Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: achibaby1974 on April 05, 2008, 06:56:51 PM
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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.
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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?