Sulfur burns in oxygen in two steps:
i.S(s) + O2(g) ->SO2(g) change of Heat=-297kJ
ii. 2SO2(g) + O2(g)->2SO3(g) change of heat=-198kJ
Calculate the total heat released if this is the final reaction: 2S(s) + 3O2(g)->2SO3(g)
Hess' law. Multiply equation i by 2 (multiply H by 2 also) and add to equation ii. That will give you the equation you want so delta H for the final reaction is the sum of 2*i + ii = final rxn.
To calculate the total heat released for the final reaction, you need to determine the heat released in each step and then add them together.
First, let's calculate the heat released in the first step (i), where sulfur reacts with oxygen to form sulfur dioxide:
S(s) + O2(g) -> SO2(g) ΔH = -297 kJ
We can see that the coefficients in this equation represent a 1:1 ratio, meaning that for every 1 mole of sulfur, 1 mole of sulfur dioxide is produced. Therefore, we don't need to do any additional calculations, and the heat released in this step is -297 kJ.
Next, let's calculate the heat released in the second step (ii), where sulfur dioxide reacts with oxygen to form sulfur trioxide:
2SO2(g) + O2(g) -> 2SO3(g) ΔH = -198 kJ
The coefficients in this equation represent a 2:1 ratio, meaning that for every 2 moles of sulfur dioxide, 1 mole of sulfur trioxide is produced. To calculate the heat released, we need to account for this ratio:
Since the given value is for 2 moles of sulfur dioxide, we need to adjust the heat released accordingly. We can do this by dividing the ΔH value for this step by 2:
Adjusted ΔH = (-198 kJ) / 2 = -99 kJ
Now we have the heat released for the second step as -99 kJ.
Finally, to find the total heat released for the overall reaction:
2S(s) + 3O2(g) -> 2SO3(g)
We need to add up the heat released in both steps:
-297 kJ + (-99 kJ) = -396 kJ
Therefore, the total heat released for the final reaction is -396 kJ.