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