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?

0.146 M

Explanation:

Equation for the reaction :

2A(aq) ↔ B(aq) + C(aq)

= 65

Molar concentration of A =

= 2.5 M

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

Initial              2.50                      0                0

Change          - 2x                      + x             + x

Equilibrium   2.50 - 2x               +x               +x

8.062(2.5 - 2x) = x

20.155 - 16.124x = x

20.155 = 16.124x+x

20.155 = 17.124x

x =

x = 1.177

[A] = 2.5 - 2x

= 2.5 - 2(1.177)

= 0.146 M

Therefore, the equilibrium concentration of A = 0.146 M

catya that's not even a b or c