Learn to use the reaction quotient, Q, to determine if a reaction is at equilibrium. This stack focuses on comparing Q to K to predict the direction in which a reaction must shift to reach its equilibrium state.
The reaction quotient, \(Q\), is a value calculated using the same formula as the equilibrium constant, \(K\), but with the concentrations or partial pressures of reactants and products at any given moment, not necessarily at equilibrium.
How does the mathematical expression for the reaction quotient (\(Q_c\)) relate to the expression for the equilibrium constant (\(K_c\))?
The expressions are identical. The only difference is that \(K_c\) uses concentrations measured *at equilibrium*, while \(Q_c\) uses concentrations measured at *any point* in the reaction.
What is the state of a reaction when its reaction quotient equals the equilibrium constant (\(Q = K\))?
The reaction is at equilibrium. There is no net shift in either the forward or reverse direction because the rates of the two reactions are equal.
If \(Q < K\), in which direction will the reaction proceed to reach equilibrium?
The reaction will shift to the right, towards the products. This consumes reactants and forms more products, increasing the value of \(Q\) until it equals \(K\).
Provide the reasoning: Why does the reaction shift right when \(Q < K\)?
When \(Q < K\), the ratio of products to reactants is too small compared to the equilibrium ratio. The forward reaction must proceed at a faster rate than the reverse reaction to increase this ratio.
If \(Q > K\), in which direction will the reaction proceed to reach equilibrium?
The reaction will shift to the left, towards the reactants. This consumes products and forms more reactants, decreasing the value of \(Q\) until it equals \(K\).
Provide the reasoning: Why does the reaction shift left when \(Q > K\)?
When \(Q > K\), the ratio of products to reactants is too large compared to the equilibrium ratio. The reverse reaction must proceed at a faster rate than the forward reaction to decrease this ratio.
For the reaction $$ 2NO_2(g) \rightleftharpoons N_2O_4(g) $$ at \(298K\), \(K_c = 170\). If a vessel contains `[NO_2] = 0.10 M` and `[N_2O_4] = 0.10 M`, calculate \(Q_c\).
Given your calculation that \(Q_c = 10\) while \(K_c = 170\) for the $$ 2NO_2 \rightleftharpoons N_2O_4 $$ system, which way will the reaction shift?
Since \(Q_c < K_c\) (\(10 < 170\)), the ratio of products to reactants is too small. The reaction will shift to the right to produce more \(N_2O_4\) and reach equilibrium.
