The overall order of a reaction is the sum of the orders with respect to each reactant. Let the rate law for the given reaction be:

$Rate = k[P]^x[Q]^y$

where x is the order with respect to P, y is the order with respect to Q, and the overall order is x + y.

Step 1: Determine the order with respect to P (x).

We are given that the time taken for 75% completion of the reaction of P is twice the time taken for 50% completion.

Let $t_{50\%}$ be the half-life of P. Let $t_{75\%}$ be the time required for 75% of P to react. This means the concentration of P becomes 25% of its initial value, i.e., $[P] = 0.25 [P]_0 = \frac{1}{4}[P]_0$.

The time taken for the concentration to drop from $[P]_0$ to $\frac{1}{2}[P]_0$ is one half-life, $t_{50\%}$. The time taken for the concentration to drop from $\frac{1}{2}[P]_0$ to $\frac{1}{4}[P]_0$ is the second half-life.

So, $t_{75\%} = t_{0 \to 50\%} + t_{50\% \to 75\%} = (1^{st} \text{ half-life}) + (2^{nd} \text{ half-life})$.

We are given $t_{75\%} = 2 \times t_{50\%}$. This implies that the first half-life is equal to the second half-life. A reaction has a constant half-life, independent of the concentration, only if it is a first-order reaction.

Let's verify this using the integrated rate law for a first-order reaction: $t = \frac{2.303}{k} \log \frac{[P]_0}{[P]_t}$ For 50% completion, $[P]_t = 0.5 [P]_0$: $t_{50\%} = \frac{2.303}{k} \log \frac{[P]_0}{0.5[P]_0} = \frac{2.303}{k} \log 2$ For 75% completion, $[P]_t = 0.25 [P]_0$: $t_{75\%} = \frac{2.303}{k} \log \frac{[P]_0}{0.25[P]_0} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \log (2^2) = 2 \times \frac{2.303}{k} \log 2$ Comparing the two expressions, we get: $t_{75\%} = 2 \times t_{50\%}$ This matches the given information. Therefore, the order of the reaction with respect to P is 1, i.e., $x = 1$.

Step 2: Determine the order with respect to Q (y).

The problem provides a graph of the concentration of Q, [Q], versus time, t. The graph is a straight line with a negative slope.

Let's recall the integrated rate laws for different orders:

• Zero order: $[A]_t = [A]_0 - kt$. A plot of $[A]$ vs. $t$ is a straight line with slope $-k$.
• First order: $\ln[A]_t = \ln[A]_0 - kt$. A plot of $\ln[A]$ vs. $t$ is a straight line.
• Second order: $\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$. A plot of $\frac{1}{[A]}$ vs. $t$ is a straight line.

The given graph shows that [Q] varies linearly with t. This corresponds to the integrated rate law for a zero-order reaction. Therefore, the order of the reaction with respect to Q is 0, i.e., $y = 0$.

Step 3: Calculate the overall order of the reaction.

The overall order is the sum of the individual orders: Overall order = $x + y$ Overall order = $1 + 0 = 1$

Thus, the overall order of the reaction is 1.