JPJC 2026 Chemical Equilibrium Tutorial Answer

2026 JPJC JC1 H2 Chemistry (9476) 1 Topic 7: Chemical Equilibrium JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) Tutorial 7: Chemical Equilibrium (Suggested Answers) Level 1 – Self–Check Questions 1 Write expressions for the equilibrium constants, Kc and Kp (if applicable), for each of the following reactions. Assume th at all pressure terms are in Pa, gi ving the units of the equilibrium constants. (a) H2(g) + I2(g) ⇌ 2HI(g) Kc I I= 2 22 [H ] [H ][ ] ; no units Kp 22 IH 2 HI PP P ; no units (b) Ni(s) + 4CO(g) ⇌ Ni(CO)4(g) Kc = 44[Ni(CO) ][CO] ; mol−3 dm9 Kp = 4Ni(CO) 4 CO P (P ) ; Pa−3 (c) CO2(aq) + H2O(l) + CaCO3(s) ⇌ Ca2+(aq) + 2HCO3−(aq) 2+ 2 3 c 2 [Ca ][HCO ]= [CO ]K − ; mol2 dm−6 Kp : NA 2 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. (a) 4X2(g) + 2Y2(g) ⇌ 4X2Y(g) (b) X2Y(g) ⇌ X2(g) + ½Y2(g) Given 2 2 c 2 22 [X Y]= [X ] [Y ]K = 5.00 mol−1 dm3 ' 4 2 c 42 22 [X Y]= [X ] [Y ]K =    22 2 2 22 [X Y] [X ] [Y ] = (Kc)2 = (5.00)2 = 25.0 mol−2 dm6 '' 1 2 22 c 2 [X ][Y ]= [X Y]K =    1 22 22 2 2 [X ] [Y ] [X Y] =    c 1 2 1 K =   1 2 1 5.00 = 0.447 31 22mol dm− 3 (a) The value of the equilibrium constant is 4.0 for the reaction below: CH3CO2H(l) + C2H5OH(l) ⇌ CH3CO2C2H5(l) + H2O(l) (i) Write an expression for the equilibrium constant, Kc, of the reverse reaction, i.e. the hydrolysis of ethyl ethanoate, stating its numerical value. Kc = 3 2 2 5 3 2 2 5 2 [CH CO H][C H OH] [CH CO C H ][H 0] = 4 1 = 0.25

2026 JPJC JC1 H2 Chemistry (9476) 2 Topic 7: Chemical Equilibrium (ii) In an experiment, 2 mol of ethyl ethanoate and 2 mol of water are mixed. Calculate the number of moles of each substance present when equilibrium is reached. CH3CO2H + C2H5OH ⇌ CH3CO2C2H5 + H2O Initial amount/mol 0 0 2 2 Change in amount/ mol +x +x -x -x Equilibrium amount/ mol x x 2-x 2-x Eqm conc/mol dm−3. x/v x/v (2-x)/v (2-x)/v Kc = =− 2 2 (x / v) ((2 x) / v) 0.25  x = 0.667 (✓) Therefore, amount of acid = amount of alcohol = 0.667 mol Amount of water = amount of ester = 1.33 mol (✓) 4 9701 J2014 P23 Q3(b) modified Nitrogen dioxide exists in equilibrium with dinitrogen tetroxide, N2O4. 2NO2(g) ⇌ N2O4(g) ∆H < 0 2.00 mol of dinitrogen tetroxide was sealed in a container at 350 K. After equilibrium has been established, the total pressure was 140 kPa and the mixture of gases contained 1.84 mol of dinitrogen tetroxide. (a) Sketch a graph showing how the rates of the forward and backward reactions change, from the time dinitrogen tetroxide starts to dissociate into nitrogen dioxide to the time the reaction reaches equilibrium. Label your two lines clearly. (b) (i) Give the expression for the equilibrium constant, Kp. 24 2 NO p 2 NO = pK p (ii) Calculate the number of moles of NO2 present at equilibrium. 2NO2(g) ⇌ N2O4(g) Initial amount/ mol 0 2.00 Change +2(0.16) = 0.32 2.00-1.84 = 0.16 Eqm amount/ mol 0.320 1.84 Time Rate backward reaction forward reaction 0

2026 JPJC JC1 H2 Chemistry (9476) 3 Topic 7: Chemical Equilibrium (iii) Calculate the total number of moles of gases which are present at equilibrium and hence the mole fraction of each gas present at equilibrium. Total amount of gases at equilibrium = 0.32 + 1.84 = 2.16 mol Mole fraction of NO2 = 0.32 2.16 = 0.148 Mole fraction of N2O4 = 1.84 2.16 = 0.852 (iv) Calculate the partial pressure of each gas at equilibrium. 2NOp = 0.148 × 140 = 20.7 kPa 24NOp = 0.852 × 140 = 119 kPa (v) Calculate the value of the equilibrium constant, Kp, at 350 K, including its units. p 2 119= (20.7)K = 0.277 kPa−1 Level 2 – Application Questions Kc and Kp calculations 5 The equilibrium mixture in a 3 dm3 vessel at 673 K and 1.5 atm contains 0.50 mol of sulfur dioxide, 0.30 mol of oxygen and 4.80 mol of sulfur trioxide. 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate the equilibrium constants, Kc and Kp. 2SO2(g) + O2(g) ⇌ 2SO3(g) Eqm amount/ mol 0.50 0.30 4.80 Eqm conc/ mol dm−3 0.50/3 = 0.167 0.30/3 = 0.100 4.80/3 = 1.60 Kc = (0.100)(0.167) 1.60 2 2 = 918 mol−1 dm3 1.55.600.500P 2SO = = 0.134 atm; 1.55.600.30P 2O = = 0.0804 atm; 3SO 4.80P = ×1.55.60 = 1.29 atm Kp = ( ) ).().( . 080401340 291 2 2 = 1150 atm−1

2026 JPJC JC1 H2 Chemistry (9476) 4 Topic 7: Chemical Equilibrium 6 The Contact process for the manufacture of sulfuric acid is as follows. 2SO2 (g) + O2(g) ⇌ 2SO3 (g) ∆H = -197 kJ mol-1 When a 2:1 mole ratio of SO2 and O2 at a total initial pressure of 3 atm is passed over a catalyst at 430 oC, the partial pressure of SO3 at equilibrium is found to be 1.9 atm. (a) Calculate the partial pressures of SO2 and O2 at equilibrium and hence the new total pressure and the percentage conversion of SO2 into SO3. New total pressure at eqm = 0.1 + 0.05 + 1.9 = 2.05 atm % conversion of SO2 to SO3 = 1.9 100 %2 = 95.0 % 2SO2(g) + O2(g) ⇌ 2SO3(g) Initial amount/mol : 2x x 0 Initial Partial Pressure/atm : 2 1 0 Change/atm : -1.9 -0.95 +1.9 Eqm Partial Pressure /atm : 0.100 0.0500 1.9 (b) Write an expression for Kp for this reaction and calculate its value. 2 2 3 22 SO p SO O = PK PP ( ) ( ) 2 2 1.9= (0.1) 0.05 = 7220 atm – 1 Kp calculations and Le Chatelier’s principle [N2022/P3/Q2(c) modified] 7 The equation for the Haber process for manufacturing ammonia is shown. 3H2(g) + N2(g) ⇌ 2NH3(g) H = -92 kJ mol-1210 kJ mol-1 A mixture of H2 and N2 in the molar ratio 3:1 is added to a sealed vessel and heated to 450 o C with an iron catalyst. At equilibrium, 40% of the nitrogen has reacted. The total pressure in the vessel is 2.80 x 104 kPa at equilibrium. (a) Write an expression for the equilibrium constant, Kp for this reaction Use your expression to calculate the value of Kp for the reaction. Include its units ( ) ( ) ( ) K = 3 22 2 NH p 3 HN P PP 3H2(g) + N2(g)  2NH3(g) Initial amount/mol 3x x − Change/mol −3(0.4) x −0.4 x +2(0.4) x Equilibrium amount/mol 1.8x 0.6 x 0.8 x Total amount at equilibrium = 1.8x + 0.6x + 0.8x = 3.2x [1] Commented [KYSJ1]: Inserted recent A level qn

2026 JPJC JC1 H2 Chemistry (9476) 5 Topic 7: Chemical Equilibrium 2HP = 41.8x ×2.80×103.2x = 1.575  104 kPa 2NP = 40.6x ×2.80×103.2x = 5.25  103 kPa 3NHP = 40.8x ×2.80×103.2x = 7.00  103 kPa ( ) ( ) ( ) 23 p 343 7.00 ×10 1.575 ×10 5.25 ×10 =K = 2.39  10−9 kPa−2 (3sf) (b) Predict, with reasoning, how the position of equilibrium and the equilibrium constant might change, if any, for the above reaction, when (i) temperature is decreased, 3H2(g) + N2(g) ⇌ 2NH3(g) Eqm (1) When temperature is decreased, by LCP, position of equilibrium in (1) will shift to the right to favour the exothermic reaction so as to produce some heat to increase temperature. Equilibrium constant increases. Important note: Kc and Kp for a reversible reaction ➢ Depends only on TEMPERATURE! ➢ When there is NO change in temperature , the value of Kc and Kp remain constant. (ii) ammonia is added. When ammonia is added, [NH3] increases, cause position of equilibrium in (1) to shift to the left to react away SOME of the ammonia that was added to decrease [NH3]. Since there is no change in temperature , equilibrium constant will remain unchange