2021 H2 Chem P1 Q13 — Kinetics Solution

The answer is B (1 and 2 only).

Statement 1 is true. [HBr] appears in the denominator of the rate equation for reaction 1, so as HBr forms and [HBr] increases, the rate decreases.

Statement 2 is true. The rate equation for the \(\text{H}_2\)/\(\text{I}_2\) reaction is first order with respect to both \(\text{H}_2\) and \(\text{I}_2\), which matches one \(\text{H}_2\) colliding with one \(\text{I}_2\) in a single step to give 2HI directly. The rate equation for the \(\text{H}_2\)/\(\text{Br}_2\) reaction contains a fractional order and the concentration of the product HBr, which a single step cannot give, so it must go by a multistep mechanism.

Statement 3 is false. Doubling the halogen concentration doubles the rate only for reaction 2, which is first order with respect to \(\text{I}_2\). For reaction 1, [HBr] = 0 at the start, so rate = \(\frac{k_1[\text{H}_2][\text{Br}_2]^{1.5}}{[\text{Br}_2]} = k_1[\text{H}_2][\text{Br}_2]^{0.5}\). Doubling \([\text{Br}_2]\) increases the rate by \(\frac{2^{1.5}}{2} = \sqrt{2} = 1.41\) times, not 2 times.

• A is wrong: it includes statement 3, which is false.
• C is wrong: it leaves out statement 1, which is true, and includes statement 3, which is false.
• D is wrong: it leaves out statement 2, which is also true.