Question

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

• 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 $t _{75 \%}=2 x$
$P \xrightarrow{ t _{1 / 2}}\frac{ P }{2} \xrightarrow{ t _{1 / 2}} \frac{ P }{4}$
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 $n =1+0=1$.