - T DS

the term TDS may assume large values at higher temperatures due to increase in multiplying factor T as well as increase in randomness. But the other factor DH does not vary appreciably with temperature. Therefore, the free energy change (overall tendency) of a process to take place is also influenced by temperature.  The effect of temperature is different for exothermic and endothermic reactions as discussed below:

a)       Exothermic Reactions

For exothermic reactions, DH is always negative and therefore , it is favourable. Now TDS can either positive or negative value.

i)         If TDS is positive, i.e., favourable, then  DG can have only negative value and the process is spontaneous at all temperatures.

ii)        If  TDS is also negative, i.e., unfavourable. The value of DG will be negative only if  DH > TDS. This fact will be more predominant at lower temperatures which makes the contributions of TDS(opposing factor) less. Hence exothermic reactions are favoured by decreasing temperature.

b)      Endothermic reactions

For endothermic reactions, DH is positive and always opposes the process. Now :

i)         If TDS is negative and opposes the process , then DG will be positive and the process is always non-spontaneous.

ii)        On the other hand, when TDS is positive (favourable) , the value of DG will be negative only if TDS > DH. This fact will be more predominant at higher temperatures which makes the contribution of TDS (favouring factor) more. This means that endothermic reactions are favoured by increasing temperature.  These results are summed in the following TABLE.

TABLE

Tendencies of reactions to occur in forward direction and effect of temperature.

Reaction	DH	Sign of TDS	DG	Behaviour
Exothermic	-	+	-	spontaneous
	-	-	-  at low T + at high T	spontaneous Non-spontaneous
Endother  -mic	+	-	+	non-spontaneous
	+	+	+ at low T	non-spontaneous
	+	+	- at high T	spontaneous

Problems

69. Consider the reaction for the dissolution of ammonium nitrate

NH₄NO₃(s) ® NH₄⁺(aq) + NO₃^-(aq)

DH = +28.1 kJ/mol : DS = 108.7 kJ mol^-1

Calculate the change in entropy of the surroundings and predict whether the reaction is spontaneous or not at 298 K ?

70.  The rusting of iron occurs as :

4 Fe(s) +  3 O₂(g) ®  2 Fe₂O₃

The enthalpy of formation of Fe₂O₃(s) is - 824.2 kJ mol^-1  and entropy change for the reaction is -549 J K^-1 . Calculate DS_sur and predict whether rusting of iron is spontaneous or not at 298 K.

71.   For the reaction :

        2 H₂O₂(g) ®  2 H₂O(g) + O₂(g)

        DH = - 211.29 kJ/mol and DS = + 131.98 J K^-1 mol^-1.

        Predict the feasibility of the reaction at 298 K.

72.   For the reaction :

        N₂(g) + O₂(g) ® 2 NO(g)

        DH = +180.5 kJ : DS = 24.6 J K^-1  at 298 K.

        CalculateDG.

73.   For the melting of ice at 298 K ,

H₂O (s) ®  2 H₂O(ℓ)

The enthalpy of fusion is 6.97 kJ mol^-1  and entropy of fusion is 25.4 J mol^-1. Calculate the free energy change and predict whether melting of ice is spontaneous at this temperature.

74. Enthalpy and entropy changes of a reaction are            40.63 kJ/mol and 108.8 kJ/mol respectively. Predict the feasibility of the reaction at 300 K.

75.    DH and DS for the reaction :

Ag₂O(s) ® 2 Ag(s) + ½ O₂(g)

are 30.56 3 kJ mol^-1and 108.8 J K^-1mol^-1 respectively. Predict  the feasibility of the reaction at 300 K.

Standard Gibb’s energy change (DG^Θ)

The standard Gibb’s energy change is the free energy for a process at 298 K and one atmospheric pressure in which the reactants in their standard states are converted to the products in their standard states.  It is denoted by DG^Θ . The standard Gibb’s energy change is related to the standard ethalpy change by the relation :

DG^o= DH^Θ - T DS^Θ

where DH^o and DS^o represent the standard enthalpy change and standard entropy change during the process respectively.

            Like DH^o  the standard free energy change (DG^Θ) of a reaction can be calculated from the standard Gibb’s energy of formation  of the products and that of the reactants.  The standard free energy of formation (D_fG^Θ ) is defined as the Gibb’s energy change taking place when 1 mole of the compound is formed from the constituent elements in the standard states.   It may be noted that the standard free energy of formation (D_fG^Θ ) of elementary substances in their most stable form is taken to be zero.

            The standard free energy of a reaction can be calculated from the knowledge of standard free energy of formation of various substances as follows :

DG⁰  = SD_fG^o (products)  -   SD_fG^o (reactants)

       = [Sum of standard free energies of formation of  products]

             - [Sum of standard free energies of formation reactants]

For example, for a reaction :

a A + b B ®  ℓ L +  m M

The standard free energy change can be calculated as :

DG^o= SD_fG^Θ (products)  -  SD_fG^Θ (reactants)

       =  [ℓ D_fG^Θ(L) + mD_fG^Θ (M)] -[a D_fG^Θ (A) + b D_fG^Θ (B)]

COUPLED REACTIONS

            There are reactions for which the value of D_rG is not negative i.e., they are non-spontaneous. However, such reactions can be made spontaneous if these are coupled (means carrying both reactions simultaneously) with reactions having very large negative free energy of reaction.

Let us consider the decomposition of iron oxide , Fe₂O₃ into iron :

2 Fe₂O₃(s)  ® 4 Fe(s)  +  3  O₂(g)

                                               : DG^Θ =+1487.0 kJ mol^-1 (a)

The positive DG shows that the reaction is non-spontaneous. This can be achieved by coupling decomposition of Fe₂O₃ with a reaction having a highly negative D_rG^Θ .

Let us consider the reaction :

2 CO(g) + O₂(g) ® 6 CO₂ (g)

                                           : D_rG^Θ = -514.4 kJ mol^-1……(b)

In Equ. (a) 3 mol of O2 is involved. Therefore , we can write equation (b) as

6 CO(g) + 3O₂(g) ® 6 CO₂(g)

                                      : D_rG^Θ = -1543.2 kJ mol^-1   ……(c)

Adding equations (a) and (c)

2 Fe₂O₃(s)  ® 4 Fe(s)  +  3  O₂(g) : DG^Θ =+1487.0 kJ mol^-1

6 CO(g) + 3O₂(g) ® 6 CO₂(g) : D_rG^Θ = -1543.2 kJ mol^-1

-----------------------------------------------------------------------------

            2 Fe₂O₃(s) + 6 CO(g) ® 4 Fe(s)  +6 CO₂(g)

: DG^Θ = -56.2 kJ mol^-1

The negative value of DG^Θ indicates that iron(III) oxide can be reduced spontaneously to free iron and carbon dioxide.  This is reaction which takes place in the lower part of blast furnace where iron ore is reduced to iron.

            The concept of coupling two reactions (one spontaneous and one non-spontaneous) is very useful in biological systems. Adenosine triphosphate , or ATP , is a large molecule containing phosphate groups. When ATP reacts with water in presence of an enzyme, it is hydrolysed to give adenosine diphosphate, ADP , and a phosphate ion.

        ATP + H₂O ® ADP + phosphate ion : D_rG^Θ = - 31 kJ mol^-1

This reaction is coupled with  various non-spontaneous reactions to carry out the necessary reactions in biological systems. For example, the biosynthesis of sucrose from glucose and fructose has DG^Θ of 23 3 kJ mol^-1. Hence , this reaction can be driven by hydrolysis of ATP.

Glucose + Fructose + ATP ® Sucrose + ADP : DG^Θ = - 8 kJ/mol

Therefore ATP is regarded as centre of activities of the cell.

Problem

76. Calculate the standard Gibb’senergy change for the formation of methane at 298 K.

C(graphite)  +  2 H₂(g) ®  CH₄(g)

D_fH^ΘCH₄(g) = - 74.81 J mol^-1. S^Θ J K^-1 mol^-1 : C(graphite) = 5.70 : H₂(g) = 130.7 : CH₄(g) = 186.3.

77.   Silane SiH₄ burns in air . The product silica is a solid.

SiH₄(g) + 2 O₂(g) ®  SiO₂(s) + 2 H₂O(g)

Standard Gibb’s energy of formation of SiH₄(g) is            +52.3 kJ mol^-1. Values  for SiO₂ and H₂O(g) are 805 and -228.6 kJ mol^-1 respectively. Calculate D_rG^Θ .

60.  Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous. For the reaction

MgO(s) + C(s) ® Mg(s) + CO(g) : D_rH^Θ= + 491.18 kJ mol^-1 and D_rS^Θ = 197.67 J K^-1 mol^-1.

78.  In a fuel cell , methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is :

CH₃OH(ℓ) + 3/2 O₂(g) ® CO₂(g) + 2 H₂O(ℓ)

Calculate the standard Gibb’s energy change for the reaction that can be converted into electrical work. If standard enthalpy change of combustion for methanol is -726 kJ/mol, calculate the efficiency of conversion of Gibb’s energy  into useful work.

Gibb’s Energy Change and Equilibrium Constant

            For chemical reactions to occur, Gibb’s energy change should be negative. During the course of the reaction, the Gibb’s energy decreases and composition of reaction mixture changes with time and finally a constat composition of mixture is obtained. A curve of the shape      (shown in Fig) is obtained.

At constant temperature and pressure, the direction of spontaneous change is towards lower Gibb’s energy. The equilibrium composition of a reaction mixture corresponds to the lowest point on the curve. In this example substatial quantities of both reactants and products are present at equilibrium and K is close to 1.

The composition at lowest point of the curve – point of minimum Gibb’s energy – corresponds to equilibrium. For a system at equilibrium , any change either in the forward reaction or reverse reaction, would lead to increase in Gibb’s energy. In other words, when system attains equilibrium, any change in forward or reverse direction is not spontaneous.

The following generalisations can be made from the above diagram :

·         When Gibb’s energy minimum lies very close to the products, the equilibrium composition strongly favours products.  In other words, the reaction goes nearly to completion i.e., K >> 1. This is shown in Fig (a)

(a) At equilibrium , the products are much more in abundance than the reactants ( K >> 1).

·         When the Gibb’s energy minimum lies very close to the reactants, the equilibrium composition favours the reactants and the reaction forms only a small amounts of products. In other words, the reaction does not proceed much i.e., K << 1. This is shown in Fig (b).

(b) Equilibrium lies in favour of reactants ( K << 1).

·         When the Gibb’s energy minimum lies approximately half-way between the reactants and products, both the reactants and products are present in almost equal concentration at equilibrium , i.e., K is close to 1.

In order to make  these ideas quantitative, we must consider Gibb’s energy change , D_rG and standard Gibb’s energy change D_rG^Θ . Here D_rG^Θ is the difference in standard Gibb’s energies of products and reactants both in their standard states. Similarly, D_rG is the reaction Gibb’s energy change at a definite, fixed composition of reaction mixture.

            The Gibb’s energy of reaction D_rG , is related to composition of the reaction mixture and to standard reaction Gibb’s energy by the equation :

            D_rG =D_rG^Θ  + R T ln Q

where Q is the reaction quotient and R is the gas constant, with the value 8.314 J/K/mol . If the species are gases, these concentrations are expressed in partial pressure and the reaction quotient will be Qp and if the species are in solution, reaction quotient will be expressed in terms of molar concentration as Qc.

At equilibrium , Q = K and therefore, above equation becomes :

                0 =  D_rG^Θ  + R T ln K

   or  D_rG^Θ  =  -R T ln K

   or  D_rG^Θ  =  - 2.303 R T log K  …. (1)

The equation can also be written as :

             
We also know that

         D_rG =  D_rH^Θ  - T DS =  - R T ln K

Since  D_rG^Θ depends on the value of D_rH^Θ, for strongly endothermic reactions the value of D_rH^Θ may be large and positive. In such case the value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, D_rH^Θ is large and negative and D_rG^Θ is likely to be large and negative too. In such cases , K will be much larger than 1. We may expect strongly exothermic reactions to have a large K , and hence can go to near completion.

            Using equation (1) we can determine the value of D_rG^Θ if the value of K is known or vice versa.

Problems

79.   For the water gas reaction:

C(s) +  H₂O(g) ⇌CO(g) + H₂(g)

The standard Gibb’s energy of reaction (at 1000 K) is               - 8.1 kJ mol^-1.  Calculate equilibrium constant.

80.   The standard Gibb’s energies for the reaction at 1773 K are given below :

C(s) + O₂(g) ® CO₂(g) : DrG^⊝= - 380 kJ mol^-1

2    C + O₂(g) ⇌ 2 CO₂(g) : DrG^⊝= - 500 kJ mol^-1

        Discuss the possibility of reducing Al₂O₃ and PbO with carbon at this temperature.

81.   Consider the following reaction :

2 NO(g) + O₂(g) ® 2 NO₂(g)

Calculate the value of standard Gibb’s energy at 298 K and predict whether the reaction is spontaneous or not.

82.   It is planned to carry the reaction :

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

a 1273 K at 1 bar.

(a)          Is this reaction spontaneous at this temperature and pressure.

(b)          Calculate the value of :

(i)        Kp at 1273 K for the reaction.

(ii)      Partial pressure of CO₂ at equilibrium.

83.   The equilibrium constant for the reaction :

PCl₅ (g) ⇌  PCl₃(g) + Cl₂(g)

at 298 K has been found to be 1.8 x 10^-7. Calculate DG^Θ for the reaction.

84.  Using the following data , calculate the value of equilibrium constant  for the following reaction at 298 K.

2 CHºCH ⇌ C₆H₆(g)

assuming ideal behaviour

D_fG^Θ (CHºCH) = 2.09 x 10⁵ J/mol

D_fG^Θ (C₆H₆) = 1.24 x 10⁵ J/mol

Can the reaction be recommended for the synthesis of benzene ?

85.   Calculate equilibrium constant for the reaction at 400 K.

2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g)  : DH^Θ  = 77.2 kJ/mol ;

                                 DS= 122 J K^-1mol^-1 at 400 K.

86.   The equilibrium constant at 298 K  for the process :

Co³⁺(aq) + 6 NH₃ ⇌ [Co(NH₃)₆]³⁺(aq)

is 2.0 x 10⁷ .  Calculate the value of DG^Θ at 298 K.

87.   The  D_rG^Θ for conversion of oxygen to ozone

3/2 O₂(g) ® O₃(g)

at 298 K , if Kp for this conversion is 2.47 x 10^-29.  

88.  Find out the value of equilibrium constant for the following reaction at 298 K.

2 NH₃(g) + CO₂(g) ⇌ NH₂CONH₂(aq) + H₂O(ℓ)

Standard Gibb’s energy change, D_rG^Θ , at the given temperature is -13.6 kJ mol^-1_. 

89.   Calculate standard free energy change for the reaction :

Zn + Cu²⁺(aq) ® Cu + Zn²⁺(aq) : E^Θ = 1.10 V

90.  Calculate at 298 K , the equilibrium constant for the reaction :

Cu(s) + 2 Ag⁺(aq) ® Cu²⁺(aq) + 2 Ag(s)

At 298 K E^Θ cell = 0.47 V.

EXERCISES

1.      What is meant by the term 'chemical energetics'?

2.      What are primary sources of energy ?

3.      What is meant by renewable and non-renewable source of energy ? Give examples.

4.      In view of the limited availability of non-renewable sources suggest suitable energy sources for future needs of the country.

5.      Explain the terms : i) Hydro-power and ii) Geothermal power.

6.      Explain the term chemical energy.

7.      Explain the term 'internal energy'.

8.      Explain the term 'internal energy change'.

9.      What is meant by term 'state property'?

10.   Explain the terms : I) Thermodynamic system and ii) Surroundings.

11.   What the origin of energy change in a chemical reaction ?

12.   Define the terms :  I) Open system  ii) Closed system  iii) isolated system  iv) extensive property   v) intensive property   vi) isothermal process   vii) adiabatic process.

13.   State and explain the Law of Conservation of energy.

14.   What are spontaneous reactions ? Give suitable examples.

15.   Define Enthalpy of a system.

16.   How is enthalpy related to the internal energy of the  system ?

17.   Mention the factors which influence the value of enthalpy change in a chemical reaction. Define standard molar enthalpy change.

18.   Define exothermic and endothermic reactions.

19.   State and explain Hess's Law of constant heat summation by taking suitable examples.

20.   Write a mathematical relationship between heat, internal energy and work done by the system.

21.   Explain the term heat of reaction. Give two examples.

22.   Explain the terms :  I) Enthalpy of combustion.  ii)  Enthalpy of formation   iii) Enthalpy of vaporisation  iv) Enthalpy of fusion   v) Enthalpy of sublimation

23.   What is bond energy ?

24.   Explain why the enthalpy of neutralisation of weak acid by a strong base is less than that of strong acid ?

25.   The heat of combustion of methane is - 890.65 kJ/mol and heats of formation of CO₂ and H₂O are - 393.5 and - 286.0 kJ/mol respectively. Calculate the heat of formation of methane (ans : - 74.85 kJ)

26.   Write notes on :

(i)       entropy of a system

(ii)      free energy of a chemical reaction

27.     Why the change in enthalpy cannot be a sole criterion for the spontaneity of a process ?

28.     In relation to DH, DS and DG values for a reaction, when could it :

(i)       proceed forward spontaneously

(ii)      be at equilibrium

29.     Why does entropy of a solid increases on fusion ?

30.     Why is entropy of ice less than that of water ?

31.     Indicate whether the entropy increases or decreases in the following changes.   Give reasons :

(i)       evaporation of water

(ii)      CaCO₃(s) ® CaO(s) + CO₂(g)

(iii)     2 SO₂(g) + O₂(g) ® 2 SO₃(g)

32.     Discuss the nature of change in entropy(increase or decrease) for the following :

(i)       H₂O(l) ® H₂O(s)

(ii)      MgCO₃(s) ® MgO(s) + CO₂(g)

(iii)     2 SO₂(g) + O₂(g) ® 2 SO₃(g)

33.     State the relationship amongst changes in free energy, enthalpy and entropy of a system at a given temperature. How does these changes , being small or large or negative of positive , affect the feasibility of a process ?

34.     Predict the feasibility of a reaction when :

(a)      both DH and DS increase ;

(b)      both DH and DS decrease

(c)      DH decreases, DS increases.

35.     Discuss the conditions for spontaneous occurrence of a process.

36.     What is meant by a ‘fuel’ ?

37.     What is meant by calorific value of a fuel ?

38.     Write a note on calorific value of foods.

39.     Will the temperature of the surrounding air increase or decrease, if :

(a)      an exothermic reaction is allowed to occur very rapidly in air ?

(b)      an endothermic reaction is allowed to occur ver rapidly in air ?

40.     Discuss the significance of the mathematical expression in which the heat absorbed by a system is related to internal energy and work done by the system.

41.     How  Gibb’s free energy is related to enthalpy, entropy and temperature of a system ? How is this used in determining the spontaneity of a process ?

42.     Why is bond energy of Cl - Cl less than that of H- H ?

43.     Predict if the following reaction :

CH₃CH₂OH(g) ® CH₂=CH₂(g) + H₂O(g)

will be exothermic or endothermic. Give reason for

your answer.

44.     Why is more energy required to break  C- H bond in acetylene compared to ethane ?

45.     Why is bond energy of ethene is less than that of   acetylene ?

46.     Explain the following terms :

(a)       zeroth  law of thermodynamics.

(b)       Reversible and irreversible process

47.     Explain two alternative energy sources for our future.

48.     Define heat capacity, specific heat and molar heat capacity. How are they related ?

49.     Define the terms:

(i)        Enthalpy of fusion.

(ii)       Enthalpy of vaporisation.

(iii)      Enthalpy of sublimation

How are these terms related ?

50.      What are extensive and intensive properties ?

MULTIPLE CHOICE QUESTIONS

1.         ‘The total  energy of an isolated system is a constant’. This statement  corresponds to :

a.            Zeroth law of thermodynamics

b.           First Law of thermodynamics

c.            Second Law of  thermodynamics

d.           Third Law of thermodynamics

e.            Dulong and Petit’s Law

2.         The enthalpy of combustion of carbon and carbon monoxide are - 394 kJ mol^-1 and - 284.5 kJ mol^-1. The enthalpy of formation of carbon monoxide is : 

a.       +109.5 kJ mol^-1        b.     - 109.5 kJ mol^-1

c.    +219    kJ mol^-1       d.    - 219.0 kJ mol^-1          e.    - 54.5 kJ mol^-1

3.         0.2 g of n-butanol was burnt in a suitable apparatus. The heat evolved was sufficient to increase the temperature of 100 g of water by 10⁰C. The heat of combustion of butanol is : 

a.       370 kcal mol^-1        b.    570 kcal mol^-1

b.       1480 kcal mol^-1       d.    37 kcal mol^-1

       e.     185  kJ mol^-1

4.         A gas absorbs 125 J of heat and also expands against 1.20 atm from volume of 0.5 L to 1.0 L.  The change in internal energy is : 

a.       + 32 J                b.    +16 J              c.   48 J

d.     96 kJ                  e.   + 64 J

5.         The energy required  for sublimation of one mole of carbon is 716.7 kJ mol^-1. The energy required to break one mole of O=O and C=O bonds are 494 kJ and 707 kJ respectively. The enthalpy  change for the reaction : 

                    C(s) +  O₂ (g) ®  CO₂(g)

a.         - 102 kJ                 b.    - 51 kJ      

c.    - 204 kJ               d.    - 104 kJ         

e.    - 252 kJ  

6.         The energy needed to raise the temperature of 10.0 g of iron from 25°C to 500°C if specific gravity of iron is 0.45 J(°C)^-1 g^-1.

a.     2.1 x 10⁵J                         b.   4.2 x 10⁴ J

c.     4.4 x 10⁵ J                        d.   6.3 x 10³ J 

e.    2.1 x 10³ J

7.         The time in minutes required to vapourise 100 g of benzene at its boiling point by  means of a   100 W electric heater, if the standard vapourisation enthalpy of benzene at its boiling point is  30.8 kJ mol^-1:

 a.   1.1                      b.   2.2       

 c.   4.44                                d.   6.6               e.    3.33

8.         The enthalpy change (DH) for the reaction ,

             N₂(g) + 3 H₂(g) ® 2 NH₃ (g)

is -92.38 kJ at 298 K. The  value ofDU at 298 K is :           

       a.  -87.4 kJ mol^-1                     b.   -92.38 kJ mol^-1

       c.   -108.2 kJ mol^-1                    d.   -46.2 kJ mol^-1

       e.    -174.8 kJ mol^-1

9.         A 1.25 g sample of octane is burned in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 K to 300.78 K. If heat capacity of the calorimeter is 8.93 kJ K¹, the heat transferred to the calorimeter is : 

a.    25 J                 b.   50 J                 c.   75 J     

d.   100 J                      e.  125 J

10.      In the graph :    

a.         I represents isobar and II represents isochore.

b.         both are isotherms

c.         I represents isotherm and II represents isochore

d.         I represents isochore and II represents isobar

e.         Both are isobars

11.      For the combustion of ethanol at 298 K , the difference between DH^o and  will be  DU^o : 

a.       - 2.48 kJ                  b.   2.48 kJ

       c. - 3.716 kJ d.   7.43 kJ         e.    9.32 kJ

12.      The standard molar enthalpy of formation of cyclohexene (ℓ)and benzene(ℓ) are - 156 and   + 49 kJ mol^-1. The standard enthalpy of hydrogenation of cyclohexane(ℓ) is                     - 119 kJ mol^-1 . The resonance energy of benzene is : 

a.   75 kJ mol^-1             b.   - 201 kJ mol^-1

c.   152 kJ mol^-1            d.      125 kJ mol^-1

        e.   840 kJ mol^-1

13.      The ratio of enthalpy of vapourisation  and normal boiling point of a liquid is approximately equals    88 J mol^-1 K^-1. This is called :     

 a.     Debye law             b.    Boyle’s law

 c.     Trouton’s law          d.    Raoult’s law

 e.     Debye law

14.      The DH for the reaction :        

           CH₂Cl₂(g) ®  C(g) + 2 H(g) + 2 Cl(g)

       The average  bond energies of C-H, C-Cl bonds are  413 kJ mol^-1 and 326 kJ  mol^-1 respectively :

a.       1478 kJ               b.   741 kJ

c.    3571 kJ               d.   341 kJ          e.    180  kJ

15.      The enthalpy per mole of a monoatomic ideal gas is :

a.       (5/2) RT                b.   (3/2) RT                c.    RT       

        d. 3 RT                    e.   (2/3)RT

16.      The enthalpy of combustion of benzoic acid at 298 K    and 1 atm.  is -3546 kJ mol^-1. The DU for the reaction is  : 

 a.  -1545 kJ mol^-1                  b.  -3544 kJ mol^-1

 c.  -3544.8 kJ mol^-1                d.  -4444.8 kJ mol^-1

         e.  -2544.8 kJ mol^-1

17.      1 g of propane was burnt in excess of oxygen in bomb calorimeter, when 52.5 kJ of heat is evolved at 25⁰C. The enthalpy change for the reaction is :  

            C₃H₈(g) + 5 O₂(g) ®  3 CO₂(g) + 4 H₂O(ℓ)

a.       -2318 kJ                          b.   -1318 kJ

        c.     -1118 kJ                       d.   -2418 kJ        

e.     -3418 kJ

18.      The law of thermodynamics that provides the basis for the determination of absolute entropy of a substance is :    

a.     zeroth                             b.   first law           

c.     second law                        d.   third law

e.      Hess’s law

19.      Match the quantities in List I with that in List II

        List I                    List II

A.   Heat change at                      1.   q  +   w

      constant  volume

B.   Heat change at                      2.    Dn R T

      constant pressure

C.    DH  -   DU                      3.    -PDV

D.   Change in                           4.    DH

     internal energy

E.   work done of                       5.   DU

      exapansion of a gas

Code

a.         A-1 ;  B-2 ;  C-3 ;  D-4 ; E-5

b.         A-5 ;  B-4 ;  C-2 ;  D-3 ; E-1

c.         A-5 ;  B-4 ;  C-2 ;  D-1 ; E-3

d.         A-2 ;  B-3 ;  C-4 ;  D-5 ; E-1

e.         A-3 ;  B-2 ;  C-5 ;  D-4 ; E-1

20.      One mole of an ideal gas is expanded freely and isothermally at 300 K from  10 L to 100 L .   If DU = 0 , the value of   DH is :  

a.       zero                 b.     10 kJ           c.   200  kJ

d.    -  200 10 kJ         e.    300 10 kJ

21.    Which of the following is an intensive property ?  

a.         surface tension

b.         temperature

c.         density

d.         pressure

e.         volume

20.      The difference between DH  and DU at 300 K for the reaction 2 A(g) ®   B₂(g)  + 3 O₂(g) in   joule  will be : 

a.         8.314  x 273 x (-2)

b.         8.314 x 300 x 2

c.         8.314 x 300 (-2)

d.         8.314 x 27 x (-2)

e.         8.314 x 300 x (1)

21.      A system which exchange  energy but prevents matter with its surroundings is called : 

a.             open system

b.             isolated system

c.             closed system

d.             adiabatic system

e.             isochoric system

22.      Which of the following is  not a state  function ? 

a.       free energy

b.       entropy

c.       enthalpy

d.       internal energy

e.       work

23.      The enthalpy of dissolution of anhydrous and hydrated copper sulphate are - 66.5 kJ mol^-1 +11.70 kJ mol^-1  respectively. The enthalpy of hydration of CuSO₄ to CuSO₄ 5 H₂O is :

a.   -78.20 kJ                  b.    + 78.20 kJ

c.   -39.10 kJ              d.    -117.20 kJ   

e.   -18.20 kJ

24.      In the graph shown : 

a.         I is isothermal and II is adiabatic.

b.         I and II represents isothermal processes.

c.         I is adiabatic and II is isothermal

d.         I and II are isobars

e.         I is  isochore and II isobar

25.      A gas expands from 1 L to 2 L at a constant pressure of 10 atmospheres. The   pressure – volume work done is : 

a.    1013.25 J             b.    - 1013.25 J

c.     2026.50 J                        d.    - 2026.50 J

e.    - 506.50 J

26.      The enthalpy of combustion glucose is -1440 kJ. Its calorific value is :

a.   8 kJ                    b.   1440 kJ           c.   16 k

d.   18 kJ                        e.   60 kJ

27.      The bond energies of C-H,  O=O , C=O, O-H  and C=C bonds  are 414, 499, 724, 460 and 619 kJ respectively. The enthalpy change for the reaction :

            CH₂=CH₂ + 3 O₂  ®   2 CO₂  + 2 H₂O

a.       964 kJ                b.   364 kJ            

c.     464 kJ                d.   964 kJ            e.   864 kJ

28.      An athlete is given 100 g of glucose of energy equivalent to 1760 kJ. He utilises 50% this gained energy in the event. In order to avoid storage of energy in the body, the mass of water he would need to perspire (enthalpy of vaporisation of water is     44  kJ mol^-1.

a.   100 g                   b.   360 g   

c.   440 g                               d.    520 g             e.   180 g

1	2		3		4		5		6		7		8
b	b		a		e		c		e		d		a
9		10		11		12		13		14		15
c		a		c		c		c		a		a
16		17		18		19		20		21		22		23
c		a		d		c		a		e		b		c
24		25		26		27		28		29		30
e		a		a		b		a		a		b