Question

The data given below are for the reaction of $A$ and $D_2$ to form product at $295\,K$. Find the correct rate expression for this reaction. $\begin{array}{lll} D_2/\text{mol L}^{-1} & A/\text{mol L}^{-1} & \text{Initial rate}/\text{mol L}^{-1}\text{s}^{-1} \\ 0.05 & 0.05 & 1\times10^{-3} \\ 0.15 & 0.05 & 3\times10^{-3} \\ 0.05 & 0.15 & 9\times10^{-3} \end{array}$

• A. \(k[D_2]^1[A]^2 \)
• B. \(k[D_2]^2[A]^1 \)
• C. \(k[D_2]^1[A]^1 \)
• D. \(k[D_2]^2[A]^2 \)
• E. \(k[D_2]^1[A]^0 \)

Hint:

If doubling concentration doubles the rate, the order is 1. If doubling concentration quadruples the rate, the order is 2.

Answer 1

The Correct Option is A

Concept: The rate law is \( \text{Rate} = k[D_2]^x[A]^y \). We determine \(x\) and \(y\) by observing how the rate changes when one concentration is varied while the other remains constant.

Step 1: Find the order with respect to \(D_2\).
Compare Row 1 and Row 2 (where \([A]\) is constant at 0.05):
• \([D_2]\) increases from 0.05 to 0.15 (3 times).
• Rate increases from \(1 \times 10^{-3}\) to \(3 \times 10^{-3}\) (3 times).
• \(3^x = 3 \implies x = 1\).

Step 2: Find the order with respect to \(A\).
Compare Row 1 and Row 3 (where \([D_2]\) is constant at 0.05):
• \([A]\) increases from 0.05 to 0.15 (3 times).
• Rate increases from \(1 \times 10^{-3}\) to \(9 \times 10^{-3}\) (9 times).
• \(3^y = 9 \implies y = 2\).

Step 3: Write the Rate Law.
\[ \text{Rate} = k[D_2]^1[A]^2 \]