Understanding the equilibrium constant for gas phase reactions is fundamental to mastering chemical thermodynamics and kinetics. Unlike reactions in solution where concentrations are typically expressed in molarity, gas phase equilibria rely on partial pressures to quantify the extent of reaction. This distinction leads to the definition of Kp, the equilibrium constant in terms of pressure, which relates directly to Kc, the concentration-based constant, through the ideal gas law. Whether you are synthesizing ammonia via the Haber process or analyzing atmospheric chemistry, a firm grasp of these principles allows you to predict reaction yields and optimize industrial conditions.
Defining the Equilibrium Constant for Gases: Kp vs. Kc
For a generic reversible gas phase reaction:
$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$
The equilibrium constant expressed in terms of partial pressures ($K_p$) is defined as:
$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$
Here, $P_X$ represents the partial pressure of gas $X$ at equilibrium, usually measured in atmospheres (atm) or bar. The exponents correspond to the stoichiometric coefficients from the balanced chemical equation.
Conversely, the concentration-based equilibrium constant ($K_c$) uses molar concentrations ($[X]$ in mol/L):
$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
The critical link between $K_p$ and $K_c$ derives from the ideal gas law, $PV = nRT$. Rearranging for concentration ($n/V = P/RT$), we substitute into the $K_c$ expression:
$K_c = \frac{(P_C/RT)^c (P_D/RT)^d}{(P_A/RT)^a (P_B/RT)^b} = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \times (RT)^{(a+b)-(c+d)}$
Recognizing the first term as $K_p$ and defining $\Delta n_{gas}$ as the change in moles of gas (moles of gaseous products $-$ moles of gaseous reactants), we arrive at the essential conversion formula:
$K_p = K_c (RT)^{\Delta n_{gas}}$
Key Implications of $\Delta n_{gas}$:
- $\Delta n_{gas} = 0$: $K_p = K_c$. The number of gas moles is conserved (e.g., $H_2 + I_2 \rightleftharpoons 2HI$).
- $\Delta n_{gas} > 0$: $K_p > K_c$ (at standard temperatures). The reaction produces more gas moles than it consumes.
- $\Delta n_{gas} < 0$: $K_p < K_c$. The reaction consumes more gas moles than it produces (e.g., ammonia synthesis).
The Role of Standard States and Dimensionless Constants
In rigorous thermodynamic treatments, the equilibrium constant is dimensionless. And this is achieved by dividing each partial pressure by the standard state pressure ($P^\circ$), typically defined as 1 bar (approx. So 0. 987 atm).
$K = \frac{(P_C/P^\circ)^c (P_D/P^\circ)^d}{(P_A/P^\circ)^a (P_B/P^\circ)^b}$
While introductory chemistry often treats $K_p$ as having units of pressure$^{\Delta n_{gas}}$, advanced work requires the dimensionless form to calculate standard Gibbs Free Energy change ($\Delta G^\circ = -RT \ln K$). Now, always verify the standard state convention used in your specific textbook or data tables (1 bar vs. 1 atm), as this affects the numerical value of $K$ when $\Delta n_{gas} \neq 0$.
Heterogeneous Equilibria Involving Gases
Many industrial gas phase reactions involve solid catalysts or produce condensed phases (liquids/solids). In heterogeneous equilibria, the concentrations (or partial pressures) of pure solids and pure liquids do not appear in the equilibrium constant expression. Their "concentrations" are constant (density/molar mass) and are incorporated into the value of $K$.
Example: Decomposition of Calcium Carbonate $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ $K_p = P_{CO_2}$ $K_c = [CO_2]$
Here, the equilibrium position depends only on the partial pressure of carbon dioxide. Adding more solid $CaCO_3$ or $CaO$ does not shift the equilibrium.
Calculating Equilibrium Partial Pressures: The ICE Table Method
Solving equilibrium problems for gas phase reactions follows a systematic approach using ICE tables (Initial, Change, Equilibrium). Because the variable is pressure, the table tracks partial pressures But it adds up..
Worked Example: Consider the reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ At 500 K, $K_p = 1.45 \times 10^{-5}$. A reaction vessel initially contains $N_2$ at 1.00 atm and $H_2$ at 3.00 atm. No $NH_3$ is present initially. Find the equilibrium partial pressures That alone is useful..
| Species | $N_2$ | $H_2$ | $NH_3$ |
|---|---|---|---|
| Initial (atm) | 1.00 | 0 | |
| Change (atm) | $-x$ | $-3x$ | $+2x$ |
| Equilibrium (atm) | $1.00 | 3.00 - x$ | $3. |
Substitute into $K_p$ expression: $K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(2x)^2}{(1.00 - x)(3.00 - 3x)^3} = 1.
Because $K_p$ is very small, $x$ will be tiny compared to initial pressures. Which means we can approximate $1. 00 - 3x \approx 3.That said, 00 - x \approx 1. Think about it: 00$ and $3. $\frac{4x^2}{(1)(27)} \approx 1.00$. 915 \times 10^{-4}$ $x^2 \approx 9.Practically speaking, 45 \times 10^{-5}$ $4x^2 \approx 3. 79 \times 10^{-5}$ $x \approx 0.
Equilibrium Pressures:
- $P_{N_2} \approx 0.99 \text{ atm}$
- $P_{H_2} \approx 2.97 \text{ atm}$
- $P_{NH_3} \approx 0.020 \text{ atm}$
Verification: The approximation is valid ($x < 5%$ of initial values). This demonstrates how a small $K_p$ favors reactants at this temperature.
Le Chatelier’s Principle in Gas Phase Systems
Predicting the response of a gas phase equilibrium to external stress is a core application of the equilibrium constant.
1. Changing Total Pressure (Volume Changes)
Increasing total pressure (by decreasing volume) shifts the equilibrium toward the side with fewer moles of gas.
- $\Delta n_{gas} < 0$ (Forward reaction reduces moles): Increasing pressure favors products. $K_p$ remains constant, but the equilibrium partial pressures of products increase relative to reactants.
2.2 Changing Total Pressure (Volume Changes) – Continued
-
Δn₍gas₎ > 0 (Forward reaction creates more moles of gas)
When the total pressure is increased (volume decreased), the system responds by shifting toward the side that contains fewer gas molecules—i.e., the reactants.
[ \text{Example: } \mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}\qquad \Delta n_{\text{gas}} = +1 ]
Raising the pressure favours the left‑hand side, increasing the partial pressure of (\mathrm{CO_2}) and decreasing that of (\mathrm{CO}). (K_p) itself does not change; only the equilibrium composition adjusts. -
Δn₍gas₎ = 0 (Equal moles on both sides)
Pressure changes have no effect on the equilibrium position because the system experiences the same number of particles on each side.
These qualitative predictions become quantitative when combined with the ICE‑table framework introduced earlier Worth keeping that in mind..
2.3 Temperature Effects
The equilibrium constant is temperature‑dependent. For a generic reaction
[ \mathrm{aA + bB \rightleftharpoons cC + dD} ]
the van ’t Hoff equation describes this dependence:
[ \frac{d\ln K}{dT}= \frac{\Delta H^\circ}{RT^{2}} ]
-
Exothermic reactions ((\Delta H^\circ<0))
Raising the temperature supplies heat, and the system counteracts this by shifting toward the reactants (the endothermic direction). As a result, (K) decreases with temperature. -
Endothermic reactions ((\Delta H^\circ>0))
Increasing temperature favours the products, raising (K).
Worked Example – Water–Gas Shift Reaction
[ \mathrm{CO(g) + H_2O(g) \rightleftharpoons
[ \mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)} ]
The water‑gas shift reaction is a classic illustration of how temperature governs the position of a gas‑phase equilibrium.
Standard thermodynamic data at 298 K give
- ΔH° ≈ ‑41 kJ mol⁻¹ (exothermic)
- ΔG° ≈ ‑28.6 kJ mol⁻¹
From ΔG° = ‑RT ln K we obtain an equilibrium constant at 298 K of
[ K_{298};=;e^{-\Delta G^\circ/(RT)};\approx;e^{28.6\times10^{3}/(8.314\times298)};\approx;1.0\times10^{5}. ]
Because the reaction is exothermic, raising the temperature supplies heat that the system counteracts by favoring the endothermic direction—here, the reverse reaction that consumes CO₂ and H₂ and regenerates CO and H₂O.
Using the van ’t Hoff relation
[ \ln\frac{K_{T_2}}{K_{T_1}}=-\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) ]
with (T_1=298;\text{K}), (T_2=800;\text{K}) and ΔH° = ‑41 kJ mol⁻¹:
[ \ln\frac{K_{800}}{K_{298}}=-\frac{-41,000}{8.314}\left(\frac{1}{800}-\frac{1}{298}\right) =+4.93\times10^{3},(-0.0021
