Question

Determine ΔH for the third reaction from the information given.
S(s) + 3/2 O2(g)

↔ SO3(g) ΔH = -395.2 kJ
2 SO2(g) + O2(g)

↔ 2 SO3(g) ΔH = -198.2 kJ
S(s) + O2(g)

↔ SO2 ΔH = ?

Key Concept: Hess's law. The enthalpy change of an overall reaction is simply the sum of the enthalpy changes for individual steps that give the overall reaction.

Answer 1

$\Delta _fH_{S(s.)} = \Delta _fH_{O_2(g.)} = 0$

For the first reaction $\Delta _rH = \Delta _fH_{SO_3(g.)}\ (*)$

For the second reaction $\Delta _rH = 2\Delta _fH_{SO_3(g.)} - 2\Delta _fH_{SO_2(g.)}\ (**)$

For the third reaction $\Delta _rH = \Delta _fH_{SO_2(g.)}\ (***)$

From $(*)$ it follows that $\Delta _fH_{SO_3(g.)} = -395.2\ kJ$

From t and $(**)$ it follows that

$-198.2\ kJ = 2*(-395.2)\ kJ - 2\Delta _fH_{SO_2(g.)}$

$592.2\ kJ = - 2\Delta _fH_{SO_2(g.)}$

$-296.1\ kJ = \Delta _fH_{SO_2(g.)}$

From this and $(***)$ it follows that

For the third reaction $\Delta _rH = -296.1\ kJ$

Answer: For the third reaction $\Delta _rH = -296.1\ kJ$