For the reaction at 298K,

2A + B → C

\( \Delta H = 400 kJ mol^{-1}\)

\( \Delta H = 0.2 kJ mol^{-1}K^{-1}\)

At what temperature will the reaction become spontaneous considering \( \Delta S\; and\; \Delta H\) to be constant over the temperature range?

Asked by Abhisek | 1 year ago |  188

1 Answer

Solution :-

Now,

\( \Delta G = \Delta H – T\Delta S\)

Let, the given reaction is at equilibrium, then \( \Delta T\) will be:

T =\( (\Delta H – \Delta G)\dfrac{1}{\Delta S}\dfrac{\Delta H}{\Delta S}\) 

(\( \Delta G\) = 0 at equilibrium)

\( \dfrac{ 400kJ mol^{-1}}{0.2kJ K^{-1}mol^{-1}}\)

Therefore, T = 2000K

Thus, for the spontaneous, \( \Delta G\) must be –ve and T > 2000K.

Answered by Pragya Singh | 1 year ago

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