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(D) For the given reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$The change in the number of moles of gaseous products and reactants is $

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No effect

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(D) For the given reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$The change in the number of moles of gaseous products and reactants is $\Delta n_g = n_{p} - n_{r} = 2 - (1 + 1) = 0$.According to Le Chatelier's principle,if $\Delta n_g = 0$,a change in pressure or volume has no effect on the position of the equilibrium.Since the pressure is halved and volume is doubled,the total number of moles of gas on both sides of the equation is equal,so the equilibrium position remains unchanged.

What is the effect of halving the pressure by doubling the volume on the following system at $500\,^{\circ}C$?
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

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What is the effect of halving the pressure by doubling the volume on the following system at $500\,^{\circ}C$?$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$