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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.

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Presentation on theme: "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."— Presentation transcript:

1 Chemistry Tutorial 5: Q4 Sit Han Yu 12S7F

2 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 – Product: H 2 SO 4

3 Part (a) a) Write a balanced equation, including state symbols, for this reaction. Solution: SO 2 (g) + ½O2 (g)+ H 2 O (l) H 2 SO 4 (aq)

4 Part (b) b) With the aid of energy cycle(s) and the data given below, calculate the enthalpy change of this reaction. – ΔH f [H 2 O(l)] = -286 kJ mol -1 – ΔH f [H 2 SO 4 (aq)] = -811 kJ mol -1 – ΔH f [SO 3 (g)] = -493 kJ mol -1 – ΔH c [SO 2 (g)] = kJ mol -1

5 Before proceeding My understanding of Hess’s Law: Vectors Initial Final F net FAFA FBFB F net = F A + F B

6 SO 2 (g) + ½O2 (g)+ H 2 O (l) H 2 SO 4 (aq) 1/8 S 8 (s) + H 2 (g) + 2O 2 (g) ΔH f [SO 2 (g)] ΔHrΔHr

7 SO 2 (g) + ½O2 (g) SO 3 (g) 1/8 S 8 (s) + H 2 (g) + 2O 2 (g) ΔH f [SO 2 (g)] By Hess’s Law, ΔH f [SO 2 (g)] + (-98.5) +0 = -493 ΔH f [SO 2 (g)] = = kJ mol -1

8 If shown as “vectors” ΔH f [SO 2 (g)] SO 2 (g) + ½O2 (g)SO 3 (g) (final) 1/8 S 8 (s) + H 2 (g) + 2O 2 (g) (initial) ΔH f [SO 2 (g)] + (-98.5) +0 = -493 ΔH f [SO 2 (g)] = = kJ mol -1

9 SO 2 (g) + ½O2 (g)+ H 2 O (l) H 2 SO 4 (aq) 1/8 S 8 (s) + H 2 (g) + 2O 2 (g) ΔHrΔHr By Hess’s Law, ΔH r + (-394.5) (-286) = -811 ΔH r = = -131 kJ mol -1 (0 dp)


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