Question

Calculate the enthalpy of the following chemical reaction: [6]
CS 2 (ℓ) + 3O 2 (g) → CO 2 (g) + 2SO 2 (g)
Given:
C(s) + O 2 (g) → CO 2 (g) ΔH = -393.5 kJ/mol
S(s) + O 2 (g) → SO 2 (g) ΔH = -296.8 kJ/mol
C(s) + 2S(s) → CS 2 (ℓ) ΔH = +87.9 kJ/mol

Answer 1

The enthalpy of the reaction can be derived by;

1st reaction enthalpy + (2 × 2nd reaction enthalpy) - 3rd reaction enthalpy

$∆H_1 + 2(∆H_2) - ∆H_3 = ∆H$

-393.5 + 2(-296.8) - 87.9 = ∆H

∆H = -1075 kJ/mol

Therefore, the enthalpy of the chemical reaction is -1075kJ/mol.