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A-Level Chemistry – Physical: Equilibria, Acids and Redox

Kc and Kp, pH and buffers, electrochemical cells.

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12 questions  ·  40 marks  ·  answers included  ·  A-Level Chemistry (Edexcel/AQA/OCR)
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This worksheet covers:

  • This worksheet covers chemical equilibria ($K_c$ and $K_p$)
  • Acids
  • Bases and buffers
  • Redox/electrochemical cells

A calculator is allowed; use $K_w = 1.0\times10^{-14}\,\text{mol}^2\,\text{dm}^{-6}$ at 298 K.

Questions

Section A – Chemical Equilibria

1

For the equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, the equilibrium concentrations in a $1\,\text{dm}^3$ container are $[\text{N}_2] = 0.20\,\text{mol dm}^{-3}$, $[\text{H}_2] = 0.60\,\text{mol dm}^{-3}$ and $[\text{NH}_3] = 0.50\,\text{mol dm}^{-3}$.(a) Write the expression for $K_c$.(b) Calculate the value of $K_c$, including units.

Show answer
(a) $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$(b) $K_c = \dfrac{0.50^2}{0.20\times0.60^3} \approx 5.79\,\text{mol}^{-2}\,\text{dm}^6$
[4]
2

The forward reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ is exothermic.(a) Predict and explain the effect of increasing the pressure on the position of equilibrium.(b) Predict and explain the effect of increasing the temperature on the value of $K_c$.

Show answer
(a) The position of equilibrium shifts to the right (towards $\text{NH}_3$), because there are fewer gas moles on the product side (2 mol vs 4 mol), so increasing pressure favours the side with fewer moles of gas.(b) $K_c$ decreases, because the reaction is exothermic, so by Le Chatelier’s principle increasing temperature shifts equilibrium in the endothermic (reverse) direction, reducing the proportion of products at equilibrium.
[4]
3

For the equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, an equilibrium mixture at total pressure $200\,\text{kPa}$ contains $0.40\,\text{mol}$ $\text{SO}_2$, $0.20\,\text{mol}$ $\text{O}_2$ and $1.40\,\text{mol}$ $\text{SO}_3$ (total $2.00\,\text{mol}$).(a) Calculate the partial pressure of each gas.(b) Write the expression for $K_p$ and calculate its value, including units.

Show answer
(a) $p_{\text{SO}_2} = 200\times\dfrac{0.40}{2.00} = 40\,\text{kPa}$; $p_{\text{O}_2} = 200\times\dfrac{0.20}{2.00} = 20\,\text{kPa}$; $p_{\text{SO}_3} = 200\times\dfrac{1.40}{2.00} = 140\,\text{kPa}$(b) $K_p = \dfrac{p_{\text{SO}_3}^2}{p_{\text{SO}_2}^2\,p_{\text{O}_2}} = \dfrac{140^2}{40^2\times20} \approx 0.613\,\text{kPa}^{-1}$
[5]
4

Explain why adding a catalyst to an equilibrium reaction changes the rate at which equilibrium is reached but does not change the value of $K_c$ or the position of equilibrium.

Show answer
A catalyst provides an alternative pathway with lower activation energy, speeding up both the forward and reverse reactions equally, so equilibrium is reached faster; because both rates increase by the same factor, the equilibrium concentrations (and therefore $K_c$) are unchanged.
[2]

Section B – Acids, Bases and Buffers

5

Sulfuric acid, $\text{H}_2\text{SO}_4$, is a strong diprotic acid that can be assumed to fully dissociate to give two moles of $\text{H}^+$ per mole of acid. Calculate the pH of a $0.020\,\text{mol dm}^{-3}$ solution of sulfuric acid.

Show answer
$[\text{H}^+] = 2\times0.020 = 0.040\,\text{mol dm}^{-3}$$\text{pH} = -\log(0.040) \approx 1.40$
[3]
6

A weak acid HA has $K_a = 1.7\times10^{-4}\,\text{mol dm}^{-3}$. Using the approximation $[\text{H}^+] \approx \sqrt{K_a c}$, calculate the pH of a $0.050\,\text{mol dm}^{-3}$ solution of HA.

Show answer
$[\text{H}^+] = \sqrt{(1.7\times10^{-4})(0.050)} \approx 2.92\times10^{-3}\,\text{mol dm}^{-3}$$\text{pH} = -\log(2.92\times10^{-3}) \approx 2.54$
[3]
7

A buffer solution is made from $0.25\,\text{mol dm}^{-3}$ propanoic acid ($K_a = 1.3\times10^{-5}\,\text{mol dm}^{-3}$) and $0.15\,\text{mol dm}^{-3}$ sodium propanoate. Use $\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$ to calculate the pH of the buffer.

Show answer
$\text{p}K_a = -\log(1.3\times10^{-5}) \approx 4.89$$\text{pH} = 4.89 + \log\left(\dfrac{0.15}{0.25}\right) = 4.89 – 0.22 \approx 4.66$
[3]
8

Explain how the propanoic acid/sodium propanoate buffer resists a change in pH when:(a) a small amount of acid is added,(b) a small amount of alkali is added.

Show answer
(a) The propanoate ions ($\text{CH}_3\text{CH}_2\text{COO}^-$) react with the added $\text{H}^+$ to reform propanoic acid, removing most of the extra $\text{H}^+$ and keeping the pH almost unchanged.(b) The propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) reacts with the added $\text{OH}^-$, donating a proton to neutralise it and forming water and propanoate ions, so the pH changes only slightly.
[3]

Section C – Redox and Electrochemical Cells

9

A cell is made from a silver half-cell ($\text{Ag}^+ + e^- \rightleftharpoons \text{Ag}$, $E^\ominus = +0.80\,\text{V}$) and a zinc half-cell ($\text{Zn}^{2+} + 2e^- \rightleftharpoons \text{Zn}$, $E^\ominus = -0.76\,\text{V}$).(a) State which half-cell is the positive electrode.(b) Calculate the standard EMF of the cell.(c) Write the overall (doubled) ionic equation for the spontaneous cell reaction.

Show answer
(a) The silver half-cell (more positive $E^\ominus$, so silver is reduced) is the positive electrode.(b) $E^\ominus_{cell} = E^\ominus_{cathode} – E^\ominus_{anode} = 0.80 – (-0.76) = 1.56\,\text{V}$(c) $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$
[5]
10

Using $E^\ominus(\text{Cu}^{2+}/\text{Cu}) = +0.34\,\text{V}$ and $E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\,\text{V}$, predict, with a calculation, whether $\text{Fe}^{3+}$ ions can oxidise copper metal to $\text{Cu}^{2+}$ under standard conditions.

Show answer
$E^\ominus_{cell} = E^\ominus_{reduced} – E^\ominus_{oxidised} = 0.77 – 0.34 = +0.43\,\text{V}$ (positive), so the reaction $\text{Cu} + 2\text{Fe}^{3+} \rightarrow \text{Cu}^{2+} + 2\text{Fe}^{2+}$ is feasible under standard conditions.
[3]
11

Chlorine reacts with cold dilute sodium hydroxide: $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}$.(a) Give the oxidation state of chlorine in $\text{Cl}_2$, in $\text{NaCl}$, and in $\text{NaClO}$.(b) State the type of redox reaction this represents, where the same element is both oxidised and reduced.

Show answer
(a) $\text{Cl}_2$: 0. $\text{NaCl}$: $-1$. $\text{NaClO}$: $+1$.(b) This is a disproportionation reaction: chlorine (oxidation state 0) is simultaneously reduced to $-1$ (in $\text{Cl}^-$) and oxidised to $+1$ (in $\text{ClO}^-$).
[3]
12

A student uses standard electrode potentials to predict that a reaction with a large positive $E^\ominus_{cell}$ will occur rapidly at room temperature. Explain why this prediction may not always be correct.

Show answer
$E^\ominus_{cell}$ values give only thermodynamic information (whether a reaction is energetically feasible), not kinetic information; a reaction with a large positive $E^\ominus_{cell}$ may still have a high activation energy and therefore proceed very slowly (or not be observed) at room temperature.
[2]

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