Question

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

Solution

Let $increment straight G$ be zero at temperature TK.
$Now comma space increment straight G space equals space increment straight H space minus space straight T increment straight S therefore space space space space space space space 0 space equals space increment straight H space minus space straight T increment straight S therefore space space space space space space space straight T space equals space fraction numerator increment straight H over denominator increment straight S end fraction space equals space fraction numerator 30.56 space kJ space mol to the power of negative 1 end exponent over denominator 0.066 space kJ space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent end fraction space space space space space space space space space equals space 463.03 space straight K$
(i) At this temperature, the reaction proceeds in either direction at the same rate and it represents the state of equilibrium.
(ii) Below this temperature,
T = 462 K (say)
$therefore space space space space space space increment straight G space equals space increment straight H space minus straight T increment straight S space space space space space space space space space space space space space space space space equals space 30.56 space kJ space mol to the power of negative 1 end exponent space minus space left parenthesis 462 space straight K right parenthesis space cross times space left parenthesis 0.066 space kJ space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis space space space space space space space space space space space space space space space space equals space 30.56 space kJ space mol to the power of negative 1 end exponent space minus space 30.942 space kJ space mol to the power of negative 1 end exponent space space space space space space space space space space space space space space space space space equals space plus 0.068 space kJ space mol to the power of negative 1 end exponent$
As ∆G is +ve below this temperature, the reaction will not be spontaneous.

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