Kp and Kc

For gases you can write $K$ with partial pressures instead of concentrations. The ideal gas law connects the two: $P = \dfrac{n}{V}RT$, and $\dfrac{n}{V}$ is the molar concentration. So each pressure is a concentration times $RT$.

Key idea

$K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ = (moles of gas on the product side) − (moles of gas on the reactant side), $R = 0.08206\ \text{L·atm/(mol·K)}$ and $T$ is in kelvins.

Why it works

Replace each pressure in $K_p$ with $[\mathrm{X}]RT$. Every product term brings a factor $RT$ to its coefficient's power on top, and every reactant term the same on the bottom. All the concentration factors together make $K_c$; the leftover factors of $RT$ give $(RT)^{c+d-a-b} = (RT)^{\Delta n}$. Solids and liquids have no partial pressure, so they do not count in $\Delta n$. The value of $R$ must match the units: pressures in atm and concentrations in mol/L call for $0.08206\ \text{L·atm/(mol·K)}$.

Each gas has its own partial pressure, set by its amount per liter and the temperature: $P = (n/V)RT$.

Worked examples

Worked example: Fewer moles of gas

For $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$, $K_c = 0.50$ at $673\ \text{K}$. Find $K_p$.

$\Delta n = 2 - (1 + 3) = -2$.
$RT = (0.08206)(673) = 55.2$.
$$K_p = 0.50 \times (55.2)^{-2} = \dfrac{0.50}{3.05 \times 10^{3}} = 1.6 \times 10^{-4}$$

Answer: $K_p = 1.6 \times 10^{-4}$.

Worked example: More moles of gas

For $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$, $K_c = 0.042$ at $523\ \text{K}$. Find $K_p$.

$\Delta n = 2 - 1 = +1$.
$$K_p = (0.042)(0.08206 \times 523) = (0.042)(42.9) = 1.8$$

Answer: $K_p = 1.8$.

Worked example: A solid in the equation

For $\mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}$, what is $\Delta n$, and when would $K_p = K_c$?

Count gases only: $\Delta n = 2 - 1 = +1$. Carbon, a solid, does not count.
$K_p = K_c$ only when $\Delta n = 0$, as in $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$; then $(RT)^0 = 1$.

Answer: $\Delta n = 1$; equal only when $\Delta n = 0$.

Worked example: Going back to Kc

At $673\ \text{K}$, $K_p = 1.6 \times 10^{-4}$ for $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$. Recover $K_c$.

Solve the relation for $K_c$: $K_c = K_p(RT)^{-\Delta n} = K_p(RT)^{2}$.
$$K_c = (1.6 \times 10^{-4})(55.2)^2 = (1.6 \times 10^{-4})(3.05 \times 10^{3}) = 0.49$$ matching the 0.50 we started from (the small gap is rounding).

Answer: $K_c \approx 0.50$.

Strictly, both constants are written with activities, so both are pure numbers: pressures are divided by the standard pressure and concentrations by $1\ \text{M}$. That is why we report $K_p$ and $K_c$ without units, even though the conversion factor $(RT)^{\Delta n}$ looks as if it carries some.

Common mistake

Counting solids and liquids in $\Delta n$, or using $R = 8.314\ \text{J/(mol·K)}$. Only gases have partial pressures, and $8.314$ belongs with energies, not with atm and liters. Check: if $\Delta n$ is negative, $K_p$ should be smaller than $K_c$ at ordinary temperatures (where $RT > 1$).

Kp and Kc

Worked examples

Practice

Builds on

Leads to

Sources and further reading