IB DP Chemistry -Reactivity 2.2 How fast? The rate of chemical change - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

The reaction between bromine and nitrogen(II) oxide follows a two-step mechanism.

\( \mathrm{Br_2(g)+NO(g)\rightarrow NOBr_2(g)} \qquad \mathrm{fast} \)

\( \mathrm{NOBr_2(g)+NO(g)\rightarrow2NOBr(g)} \qquad \mathrm{slow} \)

What is the rate equation?

(A) \( \mathrm{Rate=k[Br_2][NO]} \)
(B) \( \mathrm{Rate=k[Br_2][NO]^2} \)
(C) \( \mathrm{Rate=k[Br_2][NO]^2[NOBr_2]} \)
(D) \( \mathrm{Rate=k[NOBr]^2} \)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

The rate-determining step is the slow step:

\( \mathrm{NOBr_2+NO\rightarrow2NOBr} \)

Initially, the rate is:

\( \mathrm{Rate=k[NOBr_2][NO]} \)

The intermediate \( \mathrm{NOBr_2} \) is formed in the fast first step:

\( \mathrm{Br_2+NO\rightarrow NOBr_2} \)

Therefore, its concentration is proportional to \( \mathrm{[Br_2][NO]} \). Substituting this into the rate expression gives:

\( \mathrm{Rate=k[Br_2][NO]^2} \)

Therefore, the correct answer is (B).

Question 

Which statement about catalysts is correct?

(A) Catalysts increase the kinetic energy of reacting particles.
(B) Catalysts affect endothermic reactions more than exothermic reactions.
(C) Catalysts increase reaction rate and are used up in a reaction.
(D) Catalysts alter the mechanism of a reaction.
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

A catalyst provides an alternative reaction pathway with a lower activation energy:

\( E_{\mathrm{a,catalysed}}<E_{\mathrm{a,uncatalysed}} \)

This increases the proportion of particles able to react successfully at a given temperature and therefore increases the reaction rate.

A catalyst does not increase the kinetic energy of the particles, is not consumed overall, and does not change the enthalpy change of a reaction.

Therefore, a catalyst alters the mechanism of a reaction, so the correct answer is (D).

Question

Which calculation determines the initial rate of this reaction?
(A) \( \dfrac{0.40 – 0.19}{20 – 0} \)
(B) \( \dfrac{0.40 – 0.10}{40 – 0} \)
(C) \( \dfrac{0.40 – 0}{140 – 0} \)
(D) \( \dfrac{0.40 – 0.20}{10 – 0} \)
▶️ Answer/Explanation
Detailed solution

Markscheme: (D)

The rate of reaction can be found using:

\[ \text{rate} = -\frac{\Delta[\text{reactant}]}{\Delta t} \quad \text{or} \quad \text{rate} = \frac{\Delta[\text{product}]}{\Delta t} \]

For the initial rate, we draw a tangent to the graph at \(t = 0\).

tangent graph

From the tangent, the slope can be approximated by:

\[ \frac{0.40 – 0.20}{10 – 0} \quad \text{or} \quad \frac{0.40 – 0.00}{20 – 0} \]

Both expressions simplify to the same initial rate. The closest matching option is:

✅ Answer: (D)

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