[buckeye713]

Hey, I'm having trouble with this problem. I know I have to switch the equations around and add them together but can't get it equal what I need.


Q.

2 H2S(g) + 3 O2 ---> 2 SO2(g) + 2 H2O(g)

calculate standard enthalpy change for the above reaction given;

3 S(s) + 2 H2O(g) ---> 2 H2S(g) + SO2(g)        ^Ho = 146.9 kJ/mol

S(s) + O2(g) ---> SO2(g)                               ^Ho =  -296.4 kJ/mol



So I know I need to change the equations around and add them together to = -1036.1 kJ/mol(the correct answer)


I found this online regarding this same question but I can decipher what exactly he is doing;

"Reverse eqn 1 and add to 3x eqn 2. That gives you the equation you want. Then for dH, change the sign of eqn 1 and add to 3x dH for eqn 2.
By the way, you can't get what you want UNLESS you balance that equation you want.
2H2S + 3O2 ==> 2SO2 + 2H2O"

[Borek]

Reverse eqn 1 and add to 3x eqn 2

Sounds pretty straightforward (and correct).

[buckeye713]

you'd make an excellent teacher! lol....