Consider the following reaction and associated equilibrium constant: aA(g)+bB(g)
ID: 489438 • Letter: C
Question
Consider the following reaction and associated equilibrium constant: aA(g)+bB(g)cC(g), Kc = 3.0
part A :
Find the equilibrium concentrations of A, B, and C for a=1, b=1, and c=2. Assume that the initial concentrations of A and B are each 1.0 M and that no product is present at the beginning of the reaction.
Part B:
Find the equilibrium concentrations of A, B, and C for a=1, b=1, and c=1. Assume that the initial concentrations of A and B are each 1.0 M and that no product is present at the beginning of the reaction.
Explanation / Answer
A)
A(g)+ B(g) <-----> 2C(g)
1.0 1.0 0 (initial)
1.0-x 1.0-x 2x (at equilibrium)
Kc= [C]^2 / {[A][B]}
3.0 = (2x)^2 / (1-x)^2
sqrt(3) = 2x / (1-x)
1.732 = 2x / (1-x)
1.732 - 1.732*x = 2x
x = 0.464 M
[A] = 1.0-x = 1.0 - 0.464 = 0.536 M
[B] = 1.0-x = 1.0 - 0.464 = 0.536 M
[C] = 2x = 2* 0.464 = 0.928 M
B)
A(g)+ B(g) <-----> C(g)
1.0 1.0 0 (initial)
1.0-x 1.0-x x (at equilibrium)
Kc= [C]^2 / {[A][B]}
3.0 = X / (1-x)^2
3(1-x)^2=x
3*(1 + x^2 - 2x) = x
3 + 3*x^2 -6x = x
3*x^2 - 7*x + 3 = 0
solving above quadratic equation for positive x,
x = 1.77 M and x = 0.57 M
x can't be greater than 1 as it will make [A] negative
so, x = 0.57 M
[A] = 1.0-x = 1.0 - 0.57 = 0.43 M
[B] = 1.0-x = 1.0 - 0.57 = 0.43 M
[C] = x = 0.57 M