In the reaction PCl_{5(g)} rightleftharpoons | vedclass

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(A) At the start of the reaction,the concentration of reactants is high and products is zero,so the forward reaction rate is high and the backward rea

Vedclass · Accepted Answer

$Q > K_{eq} 	o 3, Q = K_{eq} 	o 2, Q < K_{eq} 	o 1$

Vedclass · Answer

(A) At the start of the reaction,the concentration of reactants is high and products is zero,so the forward reaction rate is high and the backward reaction rate is zero.As the reaction proceeds,the forward rate decreases and the backward rate increases until they become equal at equilibrium.$1$. $Q $2$. $Q = K_{eq}$: The system is at equilibrium. The rates of forward and backward reactions are equal (labeled $2$ in the graph).$3$. $Q > K_{eq}$: The reaction proceeds in the backward direction. This corresponds to the region where the backward rate is higher than the forward rate (labeled $3$ in the graph).Therefore,the correct mapping is $Q > K_{eq} 	o 3, Q = K_{eq} 	o 2, Q < K_{eq} 	o 1$.

In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,a graph is plotted to show the variation of the rate of forward and backward reactions against time. Which of the following is correct?

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