## 5.55b [ENDORSED]

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### 5.55b

A reaction used in the production of gaseous fuels from coal, which is mainly carbon, is C(s)+H2O(g)⇌CO(g)+H2(g)
(a) Evaluate K At 900 K, given that the standard Gibbs free energies of formation of CO(g) and H2O(g)at 900 K are −191.28kJ⋅mol−and −198.08kJ⋅mol, respectively.
(b) A sample of graphite of mass 5.20 kg and 125 g of water were placed into a 10.0-L container and heated to 900 K. What are the equilibrium concentrations?

For part b, I set up an Ice table, converting the given mass of graphite and water into their molarities and solving using the equilibrium constant I solved for in part a. I keep getting the x=0.67 when it should equal 0.364. Am I approaching this question right?

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

### Re: 5.55b  [ENDORSED]

5.20 x 10^3 g C x 1molC/12.011 g C = 433 mol C
125 g H2O x 1 mol H2O/18.016 g H2o = 6.94 mol H2O

H2O is limiting. Conc. of H2O = 6.94 mol/10 L = 0.694 mol/L

Kc=[CO][H2]/[H2O] = (x)(x)/(0.694-x)=0.403
x^2=0.280-0.403x
x^2+0.403x-0.280=0
x=+0.364 or -0.766