Answer:
a. Kc = [C]³ / [A] × [B]²
b. [A]e = 0.179 M; [B]e = 0.530 M
c. 42.7
Explanation:
There is a missing part:
For the following equilibrium reaction: A(aq) + 2B(aq) <---> 3C(aq)
a. What is the equilibrium constant expression for this reaction.
The equilibrium constant (Kc) is:
Kc = [C]³ / [A] × [B]²
b. Use an I-C-E table to determine the equilibrium concentrations for remaining species, A and B.
First, we will determine the concentrations for which we have information.
[A]i = 1.80 mol/3.50 L = 0.514 M
[B]i = 4.20 mol/3.50 L = 1.20
[C]i = 1.00 mol/3.50 L = 0.286 M
[C]e = 4.50 mol/3.50 L = 1.29 M
The I-C-E table is:
A(aq) + 2B(aq) <---> 3C(aq)
I 0.514 1.20 0.286
C -x -2x +3x
E 0.514-x 1.20-2x 0.286+3x
We know that,
0.286+3x = 1.29
x = 0.335 M
Then,
[A]e = 0.514-x = 0.514-0.335 = 0.179 M
[B]e = 1.20-2x = 1.20-2(0.335) = 0.530 M
c. Calculate the equilibrium constant for the reaction.
We will use the values calculated in b.
Kc = [C]³ / [A] × [B]²
Kc = 1.29³ / 0.179 × 0.530² = 42.7
