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AP Chemistry – Units 4–6 Practice Set

Reactions, kinetics, thermodynamics.

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12 questions  ·  34 marks  ·  answers included  ·  AP Chemistry (College Board)
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This worksheet covers:

  • Unit 4 (Chemical Reactions)
  • Unit 5 (Kinetics)
  • Unit 6 (Thermodynamics)

A calculator is permitted, as on the free-response section of the AP Chemistry exam.

Questions

Section A – Reactions and Stoichiometry (Unit 4)

1

Write the net ionic equation for the reaction between aqueous silver nitrate and aqueous sodium chloride.

Show answer
$\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \rightarrow \text{AgCl}(s)$
[2]
2

A $25.0\,\text{mL}$ sample of $0.150\,\text{M}$ HCl is titrated with $0.100\,\text{M}$ NaOH. Calculate the volume of NaOH required to reach the equivalence point.

Show answer
$n(\text{HCl})=0.0250\,\text{L}\times0.150\,\text{M}=3.75\times10^{-3}\,\text{mol}$. Since the mole ratio is 1:1, $n(\text{NaOH})=3.75\times10^{-3}\,\text{mol}$, so $V=\dfrac{3.75\times10^{-3}}{0.100}=0.0375\,\text{L}=37.5\,\text{mL}$.
[3]
3

Classify the reaction $\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$ and identify the oxidizing agent.

Show answer
This is a single-replacement (oxidation-reduction) reaction. HCl is the oxidizing agent, since $\text{H}^{+}$ is reduced to $\text{H}_2$ (Zn is oxidized and is the reducing agent).
[2]

Section B – Kinetics (Unit 5)

4

The rate law for a reaction is $\text{rate}=k[A]^2[B]$. If the concentration of A is tripled and the concentration of B is doubled, by what factor does the rate increase?

Show answer
New rate $=k(3[A])^2(2[B])=k\times9[A]^2\times2[B]=18\times k[A]^2[B]$. The rate increases by a factor of $18$.
[2]
5

For a first-order reaction, the rate constant is $k=0.0231\,\text{s}^{-1}$. Calculate the half-life of the reaction.

Show answer
For a first-order reaction, $t_{1/2}=\dfrac{\ln2}{k}=\dfrac{0.693}{0.0231}=30.0\,\text{s}$.
[3]
6

Explain, in terms of collision theory, why increasing the temperature of a reaction increases its rate.

Show answer
At higher temperature, particles have greater average kinetic energy and move faster, so they collide more frequently. More importantly, a greater proportion of collisions have energy equal to or greater than the activation energy, $E_a$, so a greater fraction of collisions are successful, increasing the rate.
[3]
7

A reaction has an activation energy of $75\,\text{kJ/mol}$. Using the Arrhenius equation qualitatively, explain the effect of adding a catalyst on the rate constant $k$.

Show answer
A catalyst provides an alternative reaction pathway with a lower activation energy, $E_a$. Since $k=Ae^{-E_a/RT}$, a smaller $E_a$ gives a larger value of $k$ (the exponent is less negative), so the rate constant, and hence the rate, increases without the catalyst itself being consumed.
[2]

Section C – Thermodynamics (Unit 6)

FRQ-style, calculator required
8

Calculate $\Delta H_{rxn}^{\circ}$ for $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, given $\Delta H_f^{\circ}$: $\text{CH}_4(g)=-74.8\,\text{kJ/mol}$, $\text{CO}_2(g)=-393.5\,\text{kJ/mol}$, $\text{H}_2\text{O}(l)=-285.8\,\text{kJ/mol}$, $\text{O}_2(g)=0$.

Show answer
$\Delta H_{rxn}^{\circ}=\sum\Delta H_f^{\circ}(\text{products})-\sum\Delta H_f^{\circ}(\text{reactants})$$=[-393.5+2(-285.8)]-[-74.8+2(0)]$$=-965.1-(-74.8)=-890.3\,\text{kJ/mol}$.
[4]
9

A reaction has $\Delta H=-120\,\text{kJ/mol}$ and $\Delta S=-250\,\text{J/(mol K)}$. Calculate $\Delta G$ at $298\,\text{K}$ and determine whether the reaction is spontaneous under these conditions.

Show answer
$\Delta G=\Delta H-T\Delta S=-120\,000\,\text{J/mol}-(298\,\text{K})(-250\,\text{J/(mol K)})=-120\,000+74\,500=-45\,500\,\text{J/mol}=-45.5\,\text{kJ/mol}$. Since $\Delta G<0$, the reaction is spontaneous at $298\,\text{K}$.
[4]
10

For the reaction in the previous question, determine the temperature above which the reaction becomes non-spontaneous.

Show answer
$\Delta G=0$ when $T=\dfrac{\Delta H}{\Delta S}=\dfrac{-120\,000}{-250}=480\,\text{K}$. Since $\Delta S<0$, $\Delta G$ becomes positive (non-spontaneous) above $T=480\,\text{K}$.
[3]
11

In a coffee-cup calorimeter, $50.0\,\text{g}$ of water at $25.0^{\circ}\text{C}$ is mixed with $50.0\,\text{g}$ of water at $85.0^{\circ}\text{C}$. Assuming no heat loss to the surroundings, calculate the final temperature of the mixture.

Show answer
Since the two samples have equal mass and specific heat capacity, heat lost by the hot water equals heat gained by the cold water: $mc(85.0-T)=mc(T-25.0) \Rightarrow 85.0-T=T-25.0 \Rightarrow T=\dfrac{85.0+25.0}{2}=55.0^{\circ}\text{C}$.
[3]
12

Justify, using entropy considerations, why the dissolution of $\text{NaCl}(s)$ in water results in an increase in the entropy of the system, despite the ordering effect of hydration shells forming around the ions.

Show answer
Dissolving breaks up the highly ordered ionic lattice into freely moving, dispersed $\text{Na}^{+}$ and $\text{Cl}^{-}$ ions in solution, greatly increasing the number of accessible microstates (positional disorder). Although water molecules become somewhat more ordered as hydration shells form around the ions (a local decrease in entropy), this decrease is smaller than the large increase from disrupting the crystal lattice, so the overall entropy of the system increases.
[3]

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