Assume the rate law

For the reaction $2A + B \to \text{products}$ let the rate law be $r = k[A]^m[B]^n$ where $m$ is the order with respect to $A$ and $n$ is the order with respect to $B$.

Use the data when only $A$ is doubled

Initial rate changes from $0.3 \to 0.6$ so the rate becomes doubled.

Since only $[A]$ is doubled, $\frac{r_2}{r_1} = \frac{k(2[A])^m[B]^n}{k[A]^m[B]^n} = 2^m$

Given $\frac{0.6}{0.3} = 2$ therefore, $2^m = 2 \implies m = 1$

So, order with respect to $A$ is 1.

Use the data when both $A$ and $B$ are doubled

Rate changes from $0.3 \to 2.4$ so the rate becomes $\frac{2.4}{0.3} = 8$ times.

When both concentrations are doubled, $\frac{r_3}{r_1} = \frac{k(2[A])^m(2[B])^n}{k[A]^m[B]^n} = 2^m \cdot 2^n = 2^{m+n}$

Hence, $2^{m+n} = 8 = 2^3$ so, $m+n = 3$

Since $m=1$, $1+n=3 \implies n=2$

So, order with respect to $B$ is 2.

Find total order

$\text{Total order} = m+n = 1+2 = 3$

Evaluate the options

• A: Total order is 4 $\to$ False
• B: Order with respect to $B$ is 2 $\to$ True
• C: Order with respect to $B$ is 1 $\to$ False
• D: Order with respect to $A$ is 2 $\to$ False

Final answer

The correct option is: $\boxed{\text{B}}$