Question

In the reaction,
$P+Q⟶R+S$
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

• A. 2
• B. 3
• C. 0
• D. 1

Answer 1

Answer: (D)

The time taken for $75\%$ reaction of $P$ is twice the time taken for $50\%$ reaction of $P$.
For $P$, if $t_{50\%}=x$ then $t_{75\%}=2x$
$P\stackrel{t_{1/2}}{\to }\frac{P}{2}\stackrel{t_{1/2}}{\to }\frac{P}{4}$
This true only for the first-order reaction.
So, order with respect to $P$ is $1$
Further, the graph shows that concentration of $Q$ decreases linearly with time. So rate with respect to $Q$, remains constant.
Hence, it is zero-order with respect to $Q$
So, the overall order is $n=1+0=1$.