Grade 12 Physical Sciences · Chemical Change

Write the Kc expression

Write the Kc expression as products over reactants with powers from the balanced equation, leaving out solids and pure liquids.

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Question 1.6 · May/June 2024 · Paper 2 · National (DBE) · 2 marks

1.6 Consider the equation below for a hypothetical reaction.

A(aq) + B(s) ⇌ C(aq) + D(aq)

If the equilibrium constant

K_{c} = 1 \times 1 0^{- 4}

, then ...(2)

A $[A] [B] < [C] [D]$
B $[A] [B] > [C] [D]$
C $[A] > [C] [D]$
D $[A] < [C] [D]$

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

More past paper questions: write the kc expression

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3. Question 6.4 · November 2023 · Paper 2 · National (DBE) · 4 marks
Consider the following hypothetical reaction reaching equilibrium in a $4$ dm $^{3}$ closed container at $150$ °C.

$2 AB(g) ⇌ A_{2} (g) + B_{2} (g)$

The graph below shows the changes in the amounts of reactants and products over time.
6.4 Calculate the equilibrium constant, $K_{c}$ , at t = $120$ s.(4)
Show the memo (4 ticks)Memo
1. ✓
  Kc expression $K_{c} = \frac{[ A _{2} ] [ B _{2} ]}{[ AB ] ^{2}}$
  1 mark · Write the Kc expression
2. ✓
  Substitute product concentrationsAt $120$ s: $n (A_{2}) = 8$ mol, $n (B_{2}) = 2$ mol, $n (AB) = 10$ mol in $4$ dm $^{3}$ .
  $K_{c} = \frac{( \frac{8}{4} ) ( \frac{2}{4} )}{( \frac{10}{4} ) ^{2}}$
  1 mark · Calculate Kc
3. ✓
  Substitute reactant concentration $[AB] = \frac{10}{4} = 2, 5$ mol·dm $^{- 3}$ , squared in the denominator: $(2, 5)^{2}$
  1 mark · Calculate Kc
4. ✓
  Final answer $K_{c} = \frac{( 2 ) ( 0 , 5 )}{( 2 , 5 ) ^{2}} = 0, 16$
  1 mark · Calculate Kc
4. 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
5. Question 6.2.2 · May/June 2022 · Paper 2 · National (DBE) · 5 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.
6.2.2 Show that the equilibrium constant for this reaction is given by $\frac{0 , 81}{( x - 1 , 62 ) ^{2}}$ .(5)
Show the memo (5 ticks)Memo
1. ✓
  Change in N2O4 $Δ n (N_{2} O_{4}) = n (N_{2} O_{4})_{e q} - n (N_{2} O_{4})_{ini} = 0, 81 - 0 = 0, 81$ mol
  1 mark · Complete an equilibrium table
2. ✓
  Mole ratioRatio $n (NO_{2}) : n (N_{2} O_{4}) = 2 : 1$
  
  Change: $NO_{2}$ $1, 62$ mol used; $N_{2} O_{4}$ $0, 81$ mol formed
  1 mark · Complete an equilibrium table
3. ✓
  Equilibrium amount of NO2 $n (NO_{2})_{e q} = n (NO_{2})_{ini} - Δ n (NO_{2}) = x - 1, 62$ mol
  1 mark · Complete an equilibrium table
4. ✓
  Divide both by 1 dm³Equilibrium concentrations: $[NO_{2}] = \frac{x - 1 , 62}{1} = x - 1, 62$ mol·dm $^{- 3}$ ; $[N_{2} O_{4}] = \frac{0 , 81}{1} = 0, 81$ mol·dm $^{- 3}$
  1 mark · Calculate concentration of a solution
5. ✓
  Correct Kc expression $K_{c} = \frac{[ N _{2} O _{4} ]}{[ NO _{2} ] ^{2}} = \frac{0 , 81}{( x - 1 , 62 ) ^{2}}$
  1 mark · Write the Kc expression
6. Question 6.6 · May/June 2024 · Paper 2 · National (DBE) · 8 marks
Consider the balanced equation for a hypothetical reaction that takes place in $2$ dm $^{3}$ sealed containers.

$P_{2} Q(g) + 3 Q(s) ⇌ 2 PQ_{2} (g)$

The graphs below, not drawn to scale, are obtained for the same reaction at two different temperatures.

Graph A is obtained at $298$ K and graph B at $398$ K.
6.6 Calculate the initial number of moles of $P_{2} Q$ .(8)
Show the memo (7 ticks)Memo
1. ✓
  Mole ratio in the change rowConcentrations (mol·dm $^{- 3}$ ):
  P $_{2}$ Q: initial $x$ ; PQ $_{2}$ : initial $0$
  Change: PQ $_{2}$ $+ 0, 35$ , so P $_{2}$ Q $- 0, 175$ (ratio P $_{2}$ Q : PQ $_{2}$ = 1 : 2)
  1 mark · Complete an equilibrium table
2. ✓
  Equilibrium concentration of P $_{2}$ QEquilibrium: P $_{2}$ Q $x - 0, 175$ ; PQ $_{2}$ $0, 35$
  1 mark · Complete an equilibrium table
3. ✓
  K $_{c}$ expression $K_{c} = \frac{[ PQ _{2} ] ^{2}}{[ P _{2} Q ]}$
  2 marks · Write the Kc expression
4. ✓
  Substitute 0,49 $0, 49 = \dots$
  1 mark · Calculate Kc
5. ✓
  Substitute the equilibrium concentrations $0, 49 = \frac{( 0 , 35 ) ^{2}}{x - 0 , 175}$
  $x = 0, 425$ mol·dm $^{- 3}$
  1 mark · Calculate Kc
6. ✓
  Substitute 2 dm $^{3}$ in n = cV $n (P_{2} Q) = c V = 0, 425 \times 2$
  1 mark · Calculate concentration of a solution
7. ✓
  Final answer $= 0, 85$ mol
  1 mark · Calculate concentration of a solution

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