For the reaction:Calculate the temperature at which it attains e

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 Multiple Choice QuestionsShort Answer Type

271.

Calculate the free energy change of the reaction straight C left parenthesis graphite right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space CO subscript 2 left parenthesis straight g right parenthesis
left parenthesis Given space increment straight H space equals space minus 300 space kJ space mol to the power of negative 1 end exponent semicolon space space increment straight S space equals space 3 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent space at space 300 space straight K right parenthesis.

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272.

For the reaction:
straight N subscript 2 left parenthesis straight g right parenthesis space plus space 3 straight H subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 2 NH subscript 3 left parenthesis straight g right parenthesis comma
increment straight H equals negative 95.4 space kJ space and space increment straight S space equals space minus 198.3 space kJ divided by straight K.

Calculate the temperature at which it attains equilibrium.



According to Gibb's Helmoltz equation,
            

which it attains equilibrium,


      
    

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 Multiple Choice QuestionsLong Answer Type

273.

At what temperature, reduction of lead oxide to lead by carbon
                                
becomes spontaneous? For this reason  are 108.4 kJ mol-1 and 190.0 JK-1 mol-1 respectively. 

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274.

For the reaction at 298 K
                  2 straight A plus straight B rightwards arrow straight C
increment straight H space equals space 400 space kJ space mol to the power of negative 1 end exponent space and space increment straight S space equals space 0.2 space kJ space straight K to the power of negative 1 end exponent mol to the power of negative 1 end exponent.

At what temperature will the reaction become spontaneous considering increment straight H space and space increment straight S to be constant over the temperature range?

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 Multiple Choice QuestionsShort Answer Type

275.

For the reaction 
    2 straight A left parenthesis straight g right parenthesis space plus space straight B left parenthesis straight g right parenthesis space space space rightwards arrow space space space 2 straight D left parenthesis straight g right parenthesis
increment straight U to the power of 0 space equals space minus 10.5 space kJ space and space increment straight S to the power of 0 space equals space minus 44.1 space JK to the power of negative 1 end exponent.

Calculate increment straight G to the power of 0 for the reaction and predict whether the reaction may occur spontaneously. 

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 Multiple Choice QuestionsLong Answer Type

276.

For the reaction
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increment straight H space equals space 30.56 space kJ space mol to the power of negative 1 end exponent space and space increment straight S space equals space 0.06 space kJK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent left parenthesis at space 1 space atm right parenthesis, calculate the temperature at which increment straight G is equal to zero. Also predict the direction of the reaction at:
(i) temperature
(ii) below this temperature

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 Multiple Choice QuestionsShort Answer Type

277.

For the melting of ice at 25 degree straight C comma
               straight H subscript 2 straight O left parenthesis straight s right parenthesis space space space rightwards harpoon over leftwards harpoon space space straight H subscript 2 straight O left parenthesis l right parenthesis comma
the enthalpy of fusion is 6.97 space kJ space mol to the power of negative 1 end exponent and entropy of fusion is 25.4 JK to the power of negative 1 end exponent mol to the power of negative 1 end exponent. Calculate the free energy change and predict whether the melting of ice is spontaneous or not at this temperature. 

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 Multiple Choice QuestionsLong Answer Type

278.

What do you understand by:
(i) Standard free energy change (∆rG0)
(ii) Standard free energy of formation (∆fG0).

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 Multiple Choice QuestionsShort Answer Type

279.

Calculate the standard free energy change for the reaction
4 NH subscript 3 left parenthesis straight g right parenthesis space plus space 5 straight O subscript 2 space space space rightwards arrow space space space 4 NO left parenthesis straight g right parenthesis space plus space 6 straight H subscript 2 straight O left parenthesis straight l right parenthesis

Given that the standard free energies of formation increment subscript straight f straight G to the power of 0 for NH subscript 3 left parenthesis straight g right parenthesis comma space space NO left parenthesis straight g right parenthesis space and space straight H subscript 2 straight O left parenthesis straight l right parenthesis are -16.8, +86.7 and -237.2 kJ mol-1 respectively. Predict the feasibility of the above reaction at the standard state.

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280. How is ∆rG0 related to the equilibrium constant K?
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