Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant Kc.

3A + 2B -><- 4C kc=1.33x10^27

If at this temperature, 2.00 mol of A and 3.80 mol of B are placed in a 1.00L container, what are the concentrations of A, B, C at equilibrium?

Answer 1

Given,

Kc = 1.33 x 10^27

This is a very large value of Kc and indicates that the reaction will favor the forward direction and nearly all the reactants will be converted to product.

Reaction: 3 A + 2 B ---------> 4C

Initial: ....2 ........3.8 ...............0

We will assume that the reaction completely moves in the forward direction.

Change: -3X.....-2X.............+4X

3X = 2

=> X= 2/3 = 0.667 M

Final: 0..........2.467 .........2.667

Now, we move the reaction in backward direction by a quantity 'y'

=> ...3y.........2.467+2y.........2.667-4y (These are the equilibrium conc.)

Kc = (2.667-4y)^4 / (2.467+2y)^2 * (3y)^3

=> 1.33 x 10^27 = (2.667-4y)^4 / (2.467+2y)^2 * (3y)^3

Neglecting 4y and 2y with respect to 2.667 and 2.467 respectively

=> 1.33 x 10^27 = (2.667)^4 / (2.467)^2 * (3y)^3

=> y = 6.14 x 10^-10 M

[A] = 3y = 1.84 x10^-9 M

[B] = 2.467 + 2y = 2.467 M

[C] = 2.667 - 4y = 2.667 M