RVHS_H2_CHEM_P2_ANS_RV Prelim

Answer key to Paper 2 1 (a) RvCO3 + 2HCl → RvCl2 + CO2 + H2O [1] (b) [2] (c) Preparation of saturated RvCO3 solution 1. Using a 100 cm 3 measuring cylinder , place about 100 cm 3 of deionised water (accept any volume between 80 – 250 cm 3) into a polystyrene cup. 2. Measure and record the initial temperature of the water using a thermometer. 3. Using a spatula, add solid RvCO 3 into the polystyrene cup and stir/swirl gently with the thermometer / stirrer. Continue to add more solid and stir, until some solid RvCO3 remains undissolved to ensure a saturated solution is obtained. 4. Record the lowest temperature reached on the thermometer. 5. Filter the saturated solution using a dry filter funnel and dry filter paper into a dry 250 cm 3 conical flask / beaker to ensure the saturated solution is not diluted. Justification of concentration of standard HCl solution For a 25.0 cm3 aliquot of saturated RvCO 3(aq), a reasonable HC l titre value should be 25.00cm3 (accept any value between 20−30 cm3). RvCO3 + 2HCl → RvCl2 + CO2 + H2O RvCO3 ≡ 2HCl Given that solubility of RvCO3 is approximately 0.25 mol dm−3, (250 cm3) beaker polystyrene cup thermometer lid (100 cm3) H2O + RvCO3

2 [HCl] required = = 0.500 mol dm−3 Titration of saturated RvCO3 solution against HCl 1. Pipette 25.0 cm 3 of saturated RvCO 3 solution into a 250 cm 3 conical flask. 2. Add 2-3 drops of methyl orange indicator. 3. Titrate with the standard 0.500 mol dm −3 HCl from the burette until the colour of the solution changes from yellow to orange. 4. Repeat the titration to obtain two consistent results within 0.10 cm 3 in difference. [6] (d) (i) nHCl used = = 1.00 x 10−3 WC mol RvCO3 ≡ 2HCl nRvCO3 in 25.0 cm3 of saturated solution = 1.00 x 10−3 WC = 5.00 x 10−4 WC mol solubility of RvCO3 = 5.00 x 10−4 WC = 0.0200 WC mol dm−3 (ii) Heat taken in by V cm3 solution = VCΔT = 4.18 V ΔT J nRvCO3 dissolved in V cm3 solution = 0.0200 WC = 2.00 x 10−5 WCV mol ΔHsoln = J mol−1 [3] [Total: 12]

3 2 (a) (i) initial nBr2 = = 6.70 x 10−4 mol At 900 K, Br2(g) → 2Br(g) initial nBr = 2 x 6.70 x 10−4 = 1.34 x 10−3 mol (ii) W(s) + 4Br(g) ⇌ WBr4(g) initial amount / mol − 1.34 x 10−3 0 change in amount / mol − −4y +y equilibrium amount / mol − 1.34 x 10−3 −4y y mass of gas at equilibrium = (1.34 x 10−3 −4y)(79.9) + (y)(184 + 4(79.9)) = 0.107 + 184y