Overall Order of Reaction for Reactants P and Q

To determine the overall order of the reaction: $P + Q \longrightarrow R + S$ let the rate law for the reaction be represented as: $\text{Rate} = k [P]^x [Q]^y$ where $x$ is the order with respect to reactant $P$ and $y$ is the order with respect to reactant $Q$.

1. Order with respect to reactant $P$ ($x$): It is given that the time taken for $75\%$ reaction of $P$ is twice the time taken for $50\%$ reaction of $P$: $t_{75\%} = 2 \times t_{50\%}$ For a reaction of order $n$:

   • For a first-order reaction ($n = 1$): $t_{50\%} = t_{1/2} = \frac{\ln 2}{k}$ $t_{75\%} = \frac{\ln(1 / 0.25)}{k} = \frac{\ln 4}{k} = \frac{2 \ln 2}{k} = 2 t_{1/2}$ Since the relationship $t_{75\%} = 2 t_{50\%}$ holds true, the order with respect to reactant $P$ is: $x = 1$

2. Order with respect to reactant $Q$ ($y$): The given plot depicts the concentration of reactant $Q$, denoted as $[Q]$, versus time. The plot is a straight line with a negative slope passing through the initial concentration $[Q]_0$: $[Q]_t = [Q]_0 - k' t$ This linear relationship between reactant concentration and time is the characteristic integrated rate law for a zero-order reaction. Therefore, the order with respect to reactant $Q$ is: $y = 0$

3. Overall order of the reaction: $\text{Overall order} = x + y = 1 + 0 = 1$

Hence, the correct option is D.