Carbonyl bromide decomposes to carbon monoxide and bromine. COBr2(g) uv CO(g) +
ID: 303883 • Letter: C
Question
Carbonyl bromide decomposes to carbon monoxide and bromine.
COBr2(g) uv CO(g) + Br2(g)
Kc is 0.190 at 73 °C. Suppose you place 0.500 mol of COBr2 in a 2.00-L flask and heat it to 73 °C (see Study Question 17). After equilibrium has been achieved, you add an additional 2.00 mol of CO.
(a) How is the equilibrium mixture affected by adding more CO?
(b) When equilibrium is reestablished, what are the new equilibrium concentrations of COBr2, CO, and Br2?
(c) How has the addition of CO affected the percentage of COBr2 that decomposed?
Explanation / Answer
a) by the addition of CO, equilibrium shift towards Left. backwards reaction takes place.
b) COBr2(g) <-----> CO(g) + Br2(g)
initial 0.5/2 = 0.25 M - -
change -0.095 +0.095 +0.095
at equil 0.155 0.095 0.095
added - 1 M -
change 0.155+x 1.095 - x - x
re-equil
Kc = [CO][Br2]/[COBr]
0.19 = x*x/(0.25-x)
x = 0.095
after re-establishing
0.19 = (1.095-x)(0.095-x)/(0.155+x)
x = 0.0563
after equilibrium re-established
[CO ] = 1.095-0.0563 = 1.0387 M
[Br2] = 0.095-0.0563 = 0.0387 M
[COBr2] = 0.155+0.0563 = 0.2113 M
c) percentage of COBr2 decreases due to reverse reaction.