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(C) The reaction is endothermic $(\Delta H > 0)$,so according to Le Chatelier's principle,increasing the temperature will shift the equilibrium in the

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Increasing temperature and decreasing pressure

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(C) The reaction is endothermic $(\Delta H > 0)$,so according to Le Chatelier's principle,increasing the temperature will shift the equilibrium in the forward direction.The change in the number of moles of gaseous products is $\Delta n_g = (2+1) - 2 = +1$.Since $\Delta n_g > 0$,decreasing the pressure will shift the equilibrium in the direction of more moles of gas,which is the forward direction.Therefore,increasing the temperature and decreasing the pressure will lead to the maximum decomposition of $H_2O_{(g)}$.

Consider the following reversible reaction at equilibrium,$2H_2O_{(g)} \rightleftharpoons 2H_{2(g)} + O_{2(g)}$; $\Delta H = 241.7 \ kJ$. Which one of the following changes in conditions will lead to maximum decomposition of $H_2O_{(g)}$?

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