Download presentation

Presentation is loading. Please wait.

Published byAja Seaman Modified about 1 year ago

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)] = -98.5 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)] -286-811 0 ΔHrΔHr

7
SO 2 (g) + ½O2 (g) SO 3 (g) 1/8 S 8 (s) + H 2 (g) + 2O 2 (g) -98.5 ΔH f [SO 2 (g)] 0-493 By Hess’s Law, ΔH f [SO 2 (g)] + (-98.5) +0 = -493 ΔH f [SO 2 (g)] = -493 + 98.5 = -394.5 kJ mol -1

8
If shown as “vectors” ΔH f [SO 2 (g)] -98.5 -493 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)] = -493 + 98.5 = -394.5 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) -394.5 -286-811 0 ΔHrΔHr By Hess’s Law, ΔH r + (-394.5) + 0 + (-286) = -811 ΔH r = -811 + 394.5 + 286 = -131 kJ mol -1 (0 dp)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google