IB Chemistry SL 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.

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Question 1

Multiple choiceHard1 mark

Magnesium reacts with hydrochloric acid according to the equation below. The hydrogen is collected at the conditions for which the molar gas volume is $24.0\ \mathrm{dm^3\,mol^{-1}}$ Use $A_\mathrm{r}(\mathrm{Mg})=24.31$

\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}

Substance	Amount
Mg	$0.120\ \mathrm{g}$
HCl(aq)	$25.0\ \mathrm{cm^3}$ of $0.200\ \mathrm{mol\,dm^{-3}}$
$H_2$ collected	$56.4\ \mathrm{cm^3}$

Which option gives the limiting reactant and the percentage yield of $H_2$

Show answer and explanation

Answer: C. Compare the available amounts using the $1:2$ mole ratio of Mg to HCl, then calculate the yield relative to the theoretical $H_2$ volume.

$n(\mathrm{Mg})=0.120/24.31=4.94\times10^{-3}\ \mathrm{mol}$ $n(\mathrm{HCl})=0.0250\times0.200=5.00\times10^{-3}\ \mathrm{mol}$ The Mg would require $9.87\times10^{-3}\ \mathrm{mol}$ HCl, so HCl is limiting.
$n(\mathrm{H_2})=5.00\times10^{-3}/2=2.50\times10^{-3}\ \mathrm{mol}$ giving a theoretical volume of $60.0\ \mathrm{cm^3}$
Percentage yield $=(56.4/60.0)\times100=94.0\%$

A: Compares the raw mole amounts to label Mg limiting, ignoring the equation’s coefficients.
B: Reverses the yield ratio, calculating $60.0/56.4\times100=106\%$
D: Treats Mg as limiting and uses its amount to calculate the theoretical $H_2$ volume, giving $118.5\ \mathrm{cm^3}$ and a $47.6\%$ yield.
Remember: compare available amounts after accounting for stoichiometric coefficients to identify the limiting reactant.

Question 2

Paper 2Hard8 marks

Aspirin (acetylsalicylic acid) can be prepared by either of the reactions shown in Figure 1. Use the atom economy equation in the data booklet and the following relative atomic masses: $A_\mathrm{r}(\mathrm C)=12.01$ $A_\mathrm{r}(\mathrm H)=1.01$ $A_\mathrm{r}(\mathrm O)=16.00$ $A_\mathrm{r}(\mathrm{Cl})=35.45$

(a)

Calculate the atom economy of Route I, to 3 significant figures.

[2]

(b)

Calculate the atom economy of Route II, to 3 significant figures.

[2]

(c)

For each route, determine the mass of reactant material not incorporated into aspirin, per mole of aspirin formed, on the basis of the balanced reaction. Use your answers to (a) and (b) to assess the claim: “Route II must generate less waste in an industrial process because it has a higher atom economy.”

[4]

Question 3

Multiple choiceHard1 mark

In a sealed vessel, the concentrations in the reversible system $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)}$ become constant while all three gases remain present. Which account correctly describes the reaction rates at this point and gives the expression for $K_c$

Show answer and explanation

Answer: B. Constant concentrations mean there is no net change, so the forward and backward rates are equal. At dynamic equilibrium, both reactions continue, making these rates non-zero. The equilibrium expression uses each concentration raised to its coefficient: $K_c=\dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$

A: takes the reciprocal, placing reactants over products.
C: treats equilibrium as static and sets both reaction rates to zero.
D: unequal rates would cause net formation of $\mathrm{NH_3}$ so concentrations would change.

Remember: At equilibrium, concentrations are constant; the rates, not the concentrations, are equal.

Question 4

Paper 2Hard12 marks

At 298 K, dinitrogen tetroxide gas dissociates to nitrogen dioxide gas in two sealed, rigid vessels. Each vessel initially contains only dinitrogen tetroxide. Figure 1 shows the structure of dinitrogen tetroxide.

The concentrations were measured over time. The final readings are treated as equilibrium concentrations for calculating $K_\mathrm{c}$

Time / s	Run A [N₂O₄] / mol dm⁻³	Run A [NO₂] / mol dm⁻³	Run B [N₂O₄] / mol dm⁻³	Run B [NO₂] / mol dm⁻³
0	0.0800	0.0000	0.1200	0.0000
20	0.0650	0.0300	0.0960	0.0480
40	0.0560	0.0480	0.0850	0.0700
60	0.0510	0.0580	0.0820	0.0760
80	0.0500	0.0600	0.0817	0.0767
100	0.0500	0.0600	0.0817	0.0767
120	0.0500	0.0600	0.0817	0.0767

(a)

Use Figure 1 to write a balanced equation, including state symbols, for the dissociation of dinitrogen tetroxide gas into nitrogen dioxide gas.

[2]

(b)

For Run A, calculate the ratio of the increase in [NO₂] to the decrease in [N₂O₄] between $t=0$ and $t=80\ \mathrm{s}$

[1]

(c)

Write the equilibrium law expression for $K_\mathrm{c}$ for the reaction in (a).

[2]

(d)

A researcher claims that $K_\mathrm{c}$ is greater in Run B because its equilibrium concentration of NO₂ is greater. Determine whether the data support this claim. Include calculations for both runs and a conclusion.

[5]

(e)

A student argues that Run B cannot be at equilibrium because the concentrations of N₂O₄ and NO₂ are unequal. Evaluate this argument using the definition of dynamic equilibrium and the time data.

[2]

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