# Prelab Hess problem

1. Apr 12, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

ΔH for the rxn 3MgO(s) + 6H+(aq) --> 3Mg2+(aq) + 3H20(l) = 409.5 kj/mol
What is ΔH for Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq)

3. The attempt at a solution
We just went over this in class I'm confused as ever. I don't know if it is a
ΔH = Ʃreactants - Ʃproducts
Or if I had to somehow manipulate the equations which is what we just went over . I don't really understand so if you point me in the right direction I would appreciate it. I'm confused because all the worksheets where we did the latter choice I mentioned we were given a target equation. Here I don't see one. Explain your thinking please.
Thanks ,
J

2. Apr 13, 2013

### Staff: Mentor

Compare both equations reagent by reagent.

3. Apr 13, 2013

### Jbreezy

Can you elaborate? I don't understand. Thank you.

4. Apr 13, 2013

### Jbreezy

Hi
I looked at it again. I thought how can I get from the first equation to the second
ΔH for the rxn 3MgO(s) + 6H+(aq) --> 3Mg2+(aq) + 3H20(l) = 409.5 kj/mol
What is ΔH for Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq)
If I divide the first equation by 3
I get MgO(s) + 2H+(aq) --> Mg2+(aq) + H20(l) = 136.5 kj/mol
Then I can flip this and I will get
Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq) = -136.5 kj/ mol
Which is the first second equation. So I just transformed the first into the second. I'm not sure if this is even proper I just kind of went with it lol. What does someone think?

5. Apr 13, 2013

### Staff: Mentor

That's exactly what you were expected to do. Note that it is a direct application of the Hess law.

6. Apr 13, 2013

### Jbreezy

Thanks dude we just went over it and it didn't sink in yet. Appreciate your help.