Grade 12 Physical Sciences · Chemical Change

Calculate Kc

Substitute equilibrium concentrations into the Kc expression to calculate Kc, or use Kc to find an unknown concentration.

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Question 1.5 · November 2023 · Paper 2 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10).

1.5 The diagram below represents a mixture of

NO_{2}

(g) and

N_{2} O_{4}

(g) molecules at equilibrium in a

1

^{3}

container at T °C.

The balanced equation for this reaction is:

2 NO_{2} (g) ⇌ N_{2} O_{4} (g)

Which ONE of the following is TRUE for the value of the equilibrium constant,

K_{c}

, for the reaction at T °C?(2) Diagram for question 1.5

A $K_{c} = 24$
B $K_{c} > 1$
C $K_{c} = 1$
D $0 < K_{c} < 1$

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: calculate kc

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3. Question 6.2 · November 2022 · Paper 2 · National (DBE) · 4 marks
Carbon, C(s), reacts with sulphur, S(g), according to the following balanced equation:

$C(s) + 2 S(g) ⇌ CS_{2} (g)$ $Δ H > 0$

The system reaches equilibrium at temperature T in a sealed $2$ dm $^{3}$ container.

The $K_{c}$ value is $9, 4$ at temperature T.
At equilibrium, $1$ mole of carbon disulphide, $CS_{2}$ (g), is present in the container.
6.2 Calculate the concentration of S(g) present at equilibrium.(4)
Show the memo (4 ticks)Memo
1. ✓
  Kc expression $K_{c} = \frac{[ CS _{2} ]}{[ S ] ^{2}}$
  1 mark · Write the Kc expression
2. ✓
  Substitute Kc value $9, 4 = \frac{[ CS _{2} ]}{[ S ] ^{2}}$
  1 mark · Calculate Kc
3. ✓
  Substitute [CS₂] $9, 4 = \frac{0 , 5}{[ S ] ^{2}}$
  1 mark · Calculate Kc
4. ✓
  Final answer $[S]^{2} = \frac{0 , 5}{9 , 4}$
  $[S] = 0, 23$ mol·dm $^{- 3}$
  1 mark · Calculate Kc
4. Question 6.2.3 · May/June 2022 · Paper 2 · National (DBE) · 6 marks
6.2 The equation below represents the reversible reaction that takes place when $NO_{2}$ (g) is converted to $N_{2} O_{4}$ (g).

$2 NO_{2} (g) ⇌ N_{2} O_{4} (g)$

Initially, $x$ mol of $NO_{2}$ (g) is sealed in a $1$ dm $^{3}$ container at $350$ K. When equilibrium is established at this temperature, $0, 81$ mol $N_{2} O_{4}$ (g) is present in the container.

Answer the questions that follow.
$0, 79$ moles of $N_{2} O_{4}$ (g) is now added to the equilibrium mixture above. When the NEW equilibrium is established at $350$ K, it is found that the amount of $NO_{2}$ (g) increased by $1, 2$ moles.
6.2.3 Calculate the value of $x$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Add 0,79 mol to N2O4Initial (new): $n (NO_{2}) = x - 1, 62$ ; $n (N_{2} O_{4}) = 0, 81 + 0, 79 = 1, 6$ mol
  1 mark · Complete an equilibrium table
2. ✓
  Mole ratio: 0,6 mol N2O4Ratio $n (NO_{2}) : n (N_{2} O_{4}) = 2 : 1$
  Change: $NO_{2}$ $+ 1, 2$ mol; $N_{2} O_{4}$ $- 0, 6$ mol
  1 mark · Complete an equilibrium table
3. ✓
  New equilibrium amounts $n (NO_{2})_{e q} = x - 1, 62 + 1, 2 = x - 0, 42$ mol
  $n (N_{2} O_{4})_{e q} = 1, 6 - 0, 6 = 1$ mol
  Concentrations ( $1$ dm $^{3}$ ): $[NO_{2}] = x - 0, 42$ ; $[N_{2} O_{4}] = 1$ mol·dm $^{- 3}$
  1 mark · Complete an equilibrium table
4. ✓
  Substitution into Kc expression $K_{c} = \frac{[ N _{2} O _{4} ]}{[ NO _{2} ] ^{2}} = \frac{1}{( x - 0 , 42 ) ^{2}}$
  1 mark · Calculate Kc
5. ✓
  Equating the two Kc expressions $\frac{0 , 81}{( x - 1 , 62 ) ^{2}} = \frac{1}{( x - 0 , 42 ) ^{2}}$
  1 mark · Calculate Kc
6. ✓
  Final answer $0, 9 (x - 0, 42) = x - 1, 62$
  $0, 1 x = 1, 242$
  $x = 12, 42$ (mol)
  1 mark · Calculate Kc
5. Question 6.4 · May/June 2025 · Paper 2 · National (DBE) · 5 marks
The reaction between powdered carbon, C(s), and carbon dioxide gas, $CO_{2}$ (g), takes place in a closed $3$ dm $^{3}$ container according to the balanced equation below.

$C(s) + CO_{2} (g) ⇌ 2 CO(g)$ $Δ H < 0$

Equilibrium is reached at temperature T °C.

The graph below, not drawn to scale, shows the curves for the mass of the reactants in the container against time.
6.4 Calculate the equilibrium constant, Kc, at T °C.(5)
Show the memo (5 ticks)Memo
1. ✓
  (a) Ratio n(CO2) : n(CO) = 1 : 2 $n (CO_{2})_{initial} = \frac{41 , 2}{44} = 0, 936$ mol
  Ratio CO $_{2}$ : CO = 1 : 2
  Initial (mol): CO $_{2}$ $0, 936$ ; CO $0$
  Change (mol): CO $_{2}$ $0, 8$ ; CO $1, 6$
  Equilibrium (mol): CO $_{2}$ $0, 14$ ; CO $1, 6$
  1 mark · Complete an equilibrium table
2. ✓
  (b) Divide by 3 dm3Equilibrium concentration (mol·dm $^{- 3}$ ): $[CO_{2}] = \frac{0 , 14}{3} = 0, 047$ ; $[CO] = \frac{1 , 6}{3} = 0, 53$
  1 mark · Calculate concentration of a solution
3. ✓
  (c) Kc expression $K_{c} = \frac{[ CO ] ^{2}}{[ CO _{2} ]}$
  1 mark · Write the Kc expression
4. ✓
  (d) Substitution $K_{c} = \frac{( 0 , 53 ) ^{2}}{0 , 047}$
  1 mark · Calculate Kc
5. ✓
  (e) Final answer $K_{c} = 5, 98$
  1 mark · Calculate Kc
6. Question 6.2 · November 2024 · Paper 2 · National (DBE) · 9 marks
Carbon monoxide gas, CO(g), reacts with water vapour, $H_{2} O$ (g), at T °C. The reaction reaches chemical equilibrium according to the balanced equation:

$CO(g) + H_{2} O(g) ⇌ CO_{2} (g) + H_{2} (g)$

Initially, $0, 6$ moles of CO(g), $0, 6$ moles of $H_{2} O$ (g), $0, 1$ moles of carbon dioxide gas, $CO_{2}$ (g), and $0, 1$ moles of hydrogen gas, $H_{2}$ (g), were mixed and sealed in a $2$ dm $^{3}$ flask.
6.2 If the equilibrium constant, $K_{c}$ , for this reaction at T °C is $4$ , calculate the mass of CO(g) present in the flask at equilibrium.(9)
Show the memo (9 ticks)Memo
1. ✓
  (a) Using ratio 1 : 1 : 1 : 1Reactants are used (x = change in amount).
  Initial amount (mol): CO $0, 6$ ; $H_{2} O$ $0, 6$ ; $CO_{2}$ $0, 1$ ; $H_{2}$ $0, 1$
  Change (mol): CO $- x$ ; $H_{2} O$ $- x$ ; $CO_{2}$ $+ x$ ; $H_{2}$ $+ x$ (ratio $1 : 1 : 1 : 1$ )
  1 mark · Complete an equilibrium table
2. ✓
  (b) Equilibrium amountsEquilibrium amount (mol): CO $0, 6 - x$ ; $H_{2} O$ $0, 6 - x$ ; $CO_{2}$ $0, 1 + x$ ; $H_{2}$ $0, 1 + x$
  1 mark · Complete an equilibrium table
3. ✓
  (c) Divide by the volume 2 dm $^{3}$ Equilibrium concentration (mol·dm $^{- 3}$ ): CO $\frac{0 , 6 - x}{2}$ ; $H_{2} O$ $\frac{0 , 6 - x}{2}$ ; $CO_{2}$ $\frac{0 , 1 + x}{2}$ ; $H_{2}$ $\frac{0 , 1 + x}{2}$
  1 mark · Calculate concentration of a solution
4. ✓
  (d) Correct Kc expression $K_{c} = \frac{[ CO _{2} ] [ H _{2} ]}{[ CO ] [ H _{2} O ]}$
  1 mark · Write the Kc expression
5. ✓
  (e) Substitute Kc value 4 $4 = \dots$
  1 mark · Calculate Kc
6. ✓
  (f) Substitute concentrations in Kc expression $4 = \frac{( \frac{0 , 1 + x}{2} ) ( \frac{0 , 1 + x}{2} )}{( \frac{0 , 6 - x}{2} ) ( \frac{0 , 6 - x}{2} )}$
  $x = 0, 37$
  1 mark · Calculate Kc
7. ✓
  (g) Substitute numerical value of x $n (CO)_{e q} = 0, 6 - 0, 37 = 0, 23$ mol
  1 mark · Calculate Kc
8. ✓
  (h) Substitute 28 in $n = \frac{m}{M}$ $n (CO)_{e q} = \frac{m}{M}$
  $0, 23 = \frac{m}{28}$
  1 mark · Calculate number of moles
9. ✓
  (i) Final answer $m (CO)_{e q} = 6, 44$ g
  1 mark · Calculate number of moles

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