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

The correct option is D $1$

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 $n=1+0=1$.