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IB Chemistry – Reactivity 2: How Much, How Fast, How Far

Stoichiometry, rates, equilibrium.

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12 questions  ·  42 marks  ·  answers included  ·  IB Chemistry SL & HL
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This worksheet covers:

  • Stoichiometry
  • Rates of reaction
  • Chemical equilibrium from IB Chemistry Reactivity 2

Use these relative atomic masses (Ar): H = 1.01, C = 12.01, O = 16.00, Na = 22.99, Mg = 24.31, Cl = 35.45. Take the molar volume of an ideal gas as $22.7\,\text{dm}^3\,\text{mol}^{-1}$ at $273\,\text{K}$ and $100\,\text{kPa}$. A calculator is required.

Questions

Section A – Stoichiometry

1

Magnesium reacts with excess hydrochloric acid: $\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$.$1.20\,\text{g}$ of magnesium is reacted with $0.500\,\text{mol dm}^{-3}$ hydrochloric acid.(a) Calculate the amount, in mol, of magnesium used.(b) Calculate the minimum volume, in cm$^3$, of hydrochloric acid required to react completely with the magnesium.(c) Calculate the volume, in cm$^3$, of hydrogen gas produced at $273\,\text{K}$ and $100\,\text{kPa}$.

Show answer
(a) $n(\text{Mg}) = \dfrac{1.20}{24.31} = 0.0494\,\text{mol}$.(b) $n(\text{HCl}) = 2 \times 0.0494 = 0.0987\,\text{mol}$; $V = \dfrac{0.0987}{0.500} = 0.197\,\text{dm}^3 = 197\,\text{cm}^3$.(c) $n(\text{H}_2) = n(\text{Mg}) = 0.0494\,\text{mol}$; $V = 0.0494 \times 22.7 = 1.12\,\text{dm}^3 = 1120\,\text{cm}^3$.
[6]
2

$5.00\,\text{g}$ of sodium is reacted with $5.00\,\text{g}$ of chlorine gas: $2\text{Na}(s) + \text{Cl}_2(g) \rightarrow 2\text{NaCl}(s)$.(a) Determine which reagent is in excess.(b) Calculate the mass of sodium chloride formed.

Show answer
(a) $n(\text{Na}) = \dfrac{5.00}{22.99} = 0.217\,\text{mol}$; $n(\text{Cl}_2) = \dfrac{5.00}{70.90} = 0.0705\,\text{mol}$. Reacting all the Na would require $0.109\,\text{mol}$ of $\text{Cl}_2$, but only $0.0705\,\text{mol}$ is available, so $\text{Cl}_2$ is the limiting reagent (Na is in excess).(b) $n(\text{NaCl}) = 2 \times n(\text{Cl}_2) = 0.141\,\text{mol}$; mass $= 0.141 \times 58.44 = 8.24\,\text{g}$.
[5]
3

The theoretical yield of a product in a synthesis reaction is $12.5\,\text{g}$. The actual mass of product obtained is $10.7\,\text{g}$. Calculate the percentage yield.

Show answer
Percentage yield $= \dfrac{10.7}{12.5} \times 100 = 85.6\%$.
[2]
4

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Determine the empirical formula of the compound.

Show answer
$n(\text{C}) = \dfrac{40.0}{12.01} = 3.33$, $n(\text{H}) = \dfrac{6.7}{1.01} = 6.63$, $n(\text{O}) = \dfrac{53.3}{16.00} = 3.33$. Dividing by the smallest value ($3.33$): $\text{C} : \text{H} : \text{O} = 1 : 2 : 1$. Empirical formula: $\text{CH}_2\text{O}$.
[4]

Section B – Rates of Reaction

5

Using collision theory, explain why increasing the temperature of a reaction mixture increases the rate of reaction.

Show answer
Increasing temperature increases the average kinetic energy of the particles, so a greater proportion of particles have energy greater than or equal to the activation energy. This increases the frequency of successful (effective) collisions per unit time, so the rate of reaction increases. (A higher temperature also increases the collision frequency, but this is a much smaller effect than the increase in the fraction of particles with sufficient energy.)
[3]
6

For a reaction, the rate equation is $\text{rate} = k[\text{A}][\text{B}]^2$.(a) State the order of reaction with respect to A and with respect to B.(b) State the overall order of reaction.(c) If the concentrations of both A and B are doubled, by what factor does the rate increase?

Show answer
(a) Order with respect to A $= 1$; order with respect to B $= 2$.(b) Overall order $= 3$.(c) Rate increases by a factor of $2^1 \times 2^2 = 8$.
[4]
7

Explain, with reference to the Maxwell-Boltzmann energy distribution, how a catalyst increases the rate of a reaction without being used up.

Show answer
A catalyst provides an alternative reaction pathway with a lower activation energy. On the Maxwell-Boltzmann distribution, a greater proportion of particles have energy exceeding this lower activation energy, so a greater fraction of collisions are successful, increasing the rate. The catalyst takes part in the reaction mechanism (e.g. forming an intermediate) but is regenerated at the end, so it is not chemically changed overall.
[3]
8

In an experiment, doubling the concentration of reactant X, while keeping the concentration of all other reactants constant, causes the initial rate of reaction to increase by a factor of four. Determine the order of reaction with respect to X.

Show answer
$2^n = 4 \Rightarrow n = 2$. The reaction is second order with respect to X.
[2]

Section C – Equilibrium

9

State Le Châtelier’s principle, and use it to predict the effect of increasing the pressure on the position of equilibrium for the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$.

Show answer
Le Châtelier’s principle states that if a system at equilibrium is subjected to a change, the position of equilibrium shifts to counteract that change. There are $4$ mol of gas on the reactant side and $2$ mol of gas on the product side, so increasing the pressure shifts the equilibrium to the right (towards fewer gas molecules), increasing the yield of ammonia.
[3]
10

For the equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $\Delta H = -198\,\text{kJ mol}^{-1}$. Predict, with a reason, the effect of increasing the temperature on the equilibrium yield of $\text{SO}_3$.

Show answer
The forward reaction is exothermic, so the reverse reaction is endothermic. Increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the extra energy, so the equilibrium yield of $\text{SO}_3$ decreases.
[3]
11

For the equilibrium $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ at a certain temperature, the equilibrium concentrations are $[\text{H}_2] = 0.020\,\text{mol dm}^{-3}$, $[\text{I}_2] = 0.020\,\text{mol dm}^{-3}$ and $[\text{HI}] = 0.152\,\text{mol dm}^{-3}$.(a) Write the expression for $K_c$.(b) Calculate the value of $K_c$ at this temperature.

Show answer
(a) $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$.(b) $K_c = \dfrac{(0.152)^2}{(0.020)(0.020)} = 57.8$ (no units, since moles of gas are equal on both sides).
[4]
12

Explain what is meant by a dynamic equilibrium, referring to the rates of the forward and backward reactions and to the macroscopic properties of the system.

Show answer
At dynamic equilibrium, the forward and backward reactions are both still occurring, but they occur at exactly the same rate, so there is no overall change in the amounts (concentrations) of reactants and products. Macroscopic properties of the system, such as colour, pressure and concentration, remain constant over time, even though the forward and backward reactions continue at a molecular level.
[3]

Related worksheets

More worksheets in IB Chemistry (SL & HL):

IB Chemistry – Structure 1: Particulate Nature of MatterIB Chemistry – Structure 2: Bonding and StructureIB Chemistry – Structure 3: Classification of MatterIB Chemistry – Reactivity 1: What Drives ReactionsIB Chemistry – Reactivity 3: Mechanisms of ChangeIB Chemistry – Data-Based Questions (Paper 1B)

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