Question

Given: H2(g) + CO2(g) <-> H2O(g) + CO(g) [H2]=0.061M [CO2]=0.012M [H2O]=3.1M [CO]=8.4M Calc.Kc. The reaction is performed at a new temperature. An initial sample of 0.23 M H2 and 0.41 M CO2 gives 0.18M H2O and 0.18 M CO at equilibrium. Determine the composition of the equilibrium mixture. Calc. Kc.

Of note, I got Kc = 3.6 e 4 for the first part and needed to know if I got the right answer, and I need to see how to do the second part, aka "The reaction is performed at a new temp"...do I need the Kc from the first part to answer the second part? Thank you.

Answer 1

For, H₂(g) + CO₂(g) H₂O(g) + CO(g)

K_c = [H₂O][CO]/[H₂][CO₂]

= (3.1)(8.4)/(0.061)(0.012)

= 3.6 x 10⁴

Yes.You got the right answer!

H₂(g) + CO₂(g) H₂O(g) + CO(g)

Initial concentration (M) 0.23 0.41 0 0

Change in concentration (M) - 0.18 - 0.18 0.18 0.18

Equilibrium concentration (M) 0.050 0.23 0.18 0.18

Thus, at equilibrium, [H₂] = 0.050 M, [CO₂] = 0.23 M, [H₂O] = 0.18 M and [CO] = 0.18 M

Now,

K_c = [H₂O][CO]/[H₂][CO₂]

= (0.18)(0.18)/(0.050)(0.23)

= 2.8

Therefore, the K_c at new temperature = 2.8