Question

Consider the reaction carried out at T(K):
\(\text{A(g) + B(g) → C(g)}\)
\(\text{The rate law for this reaction is \(r = k[A]^2[B]^2\)}\). The concentration of A in experiment 2 and rate in experiment 3 shown as \(x\) and \(z\) in the table. \(x\) and \(z\) are respectively:\[\begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]~(\text{mol L}^{-1}) & [B]~(\text{mol L}^{-1}) & \text{Initial rate (mol L}^{-1} \text{s}^{-1}) \\ \hline 1 & 0.05 & 0.05 & R \\ 2 & x & 0.05 & 2R \\ 3 & 0.20 & 0.10 & z\\ \hline \end{array}\]

• A. x = 0.10,~z = 8R
• B. x = 0.05,~z = 4R
• C. x = 0.10,~z = 64R
• D. x = 0.20,~z = 16R

Hint:

To compare rates from experimental data using the rate law, write the ratio of rates and simplify using the concentrations. Square or cube as required by the rate law powers.

Answer 1

The Correct Option is C

Rate law: \(r = k[A]^2[B]^2\)
From experiment 1 and 2: \[ \frac{2R}{R} = \frac{kx^2(0.05)^2}{k(0.05)^2(0.05)^2} \Rightarrow \frac{2R}{R} = \frac{x^2}{(0.05)^2} \Rightarrow x^2 = 2 \times (0.05)^2 = 0.005 \Rightarrow x = \sqrt{0.005} = 0.10 \] From experiment 1 and 3: \[ \frac{z}{R} = \frac{(0.20)^2(0.10)^2}{(0.05)^2(0.05)^2} = \frac{0.04 \times 0.01}{0.0025 \times 0.0025} = \frac{4 \times 10^{-4}}{6.25 \times 10^{-6}} = 64 \Rightarrow z = 64R \]