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

##### 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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