How to calculate enthalpy change of the following reaction? :

S + 2h2o >>> so2 + 2h2

given: s+o2 >>> so2 - enthalpy change = -296.8 kJ

h2 + 1/2o2 >> h20 enthalpy change= -285.8kj

I was trying to solve it myself but i got confused.. i know i have to change the second equation to opposite side, but do i also multiply any of these equations by 2?
Thank you, I appreciate your help!

Reverse equation 2 and multiply it by 2.

To calculate the enthalpy change of the reaction S + 2H2O -> SO2 + 2H2, you need to use the given enthalpy changes of the individual reactions and manipulate them to obtain the desired reaction.

First, reverse the second equation, "H2 + 1/2O2 -> H2O," to get "H2O -> H2 + 1/2O2." When you reverse an equation, the sign of the enthalpy change also changes. So, the new equation becomes:

H2O -> H2 + 1/2O2 with an enthalpy change of +285.8 kJ (opposite sign of -285.8 kJ)

Next, multiply the first equation, "S + O2 -> SO2," by 2 so that the number of sulfurs (S) on both sides of the equation match the desired reaction:

2S + 2O2 -> 2SO2 with an enthalpy change of -593.6 kJ (2 times -296.8 kJ)

Finally, add the two manipulated equations together to obtain the desired reaction:

2S + 2O2 + H2O -> 2SO2 + H2 + 1/2O2

The corresponding enthalpy changes are added together since these are both part of the desired reaction. So the enthalpy change of the desired reaction is:

-593.6 kJ + (+285.8 kJ) = -307.8 kJ

Therefore, the enthalpy change of the reaction S + 2H2O -> SO2 + 2H2 is -307.8 kJ.

To calculate the enthalpy change of the given reaction, we can use Hess's Law, which states that the enthalpy change of a chemical reaction is independent of the pathway between the initial and final states.

To apply Hess's Law, we need to manipulate the given equations and then add them together to cancel out the intermediate species. Here's how to do it step by step:

1. Reverse the reaction and the enthalpy change for equation 2:
h2 + 1/2o2 → h2o ⟶ ΔH = +285.8 kJ

2. Multiply equation 1 by 2 to balance the number of H2O molecules:
2(S) + 4(H2O) → 2(SO2) + 4(H2)

3. Multiply equation 2 by 2 to balance the number of H2O molecules:
2(h2) + o2 → 2(h2o) ⟶ ΔH = -571.6 kJ

4. Add the modified equations together:
2(S) + 4(H2O) + 2(h2) + o2 → 2(SO2) + 4(H2O) + 2(H2)

5. Cancel out the common species:
2(S) + o2 → 2(SO2) + 2(H2)

6. Add the enthalpy changes together:
-296.8 kJ + (-571.6 kJ) = -868.4 kJ

Therefore, the enthalpy change of the given reaction is -868.4 kJ.

It's important to note that when manipulating the equations, make sure to multiply both the reactions and the enthalpy changes by the same factor as needed to balance the equation and maintain proportionality.