In the reaction,P + Q longrightarrow R + S,th | vedclass

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(D) For reactant $P$,the time taken for $75 \%$ completion $(t_{75 \%})$ is twice the time taken for $50 \%$ completion $(t_{50 \%})$,i.e.,$t_{75 \%}

Vedclass · Accepted Answer

$1$

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(D) For reactant $P$,the time taken for $75 \%$ completion $(t_{75 \%})$ is twice the time taken for $50 \%$ completion $(t_{50 \%})$,i.e.,$t_{75 \%} = 2 	imes t_{50 \%}$.This relationship is characteristic of a first-order reaction.Therefore,the order of the reaction with respect to $P$ is $1$.For reactant $Q$,the graph shows that the concentration $[Q]$ decreases linearly with time. $A$ linear decrease in concentration with time indicates that the rate of reaction is independent of the concentration of $Q$.Therefore,the order of the reaction with respect to $Q$ is $0$.The overall order of the reaction is the sum of the individual orders: $1 + 0 = 1$.

In the reaction,$P + Q \longrightarrow R + S$,the time taken for $75 \%$ reaction of $P$ is twice the time taken for $50 \%$ reaction of $P$. The concentration of $Q$ varies with reaction time as shown in the figure. The overall order of the reaction is

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