JPJC 2026 Chemical Equilibrium Revision Answer

2026 JC1 H2 Chemistry (9476) Topical Revision for Physical Chemistry (Chemical Equilibria) FOUNDATION KIT Worked solutions for MCQ 2022 JPJC JC1 H2 Chem EOY P1 Q20 1 Answer: A 1CH4(g) + 2H2(g) 1CS2(g) + 4H2 Initial conc. / mol dm−3 (= n V ) 1 0.5 = 2 2 0.5 = 4 1 0.5 = 2 2 0.5 = 4 Change (follow mole ratio in eqn) −0.88 −2 (0.88) +0.88 +4(0.88) eqm conc./ mol dm−3 1.12 (given) 2.24 2.88 7.52 Value of Kc =       4 22 2 42 CS H CH H S = ( )( ) ( )( ) 4 2 2.88 7.52 1.12 2.24 = 1640 J1988/Paper III/Q9 2 Answer: B Dynamic equilibrium refers to a state of balance in a reversible reaction in which the rate of the forward reaction equals to the rate of backward reaction ; and the concentrations of all reactants and products remain constant. P(aq) + Q(aq) ⇌ 2R(aq) + S(aq) Note position of eqm shifts to MINIMISE (not remove) the change – the change is if [Q] ↑, position of equilibrium (poe) shifts to minimise this change of ↑[Q] by shifting poe to right to ↓ [Q] to re-establish a new eqm – however at the new eqm, there is still a net ↑[Q] it is just that the change is minimized so at the new eqm, [Q] > 10 mol dm-3 but less than 15 mol dm-3. 2022 JC1 H2 Chemistry Paper 1 Q22 3 Answer: A From the graph, % product at equilibrium decreases as pressure increases.  % product  implies that position of equilibrium (poe) shifts to the left.  When pressure , less moles of gas is produced to reduce pressure.  LHS of the equilibrium equation has less gas particles. (Reject Option B, C, D) % pdt at eqm p (T + 10) C T C

From the graph, % product at equilibrium increases as temperature increases.  % product  implies that poe shifts to the right.  When temperature , it favours the endothermic reaction to ↓ temperature to absorb some heat.  Forward reaction is endothermic. (Reject Option C and D) 2014 JC2 H2 Chemistry Prelim Paper 1 Q12 4 Answer: A For the 2nd set of experiment, H2(g) + I2(g) ⇌ 2HI(g); Initial amt / mol 1 1 0 Δ in amt / mol -0.6 -0.6 +1.2 Final amt / mol 0.4 0.4 1.2 Kc = (1.2)2 / (0.4 x 0.4) = 9 Hence from T1 to T2, Kc has decreased from 64 to 9. This shows that the position of equilibrium shifts to the left. Since the reaction in endothermic, a decrease in temperature would favor the exothermic reaction. Hence, T1 > T2. 2019 JPJC EOY P1 Q16 5 Answer: C 2P(g) ⇌ 3Q(g) + R(g) ∆H > 0 (A) Adding inert solid X has no effect on the p.o.e. or value of Kc. (B) When pressure increase, the system will reduce the pressure by reducing the number of gas molecules (to reduce pressure). i.e. position of equilibrium (poe) shifts to the left. (C) Increasing temperature , poe will shift to the right to favour the endothermic reaction (∆H > 0) and value of Kc will increase. (D) Adding a catalyst has no effect on poe and Kc value (unchanged). 2018 JPJC EOY P1 Q19 6 Answer: D The equilibrium system involves 2 solids and a gas. Since solids have fixed constant concentrations/ densities, and constant concentrations are not included in Kc, Kc = [CO2(g)] = = 7.00  10−3 mol dm−3 2018 JPJC EOY P1 Q21 7 Answer: B (2 only) From the graph, when temperature increases, the yield of product increases. When temperature increases, p osition of equilibrium shifts to the right to favour endothermic reaction to absorb some heat. The yield of product increase implies that the p osition of equilibrium (poe) shifts to the right. Hence the forward reaction is endothermic (i.e. reject Option 1). When pressure increases, the yield of product decreases. When pressure increases (volume decrease), the yield of product decreases implies that the poe shift s to the left. Hence, the forward reaction proceeds with an increase in no. of gas particles. (reject Option 3) 21.40 10 2 − % pdt at eqm p (T + 10) C T C

Structured questions 8 [PJC MYE 2011/P2/Q2(a)] (a) (i) C(s) + H2O(g)  CO(g) + H2(g) Initial partial pressure / atm − 4 5 10 = 8 0 1 5 10 = 2 Change /atm − −5.6 +5.6 +5.6 Equilibrium partial pressure / atm − 2.4 5.60 7.60 (ii) Kp = 2 2 CO H HO p p p  (Note: Cannot use square bracket [ ] ) 5.6 7.6= 2.4 = 17.7 atm (iii) 5.6 atm of H2O(g) reacted to reach equilibrium. Assuming H2O(g) behaves like an ideal gas (i.e. pV = nRT), Amount of H2O reacted to reach equilibrium = pV RT = ( ) ( ) −    35.6 101325 2 10 8.31 1273 = 0.1069 mol Since mole ratio of C : H2O = 1 : 1, amount of C reacted to reach equilibrium = 0.107 mol (iv) C(s) + H2O(g)  CO(g) + H2(g) ---(1) By Le Chatelier’s Principle, when the pressure of the system is decreased, the position of equilibrium of (1) shifts to the right to increase the total amount/number of gas molecules so as to increase pressure. Hence when the equilibrium is re -established, the composition of CO and H 2 will increase and the composition of C and H2O will decrease. 9 [PJC Prelim 2010/P3/Q1(a)] (a) (i) 2SO2(g) + O2(g)  2SO3(g) ---(2) At low temperature, the position of equilibrium of (2) shifts to the right to favour the exothermic reaction to evolve some heat to increase the temperature. Hence, this gives higher yield of SO3. However, at low temperature, the equilibrium will be reached at a lower rate. Hence, a moderately high temperature of 450 C, together with Fe catalyst, is used to achieve equilibrium more rapidly (or increase the rate).

10 [PJC Prelim 2012/P2/Q4] (a) (i) Assuming ideal gas behaviour (i.e. pV = nRT) average Mr = mRT pV = ( ) ( ) ( ) −   +  6 0.084 8.31 177 273 101325 37.8 10 = 82.0 (no units) (1 d.p.) (ii) Let the mole fraction of CH3COOH in mixture be x and that of (CH3COOH)2 be (1 – x). Mr of CH3COOH = 60.0; Mr of (CH3COOH)2 = 120.0 60.0x + 120.0(1 – x) = 82.0 Mole fraction of CH3COOH in mixture, x = 0.6333 = 0.633 Mole fraction of (CH3COOH)2 in mixture = 1 – 0.6333 = 0.3667 = 0.367 (iii) Using pA = A total n n  ptotal (where A total n n is the mole fraction of A in the mixture), Kp = ( ) ( ) 3 2 3 CH COOH 2 CH COOH p p = ( )   2 0.3667 1 0.6333 1 = 0.914 atm−1 (b) Dimerisation of CH3COOH is exothermic since it involves the formation of two hydrogen bonds between two CH3COOH molecules. 2CH3COOH(g)  (CH3COOH)2(g) ---(3) H < 0 (exothermic) By Le Chatelier’s Principle, when temperature is increased, the position of equilibrium of (3) shifts to the left to favour the backward endothermic reaction so as to absorb some heat to reduce the temperature. Hence, the proportion of CH3COOH in the mixture will increase, causing the average Mr of the mixture to decrease. (ii) 2SO2 (g) + O2(g)  2SO3(g) Initial partial pressure / atm 2 1.53 = 1 1 1.53 = 0.5 0 Change /atm −0.94 (1) −½(0.94)(1) +0.94(1) Eqm partial pressure / atm 0.0600 0.0300 0.940 Kp = ( ) ( ) ( ) 2 2 0.94 0.06 0.03 = 8180 atm−1 (iii) When volume is reduced, the pressure of the system is increased. By Le Chatelier’s Principle, the position of equilibrium of (2 ) shifts right to decrease the total amount/number of gas molecules so as to decrease pressure. Hence, the percentage conversion of SO2 into SO3 increases. C O O H R C O OH R + − − + •• •• − − + + hydrogen bonds H3C CH3  

11 [PJC Prelim 2013/P2/Q4(a)] (a) (i) ( ) ( )ll l  = 32 5 PC C p PC pp pK (Note: Cannot use square bracket [ ] ) (ii) Since PCl3 and Cl2 are formed in 1 : 1 mole ratio, equilibrium partial pressure of PCl3 = equilibrium partial pressure of Cl2 = 2.00 atm (