Question

Consider the reaction 2 A(g) ⇄ B(g) + C(g) K = 0.035 A mixture of 8.00 moles of B and 12.00 moles of C in a 20.0 L container is allowed to reach equilibrium. What is the equilibrium concentration of A?

Answer 1

Answer:

52.37 moles.

Explanation:

The value of reaction quotient of a reaction when it has reached to the equilibrium is known as Chemical Equilibrium.

The equation of chemical equilibrium for the reaction mentioned in the question will be as follows -

$K = \frac{[B][C]}{[A]^2}$  Where, the entities in the large bracket denote the concentrations of respective species in the reaction.

Thus following this equation, the equilibrium concentration of reactant A will be -

$A= \sqrt{\frac{[B][C]}{K} }$

$A=\sqrt{\frac{8\times12}{0.035 }$  

$A= 52.37$ moles.