For the reaction Br_2 rightleftharpoons 2Br,t | vedclass

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Question

(A) For the reaction $Br_2 ightleftharpoons 2Br$,the equilibrium constants are given as:At $T_1 = 500 \, K$,$K_1 = 1 	imes 10^{-10}$At $T_2 = 700 \

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Endothermic

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(A) For the reaction $Br_2 ightleftharpoons 2Br$,the equilibrium constants are given as:At $T_1 = 500 \, K$,$K_1 = 1 	imes 10^{-10}$At $T_2 = 700 \, K$,$K_2 = 1 	imes 10^{-5}$Since the temperature increases $(T_2 > T_1)$ and the equilibrium constant also increases $(K_2 > K_1)$,the reaction is endothermic.According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction,leading to a higher value of the equilibrium constant $K$.

For the reaction $Br_2 \rightleftharpoons 2Br$,the equilibrium constants at $500 \, K$ and $700 \, K$ are $1 \times 10^{-10}$ and $1 \times 10^{-5}$ respectively. The reaction is .......

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