Question

For the decomposition of a compound AB at 600 K, the following data were obtained: [AB] (mol dm$^{-3}$) values of 0.20, 0.40, and 0.60 correspond to rates (mol dm$^{-3}$ s$^{-1}$) of $2.75 \times 10^{-8}$, $11.0 \times 10^{-8}$, and $24.75 \times 10^{-8}$. The order for the decomposition of AB is

• A. 1.5
• B. 0
• C. 1
• D. 2

Hint:

If doubling the concentration quadruples the rate ($2^2$), the order is 2.

Answer 1

The Correct Option is D

Step 1: Set up Rate Ratio
Using the general rate law $Rate = k[AB]^{n}$. $\frac{Rate_{2}}{Rate_{1}} = \frac{k(0.40)^{n}}{k(0.20)^{n}} = \frac{11.0 \times 10^{-8}}{2.75 \times 10^{-8}}$.
Step 2: Simplify
$2^{n} = 4$.
Step 3: Find n
$2^{n} = 2^{2}$, so $n = 2$.
Final Answer: (d)