A Data Booklet may be used.
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
CommonPredict the effect of increasing pressure on the position of equilibrium in each of the following. All species are gases.
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
Source: Learners' Alley, Chemical Equilibria slides
CommonThe equation describes the equilibrium between some gases. 2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g) ΔH < 0
Which of the changes increase the Kp of the reaction?
Source: IJC 2017 Promo, via Chemical Equilibria Topical Revision Booklet, Q4
In the industrial production of nitric acid, the first step involves the oxidation of ammonia to nitrogen oxide. 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) ΔH < 0
Which of the following would favour the oxidation of ammonia?
Source: IJC 2017 Promo, via Chemical Equilibria Topical Revision Booklet, Q6
For the reaction 2X2(g) + Y2(g) ⇌ 2X2Y(g), the value of the equilibrium constant Kc is 5.00. Determine the value of Kc for each of the following equilibria, including its units.
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A6
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A7
CommonIndustrially, coal can be converted into a combustible gas by passing steam over white hot coke. H2O(g) + C(s) ⇌ H2(g) + CO(g) ΔH = +131 kJ mol−1
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A9(a) to (c)
Zinc oxide can be produced from zinc and steam. Zn(s) + H2O(g) ⇌ ZnO(s) + H2(g) ΔH < 0. 2.60 mol of zinc and 2.60 mol of steam are placed in a sealed vessel and heated. At equilibrium, 2.58 mol of zinc oxide is produced. The total pressure at equilibrium is 10 atm.
Source: 2023 RI J2 Prelim P2 Q5(e)
Source: 2024 CJC J2 Prelim P3 Q2(b)(ii)
Kc = [NH3]2[N2][H2]3 mol−2 dm6[1]
Kp = pNH32pN2 pH23 atm−2[1]
Kc = [H+]3[Fe3+] mol2 dm−6: leave out the solid and the solvent, water.[1]
Since 1.56 mol of HI is formed, the amounts of H2 and I2 each fall by 0.78 mol, leaving 0.22 mol of each.[1]
Kc = (1.56)2(0.22)(0.22) = 50.3 (no units)[1]
Rate of forward reaction equals rate of backward reaction, so there is no net change in concentrations, in a closed system.[1]
Kc = [SO3]2[SO2]2[O2], mol−1 dm3[1]
It increases the rates of the forward and backward reactions equally, so K and the POE are unchanged.[1]
A lower T gives a higher yield, but the rate is too slow[1]
450 °C gives a reasonable yield in a reasonable time.[1]
A2 only[1]
Temperature is the only factor that will change an equilibrium constant, Kc or Kp.
Aremoval of H2O[1]
To favour the oxidation is to favour the forward reaction (the position of equilibrium shifts to the right). Removal of H2O will shift the position of equilibrium to the right.
Kc′ = [X2Y]4[X2]4[Y2]2 = (Kc)2 = (5.00)2
= 25.0 mol−2 dm6[1]
Kc″ = [X2][Y2]½[X2Y] = (1Kc)½ = (15.00)½
= 0.447 mol½ dm−3/2[1]
Amount of NaOH reacted with CH3COOH = 0.200 × 2891000 = 0.0578 mol = amount of CH3COOH present[1]
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O | |
|---|---|---|---|---|
| initial amount / mol | 1.00 | 5.00 | 0 | 0 |
| change / mol | −0.9422 | −0.9422 | +0.9422 | +0.9422 |
| equilibrium amount / mol | 0.0578 | 4.0578 | 0.9422 | 0.9422 |
Let the total volume be V dm3. Kc = [H2O][CH3COOC2H5][CH3COOH][C2H5OH] = (0.9422/V)(0.9422/V)(0.0578/V)(4.0578/V)[1]
= 3.79 (no units)[1]
When pressure is increased, the position of equilibrium will shift to decrease pressure by decreasing the total number of gaseous molecules present. Since there are 2 molecules of gas on the right-hand side of the equation and one on the left, the backward reaction is favoured. The position of equilibrium shifts to the left. Hence, at equilibrium, there will be more H2O, and less H2 and CO.[1]
When temperature is increased, the position of equilibrium will shift to decrease temperature. Since the forward reaction is endothermic, the forward reaction is favoured. The position of equilibrium shifts to the right. Hence, at equilibrium, there will be more H2 and CO, and less H2O.[1]
Since the amount of H2 produced = the amount of CO produced, the partial pressures of the two gases must be the same. Partial pressure of CO = 183 kPa[1]
Kp = (183 kPa)(183 kPa)90 kPa = 372 kPa[1]
Let the new partial pressure of H2 be x kPa. The partial pressure of CO is also x kPa. Kp = (x kPa)(x kPa)150 kPa = 372 kPa, so x = √(372 × 150) = 236 kPa (the negative root is disregarded, as partial pressures cannot be negative)[1]
| Zn(s) | H2O(g) | ZnO(s) | H2(g) | |
|---|---|---|---|---|
| initial amount / mol | 2.60 | 2.60 | 0 | 0 |
| change in amount / mol | −2.58 | −2.58 | +2.58 | +2.58 |
| equilibrium amount / mol | 0.02 | 0.02 | 2.58 | 2.58 |
Partial pressure of H2 at equilibrium = (2.580.02 + 2.58) × 10 = 9.92 atm[1]
Kp = pH2pH2O[1]
= (2.58/2.60) × 10(0.02/2.60) × 10 = 129[1]
Kp = (pCH4)(pH2O)(pCO)(pH2)3[1]
Let the initial amounts of CO and H2 be 1 and 3 mol respectively.
| CO | 3H2 | CH4 | H2O | |
|---|---|---|---|---|
| initial amount / mol | 1 | 3 | 0 | 0 |
| change / mol | −0.40 | −3(0.40) | +0.40 | +0.40 |
| equilibrium amount / mol | 0.60 | 1.80 | 0.40 | 0.40 |
Total amount at equilibrium = 0.6 + 1.8 + 0.4 + 0.4 = 3.2 mol[1]
pCO = 0.63.2 × 32 = 6 atm, pH2 = 1.83.2 × 32 = 18 atm, pCH4 = pH2O = 0.43.2 × 32 = 4 atm[1]
Kp = (4)(4)(6)(18)3 = 4.57 × 10−4 atm−2[1]