For the reaction PCl_5 rightleftharpoons PCl_ | vedclass

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(D) The given reaction is $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.According to Le Chatelier's Principle,for a reaction where the number of mo

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(D) The given reaction is $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.According to Le Chatelier's Principle,for a reaction where the number of moles of gaseous products is greater than the number of moles of gaseous reactants (here,$\Delta n_g = 2 - 1 = 1 > 0$),the forward reaction is favoured by:$1$. Increasing the volume of the container (which decreases the total pressure).$2$. Introducing an inert gas at constant pressure (which increases the volume,thereby decreasing the partial pressure of the reactants).$3$. Removing the products ($PCl_3$ or $Cl_2$) or adding more reactants $(PCl_5)$ at constant volume.Since all the given options favour the forward reaction,the correct answer is $D$.

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the forward reaction at constant temperature is favoured by

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