For the reaction at 298 K, 2A + BC. = 400 kJ and = 0.2 kJ . The reaction will become spontaneous above ________ K.

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Solution

Q.
For the reaction at 298 K, 2A + B $\to$ C. $∆ H$ = 400 kJ ${mol}^{- 1}$ and $∆ S$ = 0.2 kJ ${mol}^{- 1}$ $K^{- 1}$ . The reaction will become spontaneous above ________ K. [2024]

Ans.
(2000)

For reaction to be spontaneous,

$Δ G < 0$

$Δ H - T Δ S < 0$

$400 kJ$ ${mol}^{- 1} - T \times 0.2 kJ$ ${mol}^{- 1} K^{- 1} < 0$

$400 kJ$ ${mol}^{- 1} < T \times 0.2 kJ$ ${mol}^{- 1} K^{- 1}$

$T > \frac{400}{0.2} K$

$T > 2000 K$

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