DP Chemistry (last assessment 2024)

Question 17N.2.hl.TZ0.6

Date	November 2017	Marks available	[Maximum mark: 7]	Reference code	17N.2.hl.TZ0.6
Level	hl	Paper	2	Time zone	TZ0
Command term	Calculate, Determine, State	Question number	6	Adapted from	N/A

[Maximum mark: 7]

17N.2.hl.TZ0.6

Many reactions are in a state of equilibrium.

The following reaction was allowed to reach equilibrium at 761 K.

H₂ (g) + I₂ (g) $\rightleftharpoons$ 2HI (g) ΔH^θ < 0

(a.i)

State the equilibrium constant expression, K_c , for this reaction.

[1]

Markscheme

K_c = $\frac{{{{{\text{[HI]}}}^{\text{2}}}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{][}}{{\text{I}}_{\text{2}}}{\text{]}}}}$

(a.ii)

The following equilibrium concentrations in mol dm^–3 were obtained at 761 K.

Calculate the value of the equilibrium constant at 761 K.

[1]

Markscheme

45.6

(a.iii)

Determine the value of ΔG^θ, in kJ, for the above reaction at 761 K using section 1 of the data booklet.

[1]

Markscheme

ΔG^θ = «– RT ln K = – (0.00831 kJ K⁻¹ mol⁻¹ x 761 K x ln 45.6) =» – 24.2 «kJ»

The pH of 0.010 mol dm^–3 carbonic acid, H₂CO₃ (aq), is 4.17 at 25 °C.

H₂CO₃ (aq) + H₂O (l) $\rightleftharpoons$ HCO₃^– (aq) + H₃O⁺ (aq).

(c.i)

Calculate [H₃O⁺] in the solution and the dissociation constant, K_a , of the acid at 25 °C.

[3]

Markscheme

[H₃O⁺] = 6.76 x 10^–5 «mol dm^–3»

K_a = $\frac{{{{\left( {6.76 \times {{10}^{ - 5}}} \right)}^2}}}{{\left( {0.010 - 6.76 \times {{10}^{ - 5}}} \right)}}/\frac{{{{\left( {6.76 \times {{10}^{ - 5}}} \right)}^2}}}{{0.010}}$