Reaction Quotient & Le Châtelier’s Principle

The reaction quotient or Q is the value obtained by entering all of the required concentrations into the equilibrium expression and calculating the result
For a general equation:

aA + bB $⇌$ cC + dD

Q = $\frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$

The equilibrium constant is the numerical value of K_c when the reaction is at equilibrium
If the reaction is not at equilibrium, the numerical value of the equilibrium expression is called the reaction quotient, Q

Statement	Result
If Q does not change with time, the reaction is in a state of equilibrium	Q = K_c
Reaction is in equilibrium	Q = K_c
Reaction move to the right hand side (forward) to reach equilibrium	Q < K_c
Reaction move to the left hand side (backwards) to reach equilibrium	Q > K_c

We can use these values to determine whether or not a reaction is in equilibrium

Determine whether the following reactions are in equilibrium.

If the reaction is not in equilibrium, predict whether it will proceed in the forward or backward direction:

N₂ (g) + 3H₂ (g) $⇌$ 2NH₃ (g) K_c = 9.40
[N₂] = 3.0 x 10^-2 M
[H₂] = 3.7 x 10^-2 M
[NH₃] = 1.6 x 10^-2M
2N₂O (g) $⇌$ 2N₂ (g) + O₂ (g) K_c = 7.0 x 10³⁴
[N₂O] = 1.4 x 10^-18 M
[N₂] = 0.0510 M
[O₂] = 0.0100 M

Answer 1:

$K_{c} = \frac{{[{NH}_{3}]}^{2}}{{[H_{2}]}^{3} [N_{2}]}$
$Q = \frac{}{(3.0 \times 10^{- 2}) {(3.7 \times 10^{- 2})}^{3}} = 10529.15$
The value of Q is greater than that of K_c (Q < K_c)
So, the reaction is not at equilibrium and will move to the left side to reach equilibrium

Answer 2:

$K_{c} = \frac{{[N_{2}]}^{2} [O_{2}]}{{[N_{2} O]}^{2}}$
$Q = \frac{{(0.0510)}^{2} (0.0100)}{(1.4 \times 10^{- 18})} = 1.33 \times 10^{31}$
The value of Q is less than that of K_c (Q < K_c)
So, the reaction is not at equilibrium and will move to the right side to reach equilibrium

Reaction Quotient & Le Châtelier’s Principle (College Board AP Chemistry)