For the reaction Br_2 rightleftharpoons 2Br,t | vedclass

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Question

(A) According to the van't Hoff equation,the relationship between equilibrium constant $(K)$ and temperature $(T)$ is given by: $\ln(\frac{K_2}{K_1})

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Endothermic

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(A) According to the van't Hoff equation,the relationship between equilibrium constant $(K)$ and temperature $(T)$ is given by: $\ln(\frac{K_2}{K_1}) = \frac{\Delta H^\circ}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.Given: $T_1 = 327 + 273 = 600\,K$,$T_2 = 527 + 273 = 800\,K$.$K_1 = 6.1 	imes 10^{-12}$,$K_2 = 1.0 	imes 10^{-7}$.Since $K_2 > K_1$ as temperature increases,the value of $\ln(\frac{K_2}{K_1})$ is positive.For the term $(\frac{1}{T_1} - \frac{1}{T_2})$ to be positive,$\frac{1}{600} > \frac{1}{800}$,which is true.Since both sides are positive,$\Delta H^\circ$ must be positive.Therefore,the reaction is endothermic.

For the reaction $Br_2 \rightleftharpoons 2Br$,the equilibrium constants at $327\,^oC$ and $527\,^oC$ are $6.1 \times 10^{-12}$ and $1.0 \times 10^{-7}$ respectively. The reaction is ...........

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