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Chemistry · Ch 7 — Equilibrium

Equilibrium Constant in Gaseous Systems

7.4.1

Equilibrium Constant in Gaseous Systems

From Concentration to Partial Pressure

So far, every equilibrium constant you have seen has been written using molar concentrations — the familiar square-bracket notation [A][A], with the constant called KcK_c. For reactions that involve gases, however, there is a more natural way to express the composition of the mixture: by using the partial pressure of each gas. The reason is simple: when you work with gases in a closed container, you measure pressure, not concentration. The equilibrium constant expressed in partial pressures is given the symbol KpK_p.

The bridge between concentration and pressure is the ideal gas law. For a gas:

pV=nRTpV = nRT

Rearrange it:

p=nVRTp = \frac{n}{V}RT

The quantity n/Vn/V is the concentration in moles per unit volume. If you measure concentration cc in mol/L (or mol/dm3^3) and pressure pp in bar, the relation becomes:

p=cRTp = cRT

where R=0.0831 bar L mol−1K−1R = 0.0831\ \text{bar L mol}^{-1}\text{K}^{-1}. You can also write this as p=[gas] RTp = [\text{gas}]\,RT, where [gas][\text{gas}] is the molar concentration. At a fixed temperature, RR and TT are constants, so the pressure of a gas is directly proportional to its concentration:

p∝[gas]p \propto [\text{gas}]

This proportionality is the key that lets us convert between KcK_c and KpK_p.

The First Example: H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

For this reaction, you can write the equilibrium constant in two ways. Using concentrations:

Kc=[HI(g)]2[H2(g)][I2(g)]K_c = \frac{[HI(g)]^2}{[H_2(g)][I_2(g)]}

Using partial pressures:

Kp=(pHI)2(pH2)(pI2)K_p = \frac{(p_{HI})^2}{(p_{H_2})(p_{I_2})}

Now substitute p=[gas]RTp = [\text{gas}]RT for each gas:

Kp=([HI]RT)2([H2]RT)([I2]RT)=[HI]2(RT)2[H2][I2](RT)2=[HI]2[H2][I2]=KcK_p = \frac{([HI]RT)^2}{([H_2]RT)([I_2]RT)} = \frac{[HI]^2 (RT)^2}{[H_2][I_2] (RT)^2} = \frac{[HI]^2}{[H_2][I_2]} = K_c

The (RT)(RT) factors cancel completely. For this particular reaction, Kp=KcK_p = K_c.

Note

The cancellation happens because the number of moles of gaseous products equals the number of moles of gaseous reactants — two moles on each side. When that is true, the (RT)(RT) factors always cancel.

The Second Example: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

Here the story is different. Write KpK_p:

Kp=(pNH3)2(pN2)(pH2)3K_p = \frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Substitute p=[gas]RTp = [\text{gas}]RT:

Kp=([NH3]RT)2([N2]RT)([H2]RT)3=[NH3]2(RT)2[N2][H2]3(RT)4=[NH3]2[N2][H2]3⋅(RT)−2K_p = \frac{([NH_3]RT)^2}{([N_2]RT)([H_2]RT)^3} = \frac{[NH_3]^2 (RT)^2}{[N_2][H_2]^3 (RT)^4} = \frac{[NH_3]^2}{[N_2][H_2]^3} \cdot (RT)^{-2}

So:

Kp=Kc⋅(RT)−2K_p = K_c \cdot (RT)^{-2}

Or, more neatly:

Kp=Kc(RT)−2K_p = K_c (RT)^{-2}

The exponent −2-2 is not arbitrary. Count the moles of gas on each side:

  • Products: 22 moles of NH3NH_3
  • Reactants: 1+3=41 + 3 = 4 moles
  • Difference: Δn=2−4=−2\Delta n = 2 - 4 = -2

That difference appears as the exponent of (RT)(RT).

The General Relation

For any gaseous reaction:

aA(g)+bB(g)⇌cC(g)+dD(g)aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)

Write KpK_p:

Kp=(pC)c(pD)d(pA)a(pB)bK_p = \frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b}

Substitute p=[gas]RTp = [\text{gas}]RT for every gas:

Kp=([C]RT)c([D]RT)d([A]RT)a([B]RT)b=[C]c[D]d[A]a[B]b⋅(RT)c+d(RT)a+bK_p = \frac{([C]RT)^c ([D]RT)^d}{([A]RT)^a ([B]RT)^b} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \cdot \frac{(RT)^{c+d}}{(RT)^{a+b}}

The fraction of concentrations is exactly KcK_c. The exponent of (RT)(RT) is (c+d)−(a+b)(c+d) - (a+b), which is the change in the number of moles of gas, Δn\Delta n:

Δn=(moles of gaseous products)−(moles of gaseous reactants)\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n} …

Table 6.5Equilibrium Constants, Kp for a Few Selected Reactions -- real Kp values at several temperatures, showing how strongly K depends on temperature.
ReactionTemperature/KKpK_p
N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)2986.8×1056.8 \times 10^5
40041
5003.6×10−23.6 \times 10^{-2}
2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)2984.0×10244.0 \times 10^{24}
5002.5×10102.5 \times 10^{10}
7003.0×1043.0 \times 10^{4}