In the reaction, P + Q → R + S the time taken for 75 % reaction of...

The reaction is:

$\text{P} + \text{Q} \to \text{R} + \text{S}$
$rateofthereaction = k{\left[P\right]}^{x}{\left[Q\right]}^{y}$
The concentration of Q varies with time as per the graph by the equation.
\left[\text{Q}\right] = \left[{\text{Q}_{0}\right] - \text{kt } i.e. Zero-order kinetics.

$⇒ y = 0$
With respect to reactant $P$

let time taken for $50$ completion of the reaction is $a$.

Then the time is taken for $75$ complication will be $2a .</p><p>$P\overset{{t}_{50 %}}{\to}$\dfrac{P}{2}$\overset{{t}_{75 %}}{\to}$\dfrac{P}{4}</p><p>at</p><p>$t = 0; [ P{]}_{0 %} = Pt = {t}_{50 %}; [ P{]}_{50 %} = $\dfrac{P}{2}$t = {t}_{75 %}; [ P{]}_{75 %} = $\dfrac{P}{4}<br />As we can see,$ in each time interval of 'a' min the reactant concentration is changing by factor half. therefore rate constant is independent of the concentration of the reactantP$. hence with respect of reactant$P$, order of the reaction is$1$. </p><p>$⇒ x = 1$<br />Rate law expression is$rate = k{\left[P\right]}^{1}{\left[Q\right]}^{0}$<br />The overall order of reaction$= x + y = 1 + 0 = 1$</p>$