Consider the given endothermic reaction at eq | vedclass

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(A) For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$\Delta n_g = 2 - 1 = 1 > 0$.Effect-$1$: The graph shows an instantaneous

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$P$ increase,$T$ increase

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(A) For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$\Delta n_g = 2 - 1 = 1 > 0$.Effect-$1$: The graph shows an instantaneous increase in the concentration of all species $(PCl_5, PCl_3, Cl_2)$. This corresponds to a decrease in volume,which is equivalent to an increase in pressure ($P$ increase). According to Le Chatelier's principle,increasing pressure shifts the equilibrium towards the side with fewer moles of gas (left side),so $[PCl_5]$ increases while $[PCl_3]$ and $[Cl_2]$ decrease.Effect-$2$: The graph shows a gradual change in concentrations without an instantaneous jump. This indicates a change in temperature. Since the reaction is endothermic,increasing the temperature ($T$ increase) shifts the equilibrium to the right (forward direction). Consequently,$[PCl_3]$ and $[Cl_2]$ increase,while $[PCl_5]$ decreases. This matches the observed trends in the graph.

Consider the given endothermic reaction at equilibrium,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. $A$ graph is plotted between concentration and time as shown. Effect-$1$ and Effect-$2$ are due to respectively:

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