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6.22   Calculate the entropy change in surroundings when 1.00 mol ofH_2 O (l) is formed under standard conditions. \Delta _f H ^ \ominus = -286 KJ mol ^ { -1} 

-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also. kJ/mol We know that, Entropy changes for the surrounding is  = q(surr)/ temp.                                                                 = 286000/ 298                                                                 = 959.73 J/mol/K

6.21   Comment on the thermodynamic stability of NO(g), given

1/2 N_2(g) + 1/2 O_2(g)\rightarrow NO(g) ; \Delta _ rH ^ \ominus = 90 kJ mol ^{-1}\\\\ .\: \: \: NO(g) + 1/2 O_2(g) \rightarrow NO_ 2(g) : \Delta _ r H^ \ominus = -74 kJ mol ^{-1}


 

The formation of  is unstable because  is positive it means heat is absorbed during the reaction. So,  (g) has higher energy than its reactants  and . On the other hand,  is stable because  is negative means heat is released during the formation of . It is stablised with minimum energy. Hence unstable  changes to stable .

6.20  The equilibrium constant for a reaction is 10. What will be the value of   

             \Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K  

Given values are, and equilibrium constant  = 10 It is known that,             Hence the value of  is -5.744 kJ/mol

6.19   For the reaction
           2 A (g) + B ( g ) \rightarrow 2 D ( g)
           \Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}
          Calculate \Delta G ^ \ominus  for the reaction, and predict whether the reaction may occur spontaneously.

Given reaction is  We know that,  and  here  and   substituting the given values in equations-            Hence the reaction will not occur spontaneously because the value is positive for this reaction.

6.18   For the reaction,  2 Cl (g) \rightarrow Cl_2 ( g), what are the signs of  \Delta H \: \: and \: \: \Delta S ?

The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So,  is negative.  Two moles of an atom have more randomness than the one mole of chlorine. So, spontaneity is decreased. Hence  is negative.

6.17   For the reaction at 298 K,
            2A + B \rightarrow C
          \Delta H = 400 kJ mol^ {-1 }\: \: and \: \: \Delta S = 0.2 kJ K ^{-1} mol^{-1}
At what temperature will the reaction become spontaneous considering \Delta H and \Delta S to be constant over the temperature range.

From the equation, Suppose the reaction is at equilibrium, So the change in temperature is given as;         ( at equilibrium is zero)     To reaction should be spontaneous,  should be neagtive. So, that for the given reaction T should be greater than 2000 K

6.16   For an isolated system, \Delta U = 0 , what will be \Delta S?

Since  So, the  (positive) therefore the reaction will be feasible.

6.15   Calculate the enthalpy change for the process  CCl_4(g)\rightarrow C(g) + 4 Cl(g) and calculate bond enthalpy of C-Cl in CCl_4(g)
        \Delta _{vap} H ^{\ominus }(CCl_4) = 30.5 kJ mol^{-1}.\\\\ .\: \: \: \: \: \Delta _f H ^\ominus (CCl_4) = -135.5 kJ mol^{-1}.\\\\.\: \: \: \: \Delta _ a H ^ \ominus (C) = 715.0 kJ mol ^{-1} , where \: \: \Delta_ a H ^\ominus \: \: is\: \: enthalpy \: \: of \: \: atomisation\\\\.\: \: \: \: \Delta _ a H ^\ominus (Cl_2) = 242 kJ mol ^{-1}

We have the following chemical reactions equations- ................ ........................ .................. ............  The enthalpy change for the process  by the above reaction is calculated as;                        And the bond enthalpy of C-Cl bond in  (g) = 1304/4 = 326 kJ/mol

6.14  Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

    CH_3OH (l) + 3/2 O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \Delta _r H = -726 kJ mol^{-1}\\\\ .\: \: \: \: C(graphite) + O_2(g) \rightarrow CO_2(g) ;\Delta_ cH = -393 kJ mol ^ {-1}\\\\ .\: \: \: H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l) ; \Delta_ f H = - 286 kJ mol^{-1}.

for the formation of  the reaction is , This can be obtained by the following expressions- required eq = eq (i) + 2 (eq iii) - eq (i)                                                                             

6.13  Given  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ;\Delta rH = -92.4 kJ mol ^{-1}.What is the standard enthalpy of formation of NH_3 gas?

the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements. Thus we can re-write the reaction as; Therefore, standard enthalpy of formation of ammonia is =                                                                                      = 1/2 (-92.4)                        ...

6.12  Enthalpies of formation of CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g) are –110, – 393, 81 and 9.7 KJ mol ^{-1}  respectively. Find the value of \Delta _ r H for the reaction:

           N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)

Given, Enthalpies of formation of  are –110, – 393, 81  and 9.7  We know that the (product)(reactants) For the above reaction, Substituting the given values we get,               Thus the value of  of the reaction is -777.7 kJ/mol

6.11 Enthalpy of combustion of carbon to CO_2  is -393.5 KJ mol ^{-1}  Calculate the heat  released upon formation of 35.2 g of CO_2 from carbon and dioxygen gas.

Formation of carbon dioxide from carbon and dioxygen reaction is-     .............   = -393.5 kJ/mol 1 mole of   = 44 g So, the heat released in the formation of  44g of  = -393.5 kJ/mol therefore, In 35.5 g of   the amount of heat released =                                                                                                                                                      ...

6.10  Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 \degree C  to ice at  - 10.0 \degree C.

       \Delta _{ fus }H = 6.03 KJ mol ^{-1} at 0 \degree C.

Cp [H_2O(l)] = 75.3 J mol^{-1} K^{-1}\\\\.\: \: \: \: Cp [H_2O(s)] = 36.8 J mol^ { -1}K^{-1}

Total enthalpy change is equal to the summation of all the energy required at three different stages- (i) from  to  of 1 mol of water (water water ) (ii) from  to  of 1 mol of ice (water ice) (iii) from  to  of 1 mole of ice (ice ice) So, the total enthalpy change              Hence the total enthalpy change in the transformation process is 7.151 kJ/mol

6.9  Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35 \degree C to 55 \degree C. Molar heat capacity of Al is 24 J mol ^{-1} K ^{-1}

We know that  m = mass of the substance  c = heat capacity  = change in temperature By putting all these values we get,      = 1066.7 J      = 1.07 kJ

6.8  The reaction of cyanamide,NH_2 CN (s), with dioxygen was carried out in a bomb calorimeter, and \Delta U was found to be -742.7 KJ mol ^{-1} at 298 K. Calculate enthalpy change for the reaction at 298 K.

         NH_2 CN ( g ) + \frac{3}{2} O_2 (g) \rightarrow N_2 + CO_2 ( g ) H_2 O ( l)

Given information, T = 298 K R = 8.314 (products)(reactants)            = (2 - 1.5)            = 0.5 moles The enthalpy change for the reaction is expressed as;   where,  = change in internal energy  and   change in no. of moles By putting the values we get,            

6.7  In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

The first law of Thermodynamics state that, where,  = change in internal energy for the process q = heat and W = work Given that,  q = +701 J (heat is absorbed) W = - 394 (work is done by the system) Substituting the value in equation of first law we get, So, the change in internal energy for the process is 307 J

6.6   A reaction, A + B \rightarrow C + D + q is found to have a positive entropy change. The reaction will be
         (i) possible at high temperature
        (ii) possible only at low temperature
        (iii) not possible at any temperature
         (v) possible at any temperature

For the reaction to be feasible the  should be negative According to question,  = positive unit  = negative (as heat is evolved in the reaction) Overall the  is negative. Therefore the reaction is possible at any temperature. So, the correct option is (iv)

6.5  The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.KJ mol ^{-1} , -393.5 KJ mol ^{-1} \: \: and -285 .8 KJ mol ^{-1}respectively. Enthalpy  of formation of  CH_4 (g)  will be

         (i) –74.8 kJ mol^{-1}

         (ii) –52.27 kJ mol^{-1}

         (iii) +74.8 kJ mol^{-1}

         (iv) +52.26 kJ mol^{-1}

................. -890.3 kJ/mol                                  ..................-393.5 kJ/mol                              ..................-258.8 kJ/mol So, the required equation is to get the formation of  by combining these three equations- Thus,                      = [-393.5 + 2(-285.8) + 890.3]                     = -74.8 kJ/mol therefore, the enthalpy of formation of methane...

6.4   \Delta U ^\ominusof combustion of methane is - X KJ mol ^{-1}. The value of \Delta H ^\ominus is
             (i) = \Delta U ^\ominus
             (ii)> \Delta U ^\ominus
             (iii) < \Delta U ^\ominus
             (iv) = 0

Equation of combustion of methane- Since, and from the equation so the correct option is (iii).  

6.3   The enthalpies of all elements in their standard states are:
           (i) unity
           (ii) zero
           (iii) < 0
           (iv) different for each element

The enthalpies of all elements in their standard states are Zero.  So, the correct option is (ii)
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