Chemical Equilibrium | Reaction Quotient

Reaction Quotient (Q_c)

• Reaction quotient helps in predicting the direction of reaction. It helps to decide whether the reaction is moving in forward direction or in backward direction.
• Q_c has a tendency to become equal to K_c.

Let us consider a chemical reaction:

$$n_{1}A+n_{2}B\rightleftharpoons n_{3}C+n_{4}D$$

At any time t, we can define reaction quotient as:

$$Q_c=\frac{[C{]}^{n_3}[D{]}^{n4}}{[A{]}^{n_1}[B{]}^{n_2}}$$

Case 1: If Q_c > K_c

Q_c tends to become equal to K_c. For this, its value should decrease. Hence, the concentration of C and D will decrease and that of A and B will increase.

This indicates that reaction will shift in backward direction.

Case 2: If Q_c < K_c

Q_c tends to become equal to K_c. For this, its value should increase. Hence, the concentration of C and D will increase and that of A and B will decrease.

This indicates that reaction will shift in forward direction.

Case 3: If Q_c = K_c

Reaction has already reached equilibrium.

Question

At 1400 K, K_c = 2.5 x 10^-3 for the reaction:
CH_4(g) + 2H₂S_g ⇌ CS_2(g) + 4H_2(g)
A 10 L reaction vessel at 1400 K contains 2 moles of CH₄, 3 moles of CS₂, 3 moles of H₂ and 4 moles of H₂S. Is the reaction mixture at equilibrium? If not, in which direction does the reaction proceed to reach equilibrium?

$$Q_c=\frac{[C{S}_2][H_2{]}^4}{[C{H}_4][H_{2}S{]}^2}$$

$$Q_c=\frac{(\frac{3}{10})\times (\frac{3}{10}{)}^4}{(\frac{2}{5})\times (\frac{4}{10}{)}^2}=7.5\times {10}^{-2}$$

Here Q_c > K_c. Hence, the reaction will proceed in backward direction to achieve equilibrium.