Questions on Kp: equilibria in the gas phase

Question 1

Explain, by writing Kp for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) in terms of mole fractions and the total pressure P, why raising the total pressure at constant temperature increases the proportion of ammonia at equilibrium, and why raising the temperature decreases it. ΔH = −92 kJ mol⁻¹.

Accepted Answer

Indicative content

each partial pressure is the mole fraction multiplied by the total pressure P, so Kp = x(NH₃)²P² ÷ (x(N₂)P × x(H₂)³P³)
this simplifies to Kp = (x(NH₃)² ÷ (x(N₂)x(H₂)³)) × P⁻², because there are two moles of gas on the right and four on the left
at constant temperature Kp is constant, so when P rises the ratio of mole fractions must rise by the factor P² to keep Kp the same
that ratio can rise only if nitrogen and hydrogen react to form ammonia, so the position shifts right and the mole fraction of ammonia increases
raising the temperature, by contrast, decreases the value of Kp itself, because the forward reaction is exothermic
the ratio of partial pressures is then larger than the new Kp, so the reverse reaction runs faster and the position shifts left, lowering the proportion of ammonia

Levels

Level 3 (5–6 marks) Kp is written in mole fractions and simplified to show the P⁻² factor, the constancy of Kp at fixed temperature is used to show that the ratio of mole fractions must rise with P and so ammonia must form, and the temperature effect is explained through the change in Kp itself; every step follows from the one before and the algebra is correct
Level 2 (3–4 marks) Kp is written correctly in mole fractions and the pressure effect is argued from Kp staying constant; the P⁻² simplification, or the smaller Kp at higher temperature, is not shown
Level 1 (1–2 marks) Kp is written, or higher pressure is said to favour ammonia because there are fewer moles of gas; Kp itself is usually said to increase with pressure

Question 2

Calculate the partial pressure of each gas in the Haber equilibrium mixture described in the table, and use them to calculate Kp for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Accepted Answer

M1 total amount = 10.00 mol, so the mole fractions are N₂ 0.200, H₂ 0.600 and NH₃ 0.200 M1 partial pressures: p(N₂) = 4.00 × 10³ kPa, p(H₂) = 1.20 × 10⁴ kPa, p(NH₃) = 4.00 × 10³ kPa M1 Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³) M1 Kp = (4.00 × 10³)² ÷ (4.00 × 10³ × (1.20 × 10⁴)³) = 1.60 × 10⁷ ÷ 6.91 × 10¹⁵ A1 Kp = 2.31 × 10⁻⁹ kPa⁻², to three significant figures

Question 3

Calculate Kp for CO(g) + 2H₂(g) ⇌ CH₃OH(g) from the experiment described in the table.

Accepted Answer

M1 equilibrium amounts: CO 0.600 mol, H₂ 2.00 − 0.800 = 1.20 mol, CH₃OH 0.400 mol M1 total amount = 2.20 mol, so the mole fractions are CO 0.273, H₂ 0.545 and CH₃OH 0.182 M1 partial pressures: p(CO) = 1364 kPa, p(H₂) = 2727 kPa, p(CH₃OH) = 909 kPa M1 Kp = p(CH₃OH) ÷ (p(CO) × p(H₂)²) = 909 ÷ (1364 × 2727²) A1 Kp = 8.96 × 10⁻⁸ kPa⁻², to three significant figures; error carried forward from M2

Question 4

Calculate the degree of dissociation of PCl₅ at 500 K and a total pressure of 150 kPa, given that Kp for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 63.0 kPa at 500 K.

Accepted Answer

M1 from 1 mol PCl₅: (1 − α) mol PCl₅, α mol PCl₃, α mol Cl₂, total (1 + α) mol M1 p(PCl₅) = (1 − α)P ÷ (1 + α) and p(PCl₃) = p(Cl₂) = αP ÷ (1 + α) M1 Kp = α²P ÷ (1 − α²), so 150α² ÷ (1 − α²) = 63.0 M1 213α² = 63.0, so α² = 0.296 A1 α = 0.544, with no units, so 54.4% of the PCl₅ dissociates

Question 5

Calculate Kp at 330 K for N₂O₄(g) ⇌ 2NO₂(g), given that N₂O₄ is 50.0% dissociated at 330 K and a total pressure of 100 kPa.

Accepted Answer

M1 from 1.00 mol N₂O₄: 0.500 mol N₂O₄ and 1.00 mol NO₂ at equilibrium, total 1.50 mol M1 partial pressures: p(N₂O₄) = (0.500 ÷ 1.50) × 100 = 33.3 kPa and p(NO₂) = (1.00 ÷ 1.50) × 100 = 66.7 kPa M1 Kp = p(NO₂)² ÷ p(N₂O₄) = 66.7² ÷ 33.3 A1 Kp = 133 kPa, to three significant figures

Question 6

Explain why a sealed tube containing an equilibrium mixture of N₂O₄ and NO₂ turns a deeper brown when it is warmed, and why adding a catalyst to the tube would change neither the colour nor Kp. The dissociation of N₂O₄ into NO₂ is endothermic.

Accepted Answer

B1 Kp for an endothermic forward reaction increases as the temperature rises B1 the ratio p(NO₂)² ÷ p(N₂O₄) is now smaller than the new Kp, so the forward reaction runs faster until it matches, and more brown NO₂ forms B1 a catalyst increases the rates of the forward and reverse reactions equally, so the partial pressures at equilibrium and the colour are unchanged B1 Kp depends only on temperature, so the catalyst does not change it

Question 7

Evaluate a student's claim that doubling the total pressure on PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) doubles Kp, because every partial pressure doubles.

Accepted Answer

B1 at the instant of compression every partial pressure does double B1 the ratio p(PCl₃) × p(Cl₂) ÷ p(PCl₅) then doubles, because there are two pressure terms on top and one below, so it no longer equals Kp B1 the reverse reaction runs faster and the position shifts towards PCl₅ until the ratio falls back to Kp B1 Kp depends only on temperature, so the claim is wrong: the position changes and Kp does not

Question 8

Calculate Kp at 298 K for NH₄HS(s) ⇌ NH₃(g) + H₂S(g), in atm² and in kPa², given that solid ammonium hydrogensulfide placed in an evacuated flask at 298 K gives a total pressure of 0.660 atm at equilibrium. Take 1 atm as 101 kPa.

Accepted Answer

M1 equal amounts of NH₃ and H₂S form, so p(NH₃) = p(H₂S) = 0.660 ÷ 2 = 0.330 atm M1 Kp = p(NH₃) × p(H₂S), the solid being left out, so Kp = 0.330 × 0.330 = 0.109 atm² M1 Kp has two powers of pressure, so the conversion factor is squared: Kp = 0.1089 × 101² A1 Kp = 1.11 × 10³ kPa², to three significant figures; error carried forward from M2

Question 9

Explain why the amount of carbon is not included when the mole fractions are calculated for the equilibrium C(s) + CO₂(g) ⇌ 2CO(g), and the effect of raising the total pressure at constant temperature on the proportion of carbon monoxide.

Accepted Answer

B1 carbon is a solid, so it exerts no partial pressure and does not appear in the Kp expression B1 mole fractions are used to find partial pressures, so the total amount is the total amount of gas only, CO₂ and CO B1 there is one mole of gas on the left and two on the right; the solid carbon is not counted B1 raising the pressure shifts the position to the left, so the proportion of carbon monoxide falls while Kp is unchanged

Question 10

State what is meant by the mole fraction and by the partial pressure of a gas in a mixture, and state how the partial pressures of the gases in a mixture are related to the total pressure.

Accepted Answer

B1 mole fraction is the amount in moles of that gas divided by the total amount in moles of gas in the mixture B1 partial pressure is the pressure the gas would exert if it alone occupied the whole volume; accept mole fraction × total pressure B1 the partial pressures of all the gases add up to the total pressure

Question 11

Give the expression for Kp for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), and give its units when the pressures are in kPa and when they are in Pa.

Accepted Answer

B1 Kp = p(SO₃)² ÷ (p(SO₂)² × p(O₂)); reject an expression with square brackets B1 units in kPa: kPa² ÷ kPa³ = kPa⁻¹ B1 units in Pa: Pa² ÷ Pa³ = Pa⁻¹

Question 12

Explain why increasing the total pressure on N₂O₄(g) ⇌ 2NO₂(g) at constant temperature changes the position of equilibrium but not Kp.

Accepted Answer

B1 Kp depends only on temperature, which is unchanged B1 raising the total pressure raises both partial pressures by the same factor, and because p(NO₂) is squared the ratio p(NO₂)² ÷ p(N₂O₄) becomes larger than Kp B1 the reverse reaction runs faster until the ratio equals Kp again, so the position shifts to the left, towards N₂O₄

Question 13

Outline the steps for calculating Kp from the initial amounts of reactants, one equilibrium amount and the total pressure.

Accepted Answer

B1 use the equation to find every equilibrium amount from the change in the measured one, in an ICE table B1 add the equilibrium amounts to give the total amount, and divide each amount by it to give the mole fractions B1 multiply each mole fraction by the total pressure to give the partial pressures, and substitute them into the Kp expression

Question 14

Suggest an expression for Kp for CaCO₃(s) ⇌ CaO(s) + CO₂(g), its units in kPa, and why the amount of calcium carbonate present does not affect the equilibrium pressure.

Accepted Answer

B1 Kp = p(CO₂), because solids do not appear in the expression B1 the units of Kp are kPa B1 at a given temperature the CO₂ settles at the pressure equal to Kp whatever the amount of solid present, as long as some of each solid remains

Question 15

Suggest why the total pressure need not be known to calculate Kp for H₂(g) + I₂(g) ⇌ 2HI(g) from the equilibrium amounts, and why changing the total pressure does not move this equilibrium.

Accepted Answer

B1 each partial pressure is the mole fraction multiplied by the same total pressure P B1 the powers on top and bottom are both 2, so P² cancels and Kp equals the ratio of mole fractions, with no units B1 a change in P therefore leaves the ratio equal to Kp, so there is no shift in the position of equilibrium

Question 16

Compare Kc and Kp for the same gas-phase equilibrium.

Accepted Answer

B1 Kc is written with equilibrium concentrations in mol dm⁻³, Kp with equilibrium partial pressures B1 both are built the same way from the balanced equation, with coefficients as powers, and their units come from the expression: powers of mol dm⁻³ for Kc, of kPa for Kp B1 both depend only on temperature and are unaffected by pressure, concentration or a catalyst

Question 17

Justify the statement that Kp for N₂O₄(g) ⇌ 2NO₂(g) has the unit kPa, when partial pressures are in kPa, but Kp for H₂(g) + I₂(g) ⇌ 2HI(g) has no unit.

Accepted Answer

B1 Kp = p(NO₂)² ÷ p(N₂O₄), so its units are kPa² ÷ kPa = kPa B1 Kp = p(HI)² ÷ (p(H₂) × p(I₂)) has two pressures above the line and two below, so kPa² ÷ kPa² cancels and Kp is a pure number B1 the unit depends on the difference between the numbers of moles of gas on the two sides: the first equation has one more mole of gas on the right, the second has equal numbers

Quantity	Value
N₂ at equilibrium	2.00 mol
H₂ at equilibrium	6.00 mol
NH₃ at equilibrium	2.00 mol
Total pressure	2.00 × 10⁴ kPa

Quantity	Value
CO at the start	1.00 mol
H₂ at the start	2.00 mol
CH₃OH at equilibrium	0.400 mol
Total pressure at equilibrium	5.00 × 10³ kPa