2019 TMJC H2 Chem Prelim P3 ANS

1 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Section A Answer all the questions from this section. 1 Hydrocarbons, CxHy, are used extensively as sources of fuel in our modern day civilisation. (a) 10 cm3 of a gaseous hydrocarbon C5Hy was allowed to burn in an excess of oxygen at 300 ºC and 1 atm. In the reaction, there was an expansion of volume by 20 cm3. (i) Write a balanced equation, with state symbols, for the reaction at 300 ºC and 1 atm. [2] C5Hy (g) + 5 4   y O2 (g)   5 CO2 (g) + 2 y H2O (g) (ii) Determine the value of y. [2] Using Avogadro’s Law, mole ratio  volume ratio Equation: C5Hy (g) + 5 4   y O2 (g)   5 CO2 (g) + 2 y H2O (g) Vol of gases: 10 cm3 ( 5 4   y x 10) cm3 50 cm3 ( 2 y x 10) cm3 Volume of gases used up in the reaction = 10 + ( 1050 4 y ) = 1060 4   y cm3 Volume of gases produced in the reaction = 50 + 2   y x 10 =  50 5 y cm3 Volume of gas produced – Volume of gas used up = 10 cm3  50 5 y – 1060 4   y = 20 y = 12 TAMPINES MERIDIAN JUNIOR COLLEGE 2019 JC2 H2 Chemistry Prelim Exam Paper 3 (Suggested Answers)

2 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Methane, CH4, is sometimes used in the production of hydrogen via a process known as steam reforming. CH4(g) + H2O(g) CO(g) + 3H2(g) (b) The steam reforming process is an endothermic reaction. (i) Using data from the Data Booklet , calculate a value for the enthalpy change of the forward reaction. [2] Hr = BE (reactants) – BE (products) = 410 x 4 + 460 x 2 – (1077 + 436 x 3) = +175 kJ mol –1 (ii) The actual value of the enthalpy change of reaction is found to be +206 kJ mol–1. Suggest a reason why your calculated answer in (i) differs from this value. [1] The bond energy calculation is an approximation method as the bond energy values given in the Data Booklet are average values . (c) At 600 K, t he value of the equilibrium constant, Kp, for the steam reforming reaction is 7.20 x 10–4. (i) Write the Kp expression for this reaction, giving its units. [2] Kp = )P)(P( )P)(P( OHCH 3 HCO 24 2 atm2 (accept Pa2) (ii) Gaseous CH4, H2O and CO are introduced into a closed container at 600 K and their initial partial pressures are 1.20 atm, 2.10 atm and 1.80 atm respectively. Determine the partial pressure of H2 when equilibrium is reached. (You may assume that the extent of the forward reaction is small.) [3] CH4(g) + H2O(g) CO(g) + 3H2(g) Initial pp/atm 1.20 2.10 1.80 0 Change pp/atm –x –x +x +3x Eqm pp/atm 1.20 – x 2.10 – x 1.80 + x 3x Kp = )P)(P( )P)(P( OHCH 3 HCO 24 2 7.20 x 10–4 = 3(1.80 )(3 ) (1.20 )(2.1