Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: buckeye713 on April 02, 2016, 03:41:26 PM
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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"
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Reverse eqn 1 and add to 3x eqn 2
Sounds pretty straightforward (and correct).
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you'd make an excellent teacher! lol....