Question

6A) Consider the following reaction: 2CH2Cl2(g) CH4(g) + CCl4(g) If 0.432 moles of CH2Cl2(g), 0.263 moles of CH4, and 0.590 moles of CCl4 are at equilibrium in a 17.7 L container at 576 K, the value of the equilibrium constant, Kc, is

6B)Consider the following reaction:

COCl2(g) CO(g) + Cl2(g)

If 3.59×10-3 moles of COCl2, 0.265 moles of CO, and 0.211 moles of Cl2 are at equilibrium in a 12.3 L container at 772 K, the value of the equilibrium constant, Kp, is

Answer 1

A)

[CH2Cl2] = number of mol of CH2Cl2 / volume in L

= 0.432 mol / 17.7 L

= 0.0244 M

[CH4] = number of mol of CH4 / volume in L

= 0.263 mol / 17.7 L

= 0.0149 M

[CCl4] = number of mol of CCl4 / volume in L

= 0.590 mol / 17.7 L

= 0.0333 M

Now use:

Kc = [CH4][CCl4] / [CH2Cl2]^2

= (0.0149 * 0.0333) / (0.0244)^2

= 0.833

Answer: 0.833

B)

for COCl2:

Given:

V = 12.3 L

n = 3.59e-3 mol

T = 772.0 K

use:

P * V = n*R*T

P * 12.3 L = 0.0036 mol* 0.08206 atm.L/mol.K * 772 K

P = 0.0185 atm

for CO:

Given:

V = 12.3 L

n = 0.265 mol

T = 772.0 K

use:

P * V = n*R*T

P * 12.3 L = 0.265 mol* 0.08206 atm.L/mol.K * 772 K

P = 1.3649 atm

for Cl2:

Given:

V = 12.3 L

n = 0.211 mol

T = 772.0 K

use:

P * V = n*R*T

P * 12.3 L = 0.211 mol* 0.08206 atm.L/mol.K * 772 K

P = 1.0867 atm

Now use:

Kp = p(Cl2)*p(CO)/p(COCl2)

= (1.0867 * 1.3649)/ (0.0185)

= 80.2

Answer: 80.2