IB Chemistry HL Reactivity 2: How much, how fast and how far? questions

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Multiple choice Written

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Easy Medium Hard

Question 1

Multiple choiceEasy1 mark

A process makes ethanol using the balanced reaction $\mathrm{C_2H_5Br + NaOH \rightarrow C_2H_5OH + NaBr}$ Which proposed change would increase the atom economy, rather than only the percentage yield?

Show answer and explanation

Answer: C. Atom economy is set by the balanced equation and compares the mass of desired product with the total reactant mass. A pathway that incorporates a greater proportion of reactant mass into ethanol therefore has greater atom economy.
A: recovering more ethanol increases the experimental yield, not the atom economy.
B: reducing unreacted bromoethane changes conversion, while the reaction stoichiometry remains unchanged.
D: a catalyst changes the reaction rate, not the stoichiometric mass fraction forming ethanol.
Remember: atom economy depends on the reaction equation; percentage yield depends on the amount actually obtained.

Question 2

Paper 2Medium7 marks

A student prepares magnesium hydroxide by mixing aqueous magnesium chloride with aqueous sodium hydroxide. The reaction is

\ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)}

Figure 1 shows the quantities used and the mass of solid collected after filtration and drying.

Use relative atomic masses from the data booklet where required.

(a)

Calculate the amount, in mol, of NaOH used. Give your answer to three significant figures.

[1]

(b)

Using your answer to (a) and Figure 1, determine the limiting reactant and calculate the theoretical mass, in g, of $\ce{Mg(OH)2}$ Show your working.

[3]

(c)

Calculate the percentage yield of $\ce{Mg(OH)2}$ to three significant figures.

[1]

(d)

Explain one reason why an experimental yield could be lower than the theoretical yield, and one reason why its apparent value could be higher.

[2]

Question 3

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 4

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 5

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]

Question 6

Paper 2Easy5 marks

A student prepares an aqueous solution of ethanoic acid at $298\ \mathrm{K}$ The dissociation equilibrium is

\ce{CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)}

At $298\ \mathrm{K}$ $K_\mathrm{a}=1.80\times10^{-5}$ Figure 1 shows the initial concentration of ethanoic acid.

(a)

Using Figure 1, calculate the equilibrium concentration of $\ce{H3O+}$ Assume that $[\ce{CH3COOH}]_\mathrm{equilibrium}\approx[\ce{CH3COOH}]_\mathrm{initial}$ Give your answer to three significant figures.

[2]

(b)

Use the equation in the data booklet to calculate $\Delta G^\circ$ in $\mathrm{kJ\,mol^{-1}}$ for the dissociation equilibrium at $298\ \mathrm{K}$

[2]

(c)

State which side of the equilibrium is favoured.

[1]

Question 7

Multiple choiceMedium1 mark

For the equilibrium $2\text{A}(g)+\text{B}(g)\rightleftharpoons2\text{C}(g)+\text{D}(g)$ the equilibrium concentrations at a fixed temperature are $[\text{A}]=0.50$ $[\text{B}]=0.20$ $[\text{C}]=0.10$ and $[\text{D}]=0.10\ \text{mol dm}^{-3}$ What is $K_c$ and which side is favoured?

Show answer and explanation

Answer: D. The stoichiometric coefficients are the powers in the equilibrium expression:

K_c=\frac{[\text{C}]^2[\text{D}]}{[\text{A}]^2[\text{B}]}=\frac{(0.10)^2(0.10)}{(0.50)^2(0.20)}=0.020

(dimensionless). Since $K_c<1$ reactants are favoured at equilibrium.

A: calculates $K_c$ correctly but reverses the implication of $K_c<1$
B: omits the square on $[\text{C}]$ obtaining $0.20$ this is still less than 1.
C: uses the reciprocal expression, which gives the equilibrium constant for the reverse reaction.

Remember: a value of $K_c<1$ indicates that reactants predominate at equilibrium.

Question 8

Paper 2Hard9 marks

In a sealed, fixed-volume vessel, nitrogen and oxygen form nitrogen monoxide at 1500 K:

N₂(g) + O₂(g) ⇌ 2NO(g)

At this temperature, Kc = 4.00 × 10⁻⁶. At a particular instant, the concentrations are [N₂] = 0.0500 mol dm⁻³, [O₂] = 0.0400 mol dm⁻³ and [NO] = 1.00 × 10⁻⁵ mol dm⁻³.

(a)

Calculate Qc for this mixture and predict the direction in which the reaction will proceed to reach equilibrium.

[2]

(b)

Determine the equilibrium concentrations of the three gases. Show how you check that any approximation used is justified.

[4]

(c)

After the mixture reaches equilibrium, an injection increases [NO] by 2.00 × 10⁻⁴ mol dm⁻³. Immediately after the injection, before further reaction occurs, calculate Qc and predict the direction in which the reaction will proceed.

[3]

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