Question

When H₂(g) is mixed with CO₂(g) at 2,000 K, equilibrium is achieved according to the following equation

CO₂(g) + H₂(g) « H₂O(g) + CO(g)

In one experiment, the following equilibrium concentrations were measured.

                                    [H₂] = 0.20 mol/L

                                 [CO₂] = 0.30 mol/L

                    [H₂O] = [CO] = 0.55 mol/L

1. What is the mole fraction of CO(g) in the equilibrium mixture?
2. Using the equilibrium concentrations given above, calculate the value of K_c, the equilibrium constant for the reaction.
3. Determine K_p in terms of K_c for this system.
4. When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO₂(g). Calculate the value of K_c at this lower temperature.

Answer 1

The given equilibrium reaction is

CO2(g) + H2(g) = H2O(g) + CO(g)

The equilibrium concentrations at 2000 K are

H2 = 0.20 mol/L

CO2 = 0.30 mol/L

H2O = 0.55 mol/L

col = 0.55 mol/L

Hence, the mole fraction of CO in the equilibrium mixture can be calculated as

co Xco = IC [CO] + [H0] + [CO2] + [H2] 0.55 Xco = = 0.34375 0.34 1.6 0.55 mol/L (0.55 +0.55 +0.20 +0.30) mol/L

Hence, the mole fraction of CO in the equilibrium mixture is about 0.34.

For a generic reaction , the equilibrium constant Kc is given as

aA+B=0C + dD

[C] [Dd Kc = [4]" [B]

Where the concentrations are the equilibrium values.

Hence, for our reaction, the equilibrium constant can be calculated as

[H20col 0.55 mol/L x 0.55 mol/L Ke= 25.04 [H2][CO2] 0.20 mol/L x 0.30 mol/L

Hence, the equilibrium constant Kc of the reaction is 5.04 approximately.

The equilibrium constant Kp (in terms of partial pressure) is related to Kc as follows:

K, = K(RT)In

Where

${\Delta n}$ is the difference in number of gaseous product and number of moles of gaseous reactants in the balanced reaction.

In our reaction 2 moles of gas are reacting to give two moles of other gases.

Hence,

An= 2 - 2 = 0

Hence,

K, = K (RT)An = K (RT) = K

Hence, Kp and Kc are equal for our reaction.

When the system is cooled, 30.0% of the CO is converted back to CO2. Hence, we can create the following ICE table to find the Kc at the lower temperature

	$CO_2$	$+H_2$	= H2O
Initial, ,mol/L	0.30	0.20	0.55	0.55
Change, mol/L	+x	+x	-x	-x
Equilibrium, mol/L	0.30+x	0.20+x	0.55-x	0.55-x

Note that since 30.0 % of CO is being converted, the same number of moles of CO must react with equal number of moles of H2O to from same number of moles of CO2 and H2.

Initial concentration of CO = 0.55 mol/L.

Hence, concentration change for CO after 30.0% is converted is

30.0 = x 0.55 mol/L = 0.165 mol/L 100

Hence, 0.165 mol/L of CO and H2O will combine to form 0.165 mol/L of H2 and CO2.

Hence, the equilibrium concentrations at the lower temperature will be

CO2 = 0.30 mol/L+I = 0.30 molL +0.165 mol/L = 0.465 mol/L

H2 = 0.20 mol/L +1 = 0.20 mol/L + 0.165 mol/L = 0.365 mol/L

CO = 0.55 mol/L -I = 0.55 mol L -0.165 mol/L = 0.385 mol/L

H2O = 0.55 mol/L - I = 0.55 molL -0.165 mol/L = 0.385 mol/L

Hence, the new equilibrium constant can be calculated as

[H20Co0.385 mol/L x 0.385 mol/L [H2][CO2] 0.465 mol/L x 0.365 mol/L 0.873

Hence, the equilibrium constant Kc at the lower temperature is 0.873 approximately.