IB Chemistry HL R2.2 How fast? The rate of chemical change questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Multiple choiceMedium1 mark

An elementary step has X as its only reactant. Its measured initial rate is plotted against [X] below. What are the molecularity of this step and its initial rate at [X] = 0.30 mol dm⁻³?

Show answer and explanation

Answer: C. Doubling [X] from 0.10 to 0.20 mol dm⁻³ increases the rate fourfold, so the rate varies with [X]². For an elementary step, this corresponds to two reacting X particles: the step is bimolecular. The graph gives a rate of 7.2 × 10⁻³ mol dm⁻³ s⁻¹ at [X] = 0.30 mol dm⁻³.

A: assumes a linear rate–concentration relationship, giving a unimolecular step and 2.4 × 10⁻³ mol dm⁻³ s⁻¹.
B: reads the rate correctly but counts one type of reactant species instead of two reacting particles.
D: identifies the bimolecular step but uses a linear relationship to obtain the rate.

Remember: molecularity counts reacting particles in an elementary step.

Question 2

Paper 2Medium6 marks

Nitric oxide is oxidized by oxygen to form nitrogen dioxide:

2NO(g) + O₂(g) → 2NO₂(g)

Initial rates were measured at 298 K.

Trial	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.0200	0.0100	2.40 × 10⁻⁴
2	0.0400	0.0100	9.60 × 10⁻⁴
3	0.0200	0.0200	4.80 × 10⁻⁴

At 308 K, an experiment with [NO] = 0.0200 mol dm⁻³ and [O₂] = 0.0100 mol dm⁻³ gave an initial rate of 3.60 × 10⁻⁴ mol dm⁻³ s⁻¹. Assume the rate equation is unchanged between these temperatures.

(a)

Determine the order of the reaction with respect to NO and O₂.

[2]

(b)

Write the rate equation for the reaction.

[1]

(c)

Calculate the rate constant, k, at 298 K, including its units.

[2]

(d)

State which temperature has the larger value of k, using the measurements to justify your answer.

[1]

Question 3

Paper 2Hard11 marks

During the gas-phase oxidation of nitric oxide to nitrogen dioxide, the overall reaction is:

2NO(g) + O₂(g) → 2NO₂(g)

Initial-rate data were collected at constant temperature.

Experiment	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	1.00 × 10⁻²	2.00 × 10⁻²	1.20 × 10⁻⁵
2	2.00 × 10⁻²	2.00 × 10⁻²	4.80 × 10⁻⁵
3	2.00 × 10⁻²	4.00 × 10⁻²	9.60 × 10⁻⁵

Two mechanisms have been proposed. Each step is elementary.

Mechanism I: NO(g) + O₂(g) → NO₃(g) (slow)
NO₃(g) + NO(g) → 2NO₂(g) (fast)

Mechanism II: 2NO(g) ⇌ N₂O₂(g) (fast equilibrium)
N₂O₂(g) + O₂(g) → 2NO₂(g) (slow)

For a proposed energy profile for mechanism II, the relative energies of successive stages are:

Stage along the reaction pathway	Relative energy / kJ mol⁻¹
2NO(g) + O₂(g)	0
During elementary step 1	+35
Between the two steps	+12
During elementary step 2	+60
2NO₂(g)	−114

(a)

Determine the order of reaction with respect to each reactant and write the rate equation.

[3]

(b)

Assess both proposed mechanisms using the overall equation and the initial-rate data. State which mechanism is consistent with both.

[4]

(c)

Sketch an energy profile for mechanism II using the values given. Label the reactants, products, intermediate and both transition states. Mark the activation energy of the rate-determining step, and state which step the profile indicates is rate-determining.

[4]

The best questions are still waiting

Unlock every question in this topic, then type, photograph or draw your answer and have it marked against the markscheme. Find out where you are dropping marks.

Unlock all questions Compare Student and Student Plus