90 likes | 274 Views
Chemistry Tutorial 5: Q4. Sit Han Yu 12S7F. Question. Sulfur dioxide is a major pollutant from sulfuric acid plants. The SO 2 emitted into the atmosphere is oxidised in the air , which then reacts with water to form sulfuric acid , hence causing acid rain. Reactants: SO 2 , O 2 , H 2 O
E N D
Chemistry Tutorial 5: Q4 Sit Han Yu 12S7F
Question • Sulfur dioxide is a major pollutant from sulfuric acid plants. The SO2 emitted into the atmosphere is oxidised in the air, which then reacts with water to form sulfuric acid, hence causing acid rain. • Reactants: SO2, O2, H2O • Product: H2SO4
Part (a) • a) Write a balanced equation, including state symbols, for this reaction. • Solution: SO2(g) + ½O2 (g)+ H2O (l) H2SO4(aq)
Part (b) • b) With the aid of energy cycle(s) and the data given below, calculate the enthalpy change of this reaction. • ΔHf[H2O(l)] = -286 kJ mol-1 • ΔHf [H2SO4 (aq)] = -811 kJ mol-1 • ΔHf [SO3 (g)] = -493 kJ mol-1 • ΔHc [SO2 (g)] = -98.5 kJ mol-1
Before proceeding • My understanding of Hess’s Law: Vectors FB Final Fnet FA Initial Fnet = FA + FB
ΔHr SO2 (g) + ½O2 (g)+ H2O (l) H2SO4 (aq) 0 -286 -811 ΔHf [SO2 (g)] • 1/8S8 (s) + H2 (g) + 2O2 (g)
-98.5 SO2 (g) + ½O2 (g) SO3 (g) 0 -493 ΔHf [SO2 (g)] • 1/8S8 (s) + H2 (g) + 2O2 (g) • By Hess’s Law,ΔHf [SO2 (g)] + (-98.5) +0 = -493 • ΔHf [SO2 (g)] = -493 + 98.5 • = -394.5 kJ mol-1
If shown as “vectors” -98.5 SO2 (g) + ½O2 (g) SO3 (g) (final) ΔHf [SO2 (g)] -493 • 1/8S8 (s) + H2 (g) + 2O2 (g) (initial) • ΔHf [SO2 (g)] + (-98.5) +0 = -493 • ΔHf [SO2 (g)] = -493 + 98.5 • = -394.5 kJ mol-1
ΔHr SO2 (g) + ½O2 (g)+ H2O (l) H2SO4 (aq) 0 -286 -811 -394.5 • 1/8S8 (s) + H2 (g) + 2O2 (g) By Hess’s Law, • ΔHr + (-394.5) + 0 + (-286)= -811 • ΔHr = -811 + 394.5 + 286 = -131 kJ mol-1 (0 dp)