Effect of Temperature on Chemical Reaction Equilibrium

[Tom Hardy]

here is the question (part I):

For part I, I need some assistance, I cannot figure out how to do the question. I know eventually what to do, it's just working out the equilibrium constants I'm having trouble with. So to start, I want to work out the K values at 600K and 800K, I do this by considering the total pressure:

$$
\text{Total Pressure} = 1 \text{ bar}
$$

Initially:

$$
P^{o_2} + P^{HCl} = 1
$$
Since there are 4 mols of hcl per o2 we can write:

$$
5P^{o_2} = 1 \implies P^{O_2} = 0.2 \text{bar}
$$

85% conversion means that in the end, 85% of the oxygen has reacted therefore, there is 0.2*0.15 = 0.03 bar oxygen remaining, therefore 0.12 bar HCl remaining. In addition to this, each mol of oxygen reacted results in 4 new mols being created, therefore there are (0.85*0.2*2) bar of water (and also chlorine) produced, with this information, I can calculate my equilibrium constant, is this correct?

 

[Navin A S]

$$K(600K) = \frac{P^{HCl}}{P^{O_2}} \frac{P^{H_2O}}{P^{Cl_2}} = \frac{0.12}{0.03} \frac{0.34}{0.17} = 10.67$$$$K(800K) = \frac{P^{HCl}}{P^{O_2}} \frac{P^{H_2O}}{P^{Cl_2}} = \frac{0.12}{0.03} \frac{0.53}{0.27} = 15.11$$