Reaction quotient (Q) Knowledge-Based Study

Study notes

1. In a reversible chemical reaction, the reaction quotient ($Q$) is a mathematical expression that quantifies the relative amounts of products and reactants present in a mixture at any given instant. While the equilibrium constant ($K$) is calculated strictly using equilibrium concentrations or partial pressures at a specific temperature, $Q$ can be evaluated at any stage of a reaction—whether the system has just been initiated, is evolving over time, or has already reached equilibrium.

   For a generalized reversible reaction:

   $$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$$

   The reaction quotient in terms of molar concentrations, denoted as $Q_c$, is formulated as:

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

   Here, the concentrations of reactants and products are the measured instantaneous values at that specific moment, not necessarily their values at dynamic equilibrium.

   Depending on how reaction mixtures are quantified, the reaction quotient can be expressed either in terms of molar concentrations ($Q_c$) or in terms of partial pressures ($Q_p$) for gas-phase systems.

   Concentration Quotient ($Q_c$)

   $Q_c$ is used for reactions in solution or homogeneous gas reactions expressed in moles per liter ($\text{mol/L}$).

   Pressure Quotient ($Q_p$)

   For gas-phase equilibria, it is often more convenient to express amounts in terms of partial pressures (usually in atmospheres or bar):

   $$Q_p = \frac{(P_\text{C})^c (P_\text{D})^d}{(P_\text{A})^a (P_\text{B})^b}$$

   Heterogeneous Systems

   Just as with equilibrium constant expressions, pure solids and pure liquids have constant thermodynamic activities (assigned a value of 1). Therefore, their concentrations or partial pressures are omitted from both $Q_c$ and $Q_p$ expressions.

   • For the decomposition: $\text{CaCO}_3(\text{s}) \rightleftharpoons \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$
   • The reaction quotient is simply: $Q_p = P_{\text{CO}_2}$

   The principal analytical utility of the reaction quotient lies in comparing its value to the equilibrium constant ($K$) determined at the same temperature. This comparison reveals whether a system is at dynamic equilibrium, and if not, the direction in which it must proceed to achieve equilibrium.

   • $Q < K$: The ratio of products to reactants is lower than at equilibrium. To achieve equilibrium, the numerator must increase and the denominator must decrease. Thus, the forward reaction dominates, producing more products (the reaction shifts to the right).
   • $Q = K$: The system is at dynamic equilibrium. The rates of the forward and reverse reactions are equal, and no net change in macroscopic concentrations will occur.
   • $Q > K$: The ratio of products to reactants exceeds the equilibrium value. To attain equilibrium, products must be consumed to reform reactants. Thus, the reverse reaction dominates, forming more reactants (the reaction shifts to the left).

   Consider the gas-phase synthesis of ammonia at 500 K:

   $$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad (K_c = 0.060)$$

   Suppose a rigid 2.0 L reaction vessel at 500 K contains $0.40\text{ mol}$ of $\text{N}_2$, $0.60\text{ mol}$ of $\text{H}_2$, and $0.20\text{ mol}$ of $\text{NH}_3$. To determine the direction of net reaction:

   1. Calculate instantaneous molar concentrations:
      [$\text{N}_2$] = $\frac{0.40\text{ mol}}{2.0\text{ L}} = 0.20\text{ M}$
      [$\text{H}_2$] = $\frac{0.60\text{ mol}}{2.0\text{ L}} = 0.30\text{ M}$
      [$\text{NH}_3$] = $\frac{0.20\text{ mol}}{2.0\text{ L}} = 0.10\text{ M}$
   2. Set up and evaluate $Q_c$:
      $$Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.10)^2}{(0.20)(0.30)^3} = \frac{0.010}{(0.20)(0.027)} = \frac{0.010}{0.0054} \approx 1.85$$
   3. Compare $Q_c$ with $K_c$:
      Since $Q_c$ (1.85) is greater than $K_c$ (0.060), there are proportionally more products than at equilibrium. Therefore, the system will proceed in the reverse direction (to the left), consuming $\text{NH}_3$ and producing $\text{N}_2$ and $\text{H}_2$ until equilibrium is established.