Question

At 25°C, the equilibrium constant Kc for the reaction 2A(aq) ↔ B(aq) + C(aq) is 65. If 2.50 mol of A is added to enough water to prepare 1.00 L of solution, what will the equilibrium concentration of A be?

Answer 1

Answer:

0.146 M

Explanation:

Equation for the reaction :

2A(aq) ↔ B(aq) + C(aq)

$K_c$ = 65

Molar concentration of A = $\frac{2.50 mol}{1.00 L}$

= 2.5 M

                       2A(aq)     ↔        B(aq)     +   C(aq)

Initial              2.50                      0                0

Change          - 2x                      + x             + x

Equilibrium   2.50 - 2x               +x               +x

$K_c =\frac {[B][C]}{[A]^2}$

$65 = \frac{[x][x]}{[2.5-2x]^2}$

$65 = \frac{[x]^2}{[2.5-2x]^2}$

$65 = (\frac{[x]}{[2.5-2x]})^2$

$\sqrt 65 = \sqrt {(\frac{[x]}{[2.5-2x]})^2}$

$8.062 = \frac{x}{2.5-2x}$

8.062(2.5 - 2x) = x

20.155 - 16.124x = x

20.155 = 16.124x+x

20.155 = 17.124x

x = $\frac{20.155}{17.124}$

x = 1.177

[A] = 2.5 - 2x

= 2.5 - 2(1.177)

= 0.146 M

Therefore, the equilibrium concentration of A = 0.146 M