IB Chemistry HL Reactivity 2: How much, how fast and how far? 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 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 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 6

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.

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