[achibaby1974]

Here's the second problem.

Use Hess’s law to determine the heat of the reaction for (at 1 atm and 25 C):

C(diamond) ----- C(graphite)

With the following equations:

1.)   C(diamond) + O2 (g) ------  CO2 (g)          Delta H = -395.4  kJ
2.)   2 CO2 (g) ------- 2 CO (g) + O2 (g)            Delta H = 566.0  kJ
3.)   C(graphite) + O2 (g) ------- CO2 (g)           Delta H = -393.5  kJ
4.)   2CO (g) -------- C(graphite) + CO2 (g)       Delta H = -172.5  kJ

The boxes are arrows!
 
I don’t really even know where to start. But for C(diamond) we can just keep Equation 1 as is. When we move on to C(graphite) however, there’s 2 of them. What do I do now? Every time I try to cancel something or get an equation in order it doesn’t work.

[macman104]

Hey, so here's the way I approached this problem.  I started by writing out the reactions.  Then, I knew that the C(graphite), in the 3rd reaction had to be on the right hand side, so I reversed that equation.  Remember that reversing an equation, we reverse delta H.

At this point we're at the following:

1) C(diamond) + O2 (g) ------> CO2 (g)    Delta H = -395.4
2) 2 CO2 (g) --------> 2 CO (g) + O2 (g)   Delta H = 566.0
3) CO2 (g) -------> C(graphite) + O2 (g)   Delta H = 393.5
4) 2CO (g) ------> C(graphite) + CO2 (g)   Delta H = -172.5

So, now let's count up the molecules on each of the reaction:
Starting Materials:
1 - C(diamond)
1 - O2
3 - CO2
2 - CO
0 - C(graphite)

Products:
0 - C(diamond)
2 - O2
2 - CO2
2 - CO
2 - C(graphite)

We know that once everything cancels, we should be left with equal moles of C(diamond) and C(graphite).  If we know that, then we see we need to gain:  1 C(diamond) and 1 O2 on the reactants side and 1 CO2 on the products side. 

Can you go from there to figure it out?