Equilibrium Constant Kp for Gaseous Systems

Understanding chemical equilibrium is crucial for predicting the outcomes of reactions, from industrial ammonia synthesis to the behavior of gases in your car's engine. For reactions involving gases, the equilibrium constant expressed in terms of partial pressures, Kp, provides a direct and practical measure of the position of equilibrium, directly linking the measurable quantity of pressure to the composition of a reaction mixture at a given temperature.

Partial Pressures and Mole Fractions: The Building Blocks of Kp

In a mixture of gases, each component exerts its own pressure as if it alone occupied the entire volume. This is its partial pressure. The total pressure of the mixture is the sum of all the partial pressures of the individual gases (Dalton's Law of Partial Pressures). To find a gas's partial pressure, you first need to know its mole fraction, which is the ratio of the number of moles of that gas to the total number of moles of gas in the mixture.

For a gas A in a mixture, its mole fraction ($x_A$) and partial pressure ($P_A$) are calculated as follows: $$x_A=\frac{n_A}{n_{total}}$$ $$P_A=x_A\times P_{total}$$

Think of a large box containing red, blue, and green balls. The mole fraction of red balls is simply (number of red balls) / (total balls). If the total pressure inside the box is like a constant "squeeze," the contribution to that squeeze from the red balls alone is their mole fraction times the total squeeze. For example, in a mixture of 2.0 mol $H_2$ and 1.0 mol $N_2$ at a total pressure of 600 kPa, the mole fraction of $H_2$ is $2.0/3.0=0.667$. Its partial pressure is $0.667\times 600kPa=400kPa$.

Writing the Expression for Kp

For a general homogeneous gas-phase reaction: $$aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)$$ The equilibrium constant Kp is defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to the power of its stoichiometric coefficient. Crucially, these are the equilibrium partial pressures.

The expression is written as: $$K_p=\frac{(P_C{)}^c(P_D{)}^d}{(P_A{)}^a(P_B{)}^b}$$

Notice that Kp is dimensionless. While we plug in pressure values (typically in units of bar or atm), the constant itself is a pure number because it is technically based on the ratio of pressures relative to a standard state pressure of 1 bar. It is essential to write this expression using partial pressures, not concentrations or total pressure. The magnitude of Kp immediately tells you the position of equilibrium; a Kp >> 1 favors products, while a Kp << 1 favors reactants.

Performing Kp Calculations from Experimental Data

Calculating Kp involves determining the equilibrium partial pressures of all gases and substituting them into the Kp expression. This often requires setting up an ICE (Initial, Change, Equilibrium) table using partial pressures or moles.

Worked Example: Consider the decomposition of dinitrogen tetroxide. $$N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$$ An experiment starts with pure $N_2{O}_4$ at an initial pressure of 1.00 bar. At equilibrium, the total pressure is found to be 1.20 bar. Find Kp.

1. Define initial partial pressures. $P_{N_2{O}_4,initial}=1.00bar$, $P_{N{O}_2,initial}=0$.
2. Set up the change. Let the decrease in $N_2{O}_4$ pressure be $x$ bar. According to the stoichiometry, the increase in $N{O}_2$ pressure will be $2x$ bar.
3. Write equilibrium expressions.

$P_{N_2{O}_4,eq}=1.00-x$ $P_{N{O}_2,eq}=2x$

1. Use the total pressure. At equilibrium, total pressure = $P_{N_2{O}_4,eq}+P_{N{O}_2,eq}=(1.00-x)+2x=1.20$. Solving gives $x=0.20bar$.
2. Find equilibrium partial pressures.

$P_{N_2{O}_4,eq}=1.00-0.20=0.80bar$ $P_{N{O}_2,eq}=2\times 0.20=0.40bar$

1. Substitute into Kp expression.

$$K_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=\frac{(0.40{)}^2}{0.80}=0.20$$

The Relationship Between Kp and Kc

Kp and Kc are both equilibrium constants for the same reaction, but they are expressed in different units (pressure vs. concentration). They are related by the equation derived from the ideal gas law, $PV=nRT$, which gives $P=(n/V)RT=CRT$, where C is concentration.

For the general reaction $aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)$, the relationship is: $$K_p=K_c(RT{)}^{\Delta n_g}$$ where:

• $R$ is the ideal gas constant (use 0.08314 bar L mol${}^{-1}$ K${}^{-1}$ if concentrations are in mol L${}^{-1}$ and pressures in bar).
• $T$ is the temperature in Kelvin.
• $\Delta n_g$ is the change in the number of moles of gas = (moles of gaseous products) - (moles of gaseous reactants) = $(c+d)-(a+b)$.

If $\Delta n_g=0$, then $K_p=K_c$. In the $N_2{O}_4$ example, $\Delta n_g=2-1=+1$, so Kp and Kc would have different numerical values at the same temperature.

How Changes in Conditions Affect Kp and Equilibrium Position

This is where a precise understanding prevents major errors. Kp is constant only for a given reaction at a specific temperature. It does not change with alterations in concentration or pressure. However, the equilibrium position (the individual partial pressures) can shift.

• Change in Total Pressure (at constant T): A change in total pressure, achieved by changing the volume of the container, does not change the value of Kp. However, it can shift the position of equilibrium to counteract the change, as predicted by Le Chatelier's Principle. If $\Delta n_g\ne 0$, the system will shift toward the side with fewer moles of gas when pressure is increased. You must recalculate the new equilibrium partial pressures, but they will always readjust so that the original Kp value is restored at that temperature.

• Change in Temperature: This is the only way to change the numerical value of Kp itself. For an endothermic reaction (positive $\Delta H$), increasing temperature increases Kp, favoring product formation. For an exothermic reaction (negative $\Delta H$), increasing temperature decreases Kp, favoring reactant formation. This is a quantitative change in the constant, not just a shift in position.

Common Pitfalls

1. Using total pressure instead of partial pressure in the Kp expression. This is the most frequent error. You must first calculate the equilibrium partial pressure of each gas. Remember: $P_{gas}=molefraction\times P_{total}$.
2. Incorrectly applying the stoichiometric coefficient. The partial pressure of each gas must be raised to the power of its coefficient from the balanced equation. For $2N{O}_2$, you use $(P_{N{O}_2}{)}^2$, not $2\times P_{N{O}_2}$.
3. Confusing the effects of pressure and temperature on Kp. Students often think increasing pressure increases Kp for a product-favored reaction. Remember: Kp only changes with temperature. A pressure change shifts the composition but the Kp value remains the same.
4. Mixing units or using the wrong R value. Ensure consistency. If pressures are in bar, use R = 0.08314 bar L mol${}^{-1}$ K${}^{-1}$ in the $K_p$/$K_c$ relationship. If in atm, use 0.08206 atm L mol${}^{-1}$ K${}^{-1}$.

Summary

• Kp is the equilibrium constant expressed in terms of the partial pressures of gaseous reactants and products, and is constant for a given reaction at a fixed temperature.
• Partial pressure is calculated as $P_A=x_A\times P_{total}$, where $x_A$ is the mole fraction.
• The expression for Kp is derived directly from the balanced chemical equation, with each partial pressure raised to the power of its stoichiometric coefficient.
• Kp and Kc are related by the equation $K_p=K_c(RT{)}^{\Delta n_g}$, where $\Delta n_g$ is the change in moles of gas.
• While changes in total pressure can shift the equilibrium position (if $\Delta n_g\ne 0$), only a change in temperature will alter the numerical value of Kp itself, with the direction of change determined by whether the reaction is endothermic or exothermic.