H_{2(g)} + I_{2(g)} rightleftharpoons 2HI_{(g | vedclass

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Question

(C) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta

Vedclass · Accepted Answer

Unaffected by change in pressure

Vedclass · Answer

(C) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = n_p - n_r = 2 - (1 + 1) = 0$.Since $\Delta n_g = 0$,the equilibrium position is unaffected by a change in pressure.Given $\Delta H = +q \ cal$,the reaction is endothermic.According to Le Chatelier's principle,increasing the temperature favours the forward (endothermic) reaction,while decreasing the temperature favours the backward reaction.Therefore,the formation of $HI$ is unaffected by a change in pressure.

$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$\Delta H = +q \ cal$. The formation of $HI$ is:

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