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

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Question 1

Paper 2Medium7 marks

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

\ceMgCl2(aq)+2NaOH(aq)−>Mg(OH)2(s)+2NaCl(aq)\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.

MgCl₂(aq) aliquotn(MgCl₂) = 7.50 × 10⁻³ molNaOH(aq)V = 30.0 cm³c = 0.400 mol dm⁻³mixfilter, thendrycollected solidmeasured mass =0.325 g

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 \ceMg(OH)2\ce{Mg(OH)2} Show your working.

[3]
(c)

Calculate the percentage yield of \ceMg(OH)2\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 2

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?

00.10.20.30.402468101214[X] / mol dm⁻³initial rate / (10⁻³ mol dm⁻³ s⁻¹)
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 3

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⁻³Initial rate / mol dm⁻³ s⁻¹
10.02000.01002.40 × 10⁻⁴
20.04000.01009.60 × 10⁻⁴
30.02000.02004.80 × 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 4

Multiple choiceMedium1 mark

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

Show answer and explanation

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

Kc=[C]2[D][A]2[B]=(0.10)2(0.10)(0.50)2(0.20)=0.020K_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 Kc<1K_c<1 reactants are favoured at equilibrium.

A: calculates KcK_c correctly but reverses the implication of Kc<1K_c<1
B: omits the square on [C][\text{C}] obtaining 0.200.20 this is still less than 1.
C: uses the reciprocal expression, which gives the equilibrium constant for the reverse reaction.

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

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