Consider the following equilibrium: COBr2(g) CO(g) + Br2(g) Kc = 0.190 at 73 °C
ID: 488907 • Letter: C
Question
Consider the following equilibrium:
COBr2(g) CO(g) + Br2(g) Kc = 0.190 at 73 °C
A 2.9 mol sample of COBr2 is transferred to a 2.0 L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
[COBr2] =______ M
[CO] = ________M
[Br2] =________ M
Part II
A 2.9 mol sample of COBr2 is transferred to a 1.0 L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
[COBr2] =______ M
[CO] = _______ M
[Br2] = _______M
Explanation / Answer
1)
initial concentration of COBr2 = number of mol/volume
= 2.9 mol / 2 L
= 1.45 M
COBr2 (g) <--------> CO (g) + Br2 (g)
1.45 0 0 (initial)
1.45 - x x x (at equilibrium)
Kc = [CO] [Br2]/[COBr2]
0.190 = x*x/(1.45-x)
0.276 - 0.190*x = x^2
x^2 + 0.190*x - 0.276 = 0
solving above quadratic equation for positive value of x,
x = 0.44 M
[COBr2] = 1.45 - x = 1.45 - 0.44 = 1.01 M
[CO] = x = 0.44 M
[Br2] = x = 0.44 M
II)
initial concentration of COBr2 = number of mol/volume
= 2.9 mol / 1 L
= 2.90 M
COBr2 (g) <--------> CO (g) + Br2 (g)
2.90 0 0 (initial)
2.90 - x x x (at equilibrium)
Kc = [CO] [Br2]/[COBr2]
0.190 = x*x/(2.90-x)
0.551 - 0.190*x = x^2
x^2 + 0.190*x - 0.551 = 0
solving above quadratic equation for positive value of x,
x = 0.65 M
[COBr2] = 1.45 - x = 1.45 - 0.65 = 0.80 M
[CO] = x = 0.65 M
[Br2] = x = 0.65 M