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### 5G.21

Posted: **Wed Feb 19, 2020 7:02 pm**

by **Aman Sankineni 2L**

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?

### Re: 5G.21

Posted: **Wed Feb 19, 2020 7:28 pm**

by **Nick Fiorentino 1E**

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!

### Re: 5G.21

Posted: **Sat Feb 22, 2020 2:12 pm**

by **Fatemah Yacoub 1F**

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.

### Re: 5G.21

Posted: **Fri Feb 28, 2020 8:57 pm**

by **Matthew ILG 1L**

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.

### Re: 5G.21

Posted: **Mon Mar 02, 2020 8:37 pm**

by **Kavya Immadisetty 2B**

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