Question

in problems 14 - 15, calculate δh for the reaction: so₂(g) + h₂(g) → h₂s(g) + o₂(g) 14. using the thermochemical equations 2 h₂s(g) + 3 o₂(g) → 2 so₂(g) + 2 h₂o(g) δh = - 1036 kj 2 h₂(g) + o₂(g) → 2 h₂o(g) δh = - 483.6 kj

Explanation:

Step1: Reverse the first equation

Reverse \(2H_2(g)+O_2(g)\to2H_2O(g)\) (\(\Delta H = - 483.6\ kJ\)) to get \(2H_2O(g)\to2H_2(g)+O_2(g)\) (\(\Delta H= + 483.6\ kJ\))

Step2: Use Hess's Law

We want \(SO_2(g)+H_2(g)\to H_2S(g)+O_2(g)\)
The second given equation is \(2H_2S(g)+3O_2(g)\to2SO_2(g)+2H_2O(g)\) (\(\Delta H=-1036\ kJ\))
Multiply the reversed first - equation by \(1\) and the second equation by \(\frac{1}{2}\)
For the reversed first equation: \(2H_2O(g)\to2H_2(g)+O_2(g)\), \(\Delta H_1 = + 483.6\ kJ\)
For the second equation: \(H_2S(g)+\frac{3}{2}O_2(g)\to SO_2(g)+H_2O(g)\), \(\Delta H_2=\frac{- 1036}{2}=-518\ kJ\)
Add them together:
\(2H_2O(g)+H_2S(g)+\frac{3}{2}O_2(g)\to2H_2(g)+O_2(g)+SO_2(g)+H_2O(g)\)
Simplify to \(H_2O(g)+H_2S(g)+\frac{1}{2}O_2(g)\to2H_2(g)+SO_2(g)\)
If we rewrite it as \(SO_2(g)+H_2(g)\to H_2S(g)+O_2(g)\)
\(\Delta H=\Delta H_1+\Delta H_2\)
\(\Delta H = 483.6-518\)

Answer:

\(\Delta H=+34.4\ kJ\)