Posted by Janavi on Sunday, March 23, 2014 at 3:10pm.
MO2(s) ==> M(s) + O2(g)
C(s) + O2(g) --> CO2(g)
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MO2(s) + C(s) ==> M(s) + CO2(g)
Calculate k for rxn 1. dGo = -RTlnK1
Look up dGoformation CO2 and calculate k for that reaction as k2.
Then for the sum of rxn 1 + rxn 2 (which gives the final rxn I show above as rxn 3, Krxn 3 is k1k2.
How do I calculae the K for Reaction 2?
Some one please help I really do not understand this question one bit.
The same way you calculated it for rxn 1. dGo = -RTlnK2. You get the dGo from the dG formation for CO2.
I did not calculate it for rxn 1 please please give me a step by step run down I honestly do not know how to do this problem and I have several more like it
I really need a template to work off of to do the others please help me.
It's tough to make it very much simpler but here are the same steps with some extra comments thrown in to help make the transition from bit to bit. But you won't learn by my doing the work for you; therefore, I don't intend to work the problem.
Why didn't you calculate it (k1 that is) for rxn 1. Rxn 1 is
MO2(s) ==> M(s) + O2(g). The problem tells you that dGo is 290.3 kJ/mol. I told you to use dGo = -RTlnK. You know dGo from the problem, R is a constant. I assume you know that is 8.314. T is 298 so you substitute those numbers and Solve for k which I called k1 for rxn 1.
The next part of the problems tells you that although rxn 1 doesn't happen readily, it says that if C(graphite) is added the reaction occurs spontaneously. So I wrote rxn 2 for you. It is
C(s) + O2 ==> CO2(g) and it will have a k2. The problem doesn't tell you what it is but you can calculate it the same way you did for rxn 1 IF you have dGo for it. You don't but you can look it up in tables. Usually these are found in the appendix of your text. So look up the dGo formation for CO2. That will give you dGo for rxn 2 and you use dGo = -RTlnK and calculate k which I've called k2.
Next I added equation 1 (which I called rxn 1) and equation2 (which I called rxn 2) together to give you equation 3 which I called rxn 3. That is the sum of 1 and 2. Notice that O2 cancels. The problem asks for this reaction so I've done that much for you already. To find k for rxn 3 (the sum of 1 and 2) you simply multiply k1*k2 = k3. Then dG = -RTlnk3. Knowing k(k3) allows you to solve for dG for rxn 3 and that is for the rxn at equilibrium.
I can learn by example, this is an example please, work it out so I can do the rest. This question does not even count towards my final grade but the others like it do, so please help me.
Oo the solids have a delta G of zero!!!!
All you have to do
Step 1: Set the solids equal to 0 for G
Step 2: Look up the gases (in this case just CO2)
Step 3: Once you look up thr gasses add the delta g for each equation to each other.
Step 4:Solve for Q using the Delta G= delta G not plus RTLnQ
**Remember Delta G equals 0 at equilibrium****
Step 5: Round to the correct sig figs if you need to.
Thank you Karan!