Questions on the Equilibrium law
1. Phosphorus(V) chloride decomposes when heated according to the equation:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Assume that the total volume is fixed at V dm3 and that an initial amount of phosphorus(V) chloride of a
mol gives x mol of chlorine in the equilibrium mixture. The equilibrium constant is represented by Kc.
(a) Show that the equilibrium expression for this reaction is:
Kc = x2 / (a – x)V
(b) Explain how this equation can be used to deduce the effect of increasing the pressure on the
equilibrium mixture at the same temperature.
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2. Steam reacts with methane in the gaseous state to form carbon dioxide and hydrogen.
(a) State the equation for this reaction.
(b) 54.06 g of steam and 16.04 g of methane are placed in a 1 dm3 container. When equilibrium has
been reached 4.04 g of hydrogen are present in the mixture. Determine the value of Kc for this reaction.
(c) The reaction is endothermic. State the effect on (i) the amount of hydrogen in the equilibrium
mixture and (ii) the value of the equilibrium constant if the reaction is carried out at a higher
temperature to the one above.
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3. Hydrogen and iodine react reversibly in the gaseous state to form hydrogen iodide. A particular
equilibrium mixture was found to contain 0.45 mol of iodine, 0.55 mol of hydrogen and 2.4 mol of
hydrogen iodide. Calculate the value of the equilibrium constant at this temperature.
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4. When 1.0 mol of ethanol reacts with 0.5 mol of ethanoic acid at 100 oC the mixture contains 0.42 mol
of ethyl ethanoate once equilibrium has been reached. Calculate the value of Kc at 100 oC for this
esterification reaction.
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5. (a) The standard enthalpy of formation of ethyl ethanoate at 298 K is – 480 kJ mol-1 and the standard
entropy of ethyl ethanoate at 298 K is + 259 J mol-1 K-1. Use these values and information from Section
12 of the IB data booklet to determine a value for Kc for the esterification reaction between ethanoic
acid and ethanol at 298 K.
(b) Compare your answer to the literature value of 4 at 298 K and suggest one reason which may
contribute to the fact that there is such a considerable difference.
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Answers
1. (a) Initial concentration of PCl5(g) = a/V
Equilibrium concentration of PCl5(g) = (a – x)/V
Equilibrium concentration of PCl3(g) = equilibrium concentration of Cl2(g) = x/V
Kc = [PCl3(g)][Cl2(g)] / [PCl5(g)]
= (x/V)2 / (a – x)/V = x2/V2 / (a – x)/V = x2 / (a – x)V
(b) From the gas laws increasing the pressure at the same temperature can be achieved by reducing the
volume. From the equation when V decreases x must also decrease to keep Kc constant hence increasing
the pressure causes the position of equilibrium to shift back to the phosphorus(V) chloride side.
2. (a) 2H2O(g) + CH4(g) ⇌ CO2(g) + 4H2(g)
(b) Initial concentration of H2O = 54.06 / 18.02 = 3.0 mol dm-3
Initial concentration of CH4 = 16.04 / 16.04 = 1.0 mol dm-3
Equilibrium concentration of H2 = 4.04 / 2.02 = 2.0 mol dm-3
1 mol of CO2 is formed for every 4 mol of H2 so equilibrium concentration of CO2 = 0.5 mol dm-3
2 mol of H2 gives 4 mol of steam so equilibrium concentration of steam = 2.0 mol dm-3
1 mol of CH4 gives 4 mol of steam so equilibrium concentration of methane = 0.5 mol dm-3
Kc = [CO2(g)]H2(g)]4 / [CH4(g)][H2O]2 = 0.5 x 2.04 / 0.5 x 2.02 = 2.02 = 4.0
(c) (i) the amount of hydrogen in the equilibrium increases
(ii) the value of the equilibrium constant increases
3. H2(g) + I2(g) ⇌ 2HI(g)
Let the volume of the container be V dm3
Kc = [HI(g)]2 / [H2(g)][I2(g)] = (2.4/V)2 / (0.45/V) x (0.55/V) = 2.42/0.45 x 0.55 = 23 (to 2 sig figs).
4. C2H5OH(l) + CH3COOH(l) ⇌ CH3COOC2H5(l) + H2O(l)
Let the volume of the container be V dm3
Kc = [CH3COOC2H5(l)][ H2O(l)] / [C2H5OH(l)][ CH3COOH(l)]
= (0.42/V x 0.42/V) / (1 – 0.42)/V x (0.5 – 0.42)/V = 0.422 / 0.58 x 0.08 = 3.8
5. (a) From the information given and from Section 12 of the data booklet
ΔH⦵reaction =[ – 480 + (– 286)] – [– 484 + (– 278)] = – 4 kJ mol-1
ΔS⦵reaction = (+ 259 + 70) – (+ 160 + 161) = 8 J mol-1 K-1
ΔG⦵reaction = ΔH⦵reaction – T ΔS⦵reaction = – 4 – (298 x 8/1000) = – 6.38 kJ mol-1 = – 6380 J mol-1
lnKc = – ΔG⦵reaction / RT = 6380/(8.314 x 298) = 2.58
Kc = e2.58 = 13.2
(b) This is considerably different to the literature value of 4. It is a consequence of the logarithmic part
of the expression. Data book values for ΔH⦵f and S⦵ can vary slightly from source to source. There only
needs to be a difference of 3 kJ mol-1 for the calculated value of ΔG⦵reaction to make it 3380 instead of
6380 J mol-1 which would make lnKc have the value 1.36 which then gives the literature value of 4 for Kc.