Question

Determine the enthalpy of the following reaction given the three reaction equations below:
2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g)
H2O(g) → H2(g) + O2(g) H = 241.8 kJ
H2S(g) → H2(g) + S(s) H = 20.6 kJ
S(s) + O2(g) → SO2(g) H = –296.8 kJ

Answer 1

2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g)

eqn 1---H2O(g) → H2(g) + O2(g) H = 241.8 kJ
eqn 2---H2S(g) → H2(g) + S(s) H = 20.6 kJ
eqn 3---S(s) + O2(g) → SO2(g) H = –296.8 kJ
Note the equation 1 is not balanced. It should be
..........H2O(g) ==> H2(g) + 1/2 O2(g) dH = -241.8 kJ

step1--multiply eqn 1 by 2, reverse it, change sign of original dH x 2.
step 2--multiply corrected eqn 2 by 2 and keep sign of initial dH x 2.
step 3--reverse eqn 3 and change sign dH.
Add steps 1, 2, 3 and CHECK THE RESULTING EQUATION TO MAKE SURE IT AGREES WITH THE QUESTION EQUATION. Add the new dH values. Watch the signs. The dH will be the dH for the new equation. Post your work if you get stuck.