Chemical Equilibrium

Lessons

Senior 6 · Chemistry

Chemical Equilibrium

5 lessons

Lesson 3 of 5

Kp, the Kp–Kc Relationship and the Reaction Quotient

25 min

Pressures, not concentrations

Nobody measures the concentration of a gas in an ammonia converter. The engineer reads a pressure gauge, and an analyser gives the composition as a percentage by volume, which for gases is the same as a percentage by moles. If the gas leaving a converter at 200 atm is 15 % ammonia by moles, then ammonia contributes 15 % of the pressure, 30 atm. For gas equilibria it is therefore natural to write the equilibrium constant with pressures. This lesson builds that constant, $K_{p}$ , links it to $K_{c}$ , and then uses the reaction quotient $Q$ to predict which way a mixture that is not at equilibrium will move.

Definition

Mole fraction, $x_\mathrm{A}$ : the amount of gas A divided by the total amount of gas, $x_\mathrm{A} = \frac{n_\mathrm{A}}{n_\text{total}}$ . The mole fractions of a mixture add up to 1. Example: 3.0 mol $\mathrm{H_2}$ with 1.0 mol $\mathrm{N_2}$ gives $x(\mathrm{H_2}) = 0.75$ .

Partial pressure, $p_\mathrm{A}$ : the pressure gas A would exert if it alone filled the container at the same temperature; $p_\mathrm{A} = x_\mathrm{A} \times p_\text{total}$ . By Dalton's law the partial pressures add up to the total pressure.

Equilibrium constant in terms of partial pressures, $K_{p}$ : the law-of-mass-action ratio written with equilibrium partial pressures instead of concentrations. Example: for $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ , $K_p = \frac{p_{\mathrm{NO_2}}^{2}}{p_{\mathrm{N_2O_4}}}$ .

Reaction quotient, $Q$ : the same ratio as $K$ , but calculated from the concentrations or partial pressures present at any moment, whether or not the system is at equilibrium.

Writing $K_{p}$

For $a\mathrm{A(g)} + b\mathrm{B(g)} \rightleftharpoons c\mathrm{C(g)} + d\mathrm{D(g)}$ :

$K_p = \frac{p_\mathrm{C}^{\,c}\;p_\mathrm{D}^{\,d}}{p_\mathrm{A}^{\,a}\;p_\mathrm{B}^{\,b}}$

Only gases appear. In a heterogeneous equilibrium, solids and liquids are left out: for $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ , $K_p = p_{\mathrm{CO_2}}$ , which is why the limestone in Lesson 1 always stopped at 1 atm at 900 °C.
The units are (pressure unit) $^{\Delta n}$ , with $\Delta n$ = moles of gaseous products − moles of gaseous reactants. Pa, kPa and atm are all used; say which.
Like $K_{c}$ , $K_{p}$ depends only on temperature.

How $K_{p}$ is related to $K_{c}$

For an ideal gas, $p_\mathrm{A}V = n_\mathrm{A}RT$ , so

$p_\mathrm{A} = \frac{n_\mathrm{A}}{V}RT = [\mathrm{A}]\,RT$

Substitute this for every partial pressure in $K_{p}$ :

$K_p = \frac{([\mathrm{C}]RT)^{c}([\mathrm{D}]RT)^{d}}{([\mathrm{A}]RT)^{a}([\mathrm{B}]RT)^{b}} = \frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}} \times (RT)^{(c+d)-(a+b)}$

$\boxed{K_p = K_c\,(RT)^{\Delta n}}$

When $\Delta n = 0$ , as for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ , $K_{p} = K_{c}$ . The units must match the value of $R$ :

Value of $R$	Concentrations in	Pressures come out in
$8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ (= $\mathrm{Pa\,m^{3}\,K^{-1}\,mol^{-1}}$ )	$\mathrm{mol\,m^{-3}}$ (1 mol dm⁻³ = 1000 mol m⁻³)	Pa
$0.08206\,\mathrm{dm^{3}\,atm\,K^{-1}\,mol^{-1}}$	$\mathrm{mol\,dm^{-3}}$	atm

Worked example

Problem ( $K_{p}$ from the degree of dissociation). At 25 °C and a total pressure of 100 kPa, dinitrogen tetroxide is 18.9 % dissociated: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ . Calculate $K_{p}$ .

Solution. Start from 1 mol of $\mathrm{N_2O_4}$ and let the fraction dissociated be $\alpha = 0.189$ .

	$\mathrm{N_2O_4}$	$\mathrm{NO_2}$	total
Equilibrium / mol	$1 - \alpha$	$2\alpha$	$1 + \alpha$
Mole fraction	$\frac{1-\alpha}{1+\alpha} = 0.682$	$\frac{2\alpha}{1+\alpha} = 0.318$	1.000
Partial pressure / kPa	68.2	31.8	100

$K_p = \frac{p_{\mathrm{NO_2}}^{2}}{p_{\mathrm{N_2O_4}}} = \frac{(31.8\,\mathrm{kPa})^2}{68.2\,\mathrm{kPa}} = 14.8\,\mathrm{kPa}$

In general, with total pressure $p$ :

$K_p = \frac{\left(\frac{2\alpha}{1+\alpha}\right)^2 p^2}{\left(\frac{1-\alpha}{1+\alpha}\right) p} = \frac{4\alpha^2 p}{1 - \alpha^2}$

which gives the same value.

Answer: $K_p = 14.8\,\mathrm{kPa}$ at 25 °C. Raising the total pressure to 200 kPa at the same temperature and solving $4\alpha^2 \times 200 = 14.8\,(1-\alpha^2)$ gives $\alpha = 0.135$ : at higher pressure less $\mathrm{N_2O_4}$ dissociates, although $K_{p}$ has not changed. Lesson 4 explains why.

Worked example

Problem (converting $K_{c}$ to $K_{p}$ ). For $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at 1000 K, $K_c = 2.80 \times 10^{2}\,\mathrm{mol^{-1}\,dm^{3}}$ (Lesson 2). Calculate $K_{p}$ in $\mathrm{Pa^{-1}}$ and in $\mathrm{atm^{-1}}$ .

Solution. $\Delta n = 2 - 3 = -1$ , so $K_p = K_c (RT)^{-1} = \frac{K_c}{RT}$ .

In SI units, $K_c = 2.80 \times 10^{2}\,\mathrm{mol^{-1}\,dm^{3}} = 2.80 \times 10^{2} \times 10^{-3}\,\mathrm{mol^{-1}\,m^{3}} = 0.280\,\mathrm{mol^{-1}\,m^{3}}$ :

$K_p = \frac{0.280}{8.314 \times 1000} = 3.37 \times 10^{-5}\,\mathrm{Pa^{-1}}$

With $R = 0.08206\,\mathrm{dm^{3}\,atm\,K^{-1}\,mol^{-1}}$ :

$K_p = \frac{2.80 \times 10^{2}}{0.08206 \times 1000} = 3.41\,\mathrm{atm^{-1}}$

Answer: $K_p = 3.37 \times 10^{-5}\,\mathrm{Pa^{-1}} = 3.41\,\mathrm{atm^{-1}}$ . Check: $3.37 \times 10^{-5}\,\mathrm{Pa^{-1}} \times 101\,325\,\mathrm{Pa\,atm^{-1}} = 3.41\,\mathrm{atm^{-1}}$ .

Which way will it go? The reaction quotient

Suppose you mix gases in proportions that are not the equilibrium ones. Calculate $Q$ exactly as you would calculate $K$ , but with the concentrations (or partial pressures) you actually have, and compare it with $K$ at that temperature (Figure 3.1).

Figure 3.1: Comparing the reaction quotient $Q$ with $K$ predicts the direction of net reaction.

$Q < K$ : there are too few products for equilibrium, so the net reaction goes forwards; $Q$ rises until it equals $K$ .
$Q > K$ : there are too many products, so the net reaction goes backwards; $Q$ falls until it equals $K$ .
$Q = K$ : the system is already at equilibrium.

For example, a flask at 430 °C contains $[\mathrm{H_2}] = 0.0050$ , $[\mathrm{I_2}] = 0.0080$ and $[\mathrm{HI}] = 0.0400\,\mathrm{mol\,dm^{-3}}$ . Then $Q = \frac{0.0400^2}{0.0050 \times 0.0080} = 40$ . Because $40 < 54.3$ , more HI forms, and the violet colour of iodine fades a little further before it stops changing.

Worked example

Problem. A 1.00 dm³ vessel at 1000 K contains 0.0500 mol $\mathrm{SO_2}$ , 0.0500 mol $\mathrm{O_2}$ and 0.200 mol $\mathrm{SO_3}$ . Is the mixture at equilibrium? If not, in which direction will it react? ( $K_c = 2.80 \times 10^{2}\,\mathrm{mol^{-1}\,dm^{3}}$ .)

Solution.

$Q_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} = \frac{(0.200)^2}{(0.0500)^2 \times 0.0500} = 3.20 \times 10^{2}\,\mathrm{mol^{-1}\,dm^{3}}$

$Q_{c} = 320 > K_{c} = 280$ .

Answer: The mixture is not at equilibrium. It contains too much $\mathrm{SO_3}$ , so some $\mathrm{SO_3}$ decomposes (net backward reaction) until $Q_{c}$ falls to 280.

Warning

Watch the units and the gases

$K_p = K_c(RT)^{\Delta n}$ works only if $R$ matches the units: $R = 8.314$ needs concentrations in $\mathrm{mol\,m^{-3}}$ and gives pascals. Forgetting the factor of 1000 between $\mathrm{dm^{3}}$ and $\mathrm{m^{3}}$ is the commonest error.
$T$ must be in kelvin.
$\Delta n$ counts gaseous moles only: for $\mathrm{C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g)}$ , $\Delta n = 2 - 1 = +1$ .
$Q$ uses the values present now; $K$ uses equilibrium values. Never put initial values into $K$ .

Tip

To remember the direction: think of $Q$ as "where you are" and $K$ as "where you are going". If $Q$ is smaller than $K$ , the top of the fraction (products) must grow, so the reaction moves to the right. A quick check on any $K_{p}$ answer: it should have pressure units to the power $\Delta n$ .

Chemistry Student: Kp, Equilibrium Constant (A-level IB Chemistry). Watch how mole fractions are turned into partial pressures, how the units of Kp are worked out, and how a change in total pressure affects the position of equilibrium.

✦

Practice

Write $K_{p}$ , with units in kPa, for: (a) $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ ; (b) $\mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}$ ; (c) $\mathrm{H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)}$ .
An equilibrium mixture at a total pressure of 250 kPa contains 0.40 mol $\mathrm{PCl_5}$ , 0.30 mol $\mathrm{PCl_3}$ and 0.30 mol $\mathrm{Cl_2}$ . Calculate $K_{p}$ for $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ .
For $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ at 375 °C, $K_c = 1.2\,\mathrm{mol^{-2}\,dm^{6}}$ . Calculate $K_{p}$ in $\mathrm{atm^{-2}}$ .
Use $K_p = 14.8\,\mathrm{kPa}$ for $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ at 25 °C to find the percentage dissociation at a total pressure of 50.0 kPa.
A mixture at 430 °C has $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.0020$ and $[\mathrm{HI}] = 0.0200\,\mathrm{mol\,dm^{-3}}$ . Is it at equilibrium ( $K_{c} = 54.3$ )? If not, which way does it react?
Show that for $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ , $K_{p} = K_{c} R T$ , and calculate $K_{c}$ at 900 °C, where $K_p = 1.0\,\mathrm{atm}$ .
A student says: "Since $K_p = K_c(RT)^{\Delta n}$ , $K_{p}$ for the Haber process must increase with temperature, because $T$ appears in the formula." Evaluate this claim.

Show answersHide answers

(a) $\frac{p_{\mathrm{NH_3}}^2}{p_{\mathrm{N_2}}\,p_{\mathrm{H_2}}^3}$ , $\mathrm{kPa^{-2}}$ ; (b) $\frac{p_{\mathrm{CO}}^2}{p_{\mathrm{CO_2}}}$ , kPa (carbon is solid); (c) $\frac{p_{\mathrm{HCl}}^2}{p_{\mathrm{H_2}}\,p_{\mathrm{Cl_2}}}$ , no units.
Total 1.00 mol; partial pressures $\mathrm{PCl_5}$ 100 kPa, $\mathrm{PCl_3}$ 75 kPa, $\mathrm{Cl_2}$ 75 kPa; $K_p = \frac{75 \times 75}{100} = 56\,\mathrm{kPa}$ .
$\Delta n = -2$ , $T = 648\,\mathrm{K}$ : $K_p = 1.2 \times (0.08206 \times 648)^{-2} = 4.2 \times 10^{-4}\,\mathrm{atm^{-2}}$ .
$\frac{4\alpha^2 \times 50.0}{1-\alpha^2} = 14.8$ gives $\alpha^2 = \frac{14.8}{214.8}$ , $\alpha = 0.262$ : 26.2 % dissociated.
$Q_c = \frac{0.0200^2}{0.0020 \times 0.0020} = 100 > 54.3$ . Not at equilibrium; net backward reaction: some HI decomposes and the violet colour deepens.
$K_p = p_{\mathrm{CO_2}} = [\mathrm{CO_2}]RT = K_c RT$ ( $\Delta n = +1$ ). $K_c = \frac{1.0}{0.08206 \times 1173} = 1.0 \times 10^{-2}\,\mathrm{mol\,dm^{-3}}$ .
Wrong reasoning. The formula only converts between two ways of expressing the same equilibrium at one temperature; $K_{c}$ itself also changes with $T$ . The Haber reaction is exothermic, so both $K_{c}$ and $K_{p}$ decrease as temperature rises (Lesson 4).

Sources and credits

$K_{p}$ for $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ at 25 °C was calculated by Ganzaa from the standard Gibbs energies of formation of $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ in the CRC Handbook of Chemistry and Physics ( $K = 0.148$ with a standard pressure of 1 bar). $K_{c}$ values for $\mathrm{SO_3}$ (1000 K) and $\mathrm{NH_3}$ (375 °C) are as quoted in Raymond Chang, Chemistry (McGraw-Hill).
Video: Kp, Equilibrium Constant (A-level IB Chemistry), Chemistry Student, embedded from YouTube.
Written by Ganzaa: the text, diagram, tables, worked examples and Practice items.

1Pressures, not concentrations

2Writing Kp K_p Kp​

3How Kp K_p Kp​ is related to Kc K_c Kc​

4Which way will it go? The reaction quotient

Pressures, not concentrations

Writing $K_{p}$

How $K_{p}$ is related to $K_{c}$

Which way will it go? The reaction quotient