## 5G.21

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

### 5G.21

5G.21 Calculate the equilibrium constant at 25 8C for each of the
following reactions, by using data in Appendix 2A:
(a) the combustion of hydrogen: 2 H2(g) 1 O2(g) Δ 2 H2O(g)
(b) the oxidation of carbon monoxide: 2 CO(g) 1O2(g) Δ
2 CO2(g)
(c) the decomposition of limestone: CaCO3(s) Δ CaO(s) 1
CO2(g)

How would you go about solving this problem?

Nick Fiorentino 1E
Posts: 102
Joined: Wed Sep 18, 2019 12:16 am

### Re: 5G.21

You’d use the equation G=-RTlnK. You want to solve for K so you’d rearrange the equation to be -G/RT=lnK. You have all the values needed besides G which you can get from appendix 2A, and then you solve for K!

Fatemah Yacoub 1F
Posts: 114
Joined: Thu Jul 11, 2019 12:16 am

### Re: 5G.21

Everything given to you in the problem is either a constant or experimentally determined, so when you rearrange the equation to solve for K, you exponentiate the G/-RT, to give you the answer.

Matthew ILG 1L
Posts: 112
Joined: Sat Aug 17, 2019 12:15 am

### Re: 5G.21

I have tried part A several times but I keep coming up with K=1.096. How does the book manage to get 1.0 x 10^80. Does it have something to do with the 2 in front of H20 in the balanced equation?
The only Standard Gibbs free energy of formation given by the appendix is -228.57 for H2O I believe.

Posts: 106
Joined: Sat Aug 24, 2019 12:17 am
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### Re: 5G.21

Matthew ILG 1L wrote:I have tried part A several times but I keep coming up with K=1.096. How does the book manage to get 1.0 x 10^80. Does it have something to do with the 2 in front of H20 in the balanced equation?
The only Standard Gibbs free energy of formation given by the appendix is -228.57 for H2O I believe.

Make sure that you multiply the gibbs free energy of the equatoin by 10^3 to convert from kJ to J