Answer to Question #138518 in General Chemistry for Allie Privitt

Question

1. Consider the equilibrium reaction:
CO(g) + Cl2(g) ⇌ COCl2(g)
In a particular experiment at equilibrium, a 0.250 L container contained 0.045 mol CO, 0.062 mol Cl2, and
0.108 mol COCl2. What is the value of the equilibrium constant K?

2. A chemist places 1.00 mol of each hydrogen and carbondioxidein a 1.25-L container at 800oC. This
reaction occurs:
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
When equilibrium is reached, 0.49 mol of gaseous carbon monoxide is in the container. Based on this
information, what is the value of the equilibrium constant, K?

3. Carbonyl bromide decomposes to carbon monoxide and bromine:
COBr2 (g) ⇌ CO(g) + Br2 (g) Kc = 0.190 at 73oC
A 1.50 L flask is filled with 0.500 mol of COBr2 and heated to 73oC. What are the equilibrium concentrations
of each species?

Answer 1

1). CO (g) + Cl2 (g) COCl2 (g)

Kc = 

[COCl2]/[CO][Cl2] = 0.14

0.012 x 0.054 = 220

Kp = Kc(RT)∆n

∆n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

3.[COBr2] = 0.000178 M

 [CO] = 0.00582 M

 [Br2] = 0.00582 M

The ICE table approach works well here. First we need to get things in terms of concentration: 

0.0150 mol Br2 = 0.00600 M COBr2

2.5L 

 COBr2 CO + Br2

I 0.00600 0 0 

C -x +x +x 

E 0.00600-x x x 

Inserting into Kc gives: 

Kc = [CO][Br2] = (x)(x)

 [COBr2] 0.00600-x

Now some algebra: 

(0.00600-x)Kc = x2

0 = x2

 + Kcx – 0.00600Kc = (0.190)2

 + 0.190x – 0.00114 

From the quadratic formula, we find x = 0.005822 or -0.196 

Since x represents the equilibrium concentration of CO and Br2, a negative value makes no chemical 

sense, therefore, the value x = 0.005822 is the reasonable result. 

Now to find the equilibrium concentrations of all species: 

[CO] = [Br2] = x = 0.00582 M 

[COBr2] = 0.00600 – x = 0.000178 M