Question

# Consider these equations: 2S (s) + 3O2 (g)→2SO3 (g), ΔH = −792 kJ 2S (s) + 2O2 (g)→2SO2 (g), ΔH = −594 kJ 2SO2 (g) + O2 (g)→2SO3 (g), ΔH =? What is the missing ΔH? a. −294 kJ b. −198 kJ c. +198 kJ d. +294 kJ

Answer

-990 kJ

- To find the missing ΔH value, we need to use Hess's Law.
- Hess's Law states that the ΔH for a reaction is the sum of the ΔH values for any set of reactions that add up to the original reaction.
- We can use the first two equations to find the missing ΔH for the third equation.
- First, we need to reverse the second equation.
- 2SO2 (g)→2S (s) + 2O2 (g), ΔH = 594 kJ
- Next, we need to multiply the first equation by 2 and the second equation by 1.
- 4S (s) + 6O2 (g)→4SO3 (g), ΔH = −1584 kJ
- 2S (s) + 2O2 (g)→2SO2 (g), ΔH = −594 kJ
- Now, we can add the two equations together to get the third equation.
- 2SO2 (g) + O2 (g)→2SO3 (g), ΔH = -990 kJ