Question

1. Calculate the enthalpy change for the following chemical reaction:

CS₂(1) + 3O₂(g) ---> CO₂(g) + 2SO₂(g)

1.C(s) + O₂(g) ---> CO₂(g) ∆H° = - 393.5 kJ

2.S(s) + O₂(g) ---> SO₂(g) ∆H° = - 296.8 kJ

4.C(s) + 2S(s) ---> CS₂(1) ∆H° = + 87.9 kJ

Answer 1

The enthalpy change of the given reaction can be deduced from the enthalpies given for reactions 1, 2 and 4. By combining these reactions together, the given one might be deduced:

reaction 1 + 2*reaction 2 +reverse reaction 4

The enthalpy of the given reaction is:

$\Delta H = \Delta H^o_1+2 \times \Delta H^o_2 - \Delta H^o_4$

$\Delta H = -393.5 + 2\times (-296.8) - 87.9 = -1075.0 kJ$