Question

Consider the reaction: C(s) + CO₂(g) = 2CO(g) K_p=168 at 1273 K

A). A system containing these ingredients is at equilibrium, and the partial pressure of CO₂(g) is found to be 0.10 atm. What is the pressure of CO(g) under these conditions?

B). What is the value of K_c for the reaction at 1273 K?

Answer 1

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

this applies for pressure

Kp = PC^C * PD^d / (PAâ)(PB^b)

Kp = CO^2 / CO2

a)

P-CO2 = 0.10

find PCO

from

Kp = CO^2 / CO2

substitut edata in equilbirum

168 = (PCO)^2 /(0.10)

(PCO)^2 = (168*0.1) = 16.8

PCO = (16.8)^0.5 = 4.098 atm

B)

Kc can be calcualted via

Kp = Kc*(RT)^dn

dn = mol of gas in product - reactatns = 2-1 = 1

Kp = Kc(R)T

Kc = Kp/(RT) = 168 / (0.082*1273) = 1.609

Kc = 1.609