Calculating equilibrium constant Kp using partial pressures (article) | Khan Academy (2023)

• The equilibrium constant, $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript, describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.

• For a gas-phase reaction, $\text{aA}(g)+\text{bB}(g) \rightleftharpoons \text{cC}(g)+\text{dD}(g)$aA(g)+bB(g)⇌cC(g)+dD(g)start text, a, A, end text, left parenthesis, g, right parenthesis, plus, start text, b, B, end text, left parenthesis, g, right parenthesis, \rightleftharpoons, start text, c, C, end text, left parenthesis, g, right parenthesis, plus, start text, d, D, end text, left parenthesis, g, right parenthesis, the expression for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript is

$K_\text p =\dfrac{(\text P_{\text C})^c (\text P_{\text D})^d}{(\text P_{\text A})^a (\text P_{\text B})^b}$Kp​=(PA​)a(PB​)b(PC​)c(PD​)d​K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, start text, P, end text, start subscript, start text, C, end text, end subscript, right parenthesis, start superscript, c, end superscript, left parenthesis, start text, P, end text, start subscript, start text, D, end text, end subscript, right parenthesis, start superscript, d, end superscript, divided by, left parenthesis, start text, P, end text, start subscript, start text, A, end text, end subscript, right parenthesis, start superscript, a, end superscript, left parenthesis, start text, P, end text, start subscript, start text, B, end text, end subscript, right parenthesis, start superscript, b, end superscript, end fraction

• $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript, by the equation below:

$K_\text p = K_\text c(\text{RT})^{\Delta \text n}$Kp​=Kc​(RT)ΔnK, start subscript, start text, p, end text, end subscript, equals, K, start subscript, start text, c, end text, end subscript, left parenthesis, start text, R, T, end text, right parenthesis, start superscript, delta, start text, n, end text, end superscript

where $\Delta \text n$Δndelta, start text, n, end text is

$\Delta \text n=\text{mol of product gas}-\text{mol of reactant gas}$Δn=molofproductgas−molofreactantgasdelta, start text, n, end text, equals, start text, m, o, l, space, o, f, space, p, r, o, d, u, c, t, space, g, a, s, end text, minus, start text, m, o, l, space, o, f, space, r, e, a, c, t, a, n, t, space, g, a, s, end text

When a reaction is at equilibrium, the forward reaction and reverse reaction have the same rate. The concentrations of the reaction components stay constant at equilibrium, even though the forward and backward reactions are still occurring.

Equilibrium constants are used to define the ratio of concentrations at equilibrium for a reaction at a certain temperature. In general, we use the symbol $K$KK or $K_\text{c}$Kc​K, start subscript, start text, c, end text, end subscript to represent equilibrium constants. When we use the symbol $K_\text{c}$Kc​K, start subscript, start text, c, end text, end subscript, the subscript c means that all concentrations are being expressed in terms of molar concentration, or $\dfrac{\text{mol solute}}{\text {L of solution}}$Lofsolutionmolsolute​start fraction, start text, m, o, l, space, s, o, l, u, t, e, end text, divided by, start text, L, space, o, f, space, s, o, l, u, t, i, o, n, end text, end fraction.

For example, let's say we have the generic balanced gas-phase reaction below:

$\text{aA}(g)+\text{bB}(g) \rightleftharpoons \text{cC}(g)+\text{dD}(g)$aA(g)+bB(g)⇌cC(g)+dD(g)start text, a, A, end text, left parenthesis, g, right parenthesis, plus, start text, b, B, end text, left parenthesis, g, right parenthesis, \rightleftharpoons, start text, c, C, end text, left parenthesis, g, right parenthesis, plus, start text, d, D, end text, left parenthesis, g, right parenthesis

In this equation, $\text a$astart text, a, end text moles of reactant $\text A$Astart text, A, end text react with $\text b$bstart text, b, end text moles of reactant $\text B$Bstart text, B, end text to make $\text c$cstart text, c, end text moles of product $\text C$Cstart text, C, end text and $\text d$dstart text, d, end text moles of product $\text D$Dstart text, D, end text.

If we know the partial pressures for each component at equilibrium, where the partial pressure of $\text A(g)$A(g)start text, A, end text, left parenthesis, g, right parenthesis is abbreviated as $\text P_\text A$PA​start text, P, end text, start subscript, start text, A, end text, end subscript, then the expression for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for this reaction is

$K_\text p =\dfrac{(\text P_{\text C})^c (\text P_{\text D})^d}{(\text P_{\text A})^a (\text P_{\text B})^b}$Kp​=(PA​)a(PB​)b(PC​)c(PD​)d​K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, start text, P, end text, start subscript, start text, C, end text, end subscript, right parenthesis, start superscript, c, end superscript, left parenthesis, start text, P, end text, start subscript, start text, D, end text, end subscript, right parenthesis, start superscript, d, end superscript, divided by, left parenthesis, start text, P, end text, start subscript, start text, A, end text, end subscript, right parenthesis, start superscript, a, end superscript, left parenthesis, start text, P, end text, start subscript, start text, B, end text, end subscript, right parenthesis, start superscript, b, end superscript, end fraction

Remember the following important points when calculating $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript:

• Make sure the reaction is balanced! Otherwise, the stoichiometric coefficients and the exponents in the equilibrium constant will be incorrect.

• Pure liquids or solids have a concentration of $1$11 in the equilibrium expression. This is the same as when calculating $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript.

• $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript is often written without units. Since the value of $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript depends on the units used for the partial pressure, you will need to check the pressure units used in your textbook when solving a $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript problem.

• All the partial pressures used for calculating $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript should have the same units.

• We can write $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for reactions that include solids and pure liquids since they do not appear in the equilibrium expression.

We can convert between gas concentration—in units of $\text M$Mstart text, M, end text or $\dfrac{\text {mol}}{\text L}$Lmol​start fraction, start text, m, o, l, end text, divided by, start text, L, end text, end fraction—and partial pressure using the ideal gas equation. Since molar concentration is the number of moles of gas per volume, or $\dfrac{\text n}{\text V}$Vn​start fraction, start text, n, end text, divided by, start text, V, end text, end fraction, we can rearrange the ideal gas equation to get the relationship between $\text P$Pstart text, P, end text and $\dfrac{\text n}{\text V}$Vn​start fraction, start text, n, end text, divided by, start text, V, end text, end fraction as follows:

$\begin{aligned}\text{PV} &= \text{nRT}~~~~~~~~~\text{Divide both sides by V.}\\\\ \text P &= (\dfrac{\text n}{\text V})\text{RT}\end{aligned}$PVP​=nRTDividebothsidesbyV.=(Vn​)RT​

We can use this relationship to derive an equation to convert directly between $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript and $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript at temperature $\text T$Tstart text, T, end text, where $\text R$Rstart text, R, end text is the gas constant:

$K_\text p = K_\text c(\text{RT})^{\Delta \text n}$Kp​=Kc​(RT)ΔnK, start subscript, start text, p, end text, end subscript, equals, K, start subscript, start text, c, end text, end subscript, left parenthesis, start text, R, T, end text, right parenthesis, start superscript, delta, start text, n, end text, end superscript

The symbol $\Delta \text n$Δndelta, start text, n, end text is the number of moles of gas on the product side minus the number of moles of gas on the reactant side in the balanced reaction:

$\Delta \text n=\text{mol of product gas}-\text{mol of reactant gas}$Δn=molofproductgas−molofreactantgasdelta, start text, n, end text, equals, start text, m, o, l, space, o, f, space, p, r, o, d, u, c, t, space, g, a, s, end text, minus, start text, m, o, l, space, o, f, space, r, e, a, c, t, a, n, t, space, g, a, s, end text

Let's practice using these equations in some examples!

Let's try finding $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for the following gas-phase reaction:

$2\text N_2 \text O_5(g) \rightleftharpoons \text O_2(g)+4\text{NO}_2(g)$2N2​O5​(g)⇌O2​(g)+4NO2​(g)2, start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 5, end subscript, left parenthesis, g, right parenthesis, \rightleftharpoons, start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, 4, start text, N, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis

We know the partial pressures for each component at equilibrium for some temperature $\text T$Tstart text, T, end text:

At temperature $\text T$Tstart text, T, end text, what is $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for this reaction?

First we can write the $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript expression for our balanced equation:

$K_\text p=\dfrac{(\text P_{\text O_2}) (\text P_{\text {NO}_2})^4}{(\text P_{\text {N}_2 \text O_5})^2 }$Kp​=(PN2​O5​​)2(PO2​​)(PNO2​​)4​K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, start text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, right parenthesis, left parenthesis, start text, P, end text, start subscript, start text, N, O, end text, start subscript, 2, end subscript, end subscript, right parenthesis, start superscript, 4, end superscript, divided by, left parenthesis, start text, P, end text, start subscript, start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 5, end subscript, end subscript, right parenthesis, squared, end fraction

We can now solve for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript by plugging in the equilibrium partial pressures in the equilibrium expression:

$K_\text p=\dfrac{(0.296) (1.70)^4}{(2.00)^2}=0.618$Kp​=(2.00)2(0.296)(1.70)4​=0.618K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, 0, point, 296, right parenthesis, left parenthesis, 1, point, 70, right parenthesis, start superscript, 4, end superscript, divided by, left parenthesis, 2, point, 00, right parenthesis, squared, end fraction, equals, 0, point, 618

Now let's look at a different reversible reaction:

$\text N_2(g) + 3\text H_2(g) \rightleftharpoons 2\text{NH}_3 (g)$N2​(g)+3H2​(g)⇌2NH3​(g)start text, N, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, 3, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, \rightleftharpoons, 2, start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, g, right parenthesis

If $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript for this reaction is $4.5 \times 10^4$4.5×1044, point, 5, times, 10, start superscript, 4, end superscript at $400\,\text K$400K400, start text, K, end text, what is the equilibrium constant, $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript, at the same temperature?

Use the gas constant that will give $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for partial pressure units of bar.

To solve this problem, we can use the relationship between the two equilibrium constants:

$K_\text p = K_\text c(\text{RT})^{\Delta \text n}$Kp​=Kc​(RT)ΔnK, start subscript, start text, p, end text, end subscript, equals, K, start subscript, start text, c, end text, end subscript, left parenthesis, start text, R, T, end text, right parenthesis, start superscript, delta, start text, n, end text, end superscript

To find $\Delta \text n$Δndelta, start text, n, end text, we compare the moles of gas from the product side of the reaction with the moles of gas on the reactant side:

$\begin{aligned}\Delta \text n&= \text{mol of product gas}-\text{mol of reactant gas}\\\\&= 2\,\text{mol NH}_3 - (1\,\text{mol N}_2+3\,\text{mol H}_2) \\\\&= -2\,\text {mol gas}\end{aligned}$Δn​=molofproductgas−molofreactantgas=2molNH3​−(1molN2​+3molH2​)=−2molgas​

We can now substitute in our values for $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript, $\text T$Tstart text, T, end text, and $\Delta \text n$Δndelta, start text, n, end text to find $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript. We will want to keep track of the units of the gas constant $\text R$Rstart text, R, end text in our equation since that will determine if we are calculating $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for partial pressures of bar or atm. Since we want to calculate $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for when partial pressure has units of bar, we will use $\text R=0.08314\,\dfrac{\text L \cdot \text{bar}}{\text K \cdot \text{mol}}$R=0.08314K⋅molL⋅bar​start text, R, end text, equals, 0, point, 08314, start fraction, start text, L, end text, dot, start text, b, a, r, end text, divided by, start text, K, end text, dot, start text, m, o, l, end text, end fraction.

Note that if we had used a gas constant defined in terms of atm, we would have gotten a different value for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript.

Finally, let's consider the equilibrium reaction for the decomposition of water:

$2\text H_2 \text O(l) \rightleftharpoons 2 \text H_2(g)+\text O_2(g)$2H2​O(l)⇌2H2​(g)+O2​(g)2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, 2, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis

Assume that initially there is no hydrogen or oxygen gas present. As the reaction proceeds to equilibrium, however, the total pressure increases by 2.10atm.

Based on this information, what is $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript for the reaction?

Note that we don't include pure liquids in our calculations for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript; the table only includes partial pressure information for the two gaseous products. Since initially there are no products in our system, we can fill in the first row of our table with zeros.

Next, we look at the balanced equation to describe how the partial pressures change when the reaction reaches equilibrium. Based on the stoichiometric coefficients, we know that if the value for $\text P_{\text{O}_2}$PO2​​start text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript increases by $x$xx, the change for $\text P_{\text{H}_2}$PH2​​start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript will be twice that much, $2x$2x2, x. The third row in the table sums up the expressions in the first two rows to describe the partial pressures at equilibrium.

At this point, Dalton's Law can help us solve for $x$xx. We know from Dalton's Law that the total pressure of a system, $\text P_\text{total}$Ptotal​start text, P, end text, start subscript, start text, t, o, t, a, l, end text, end subscript, is equal to the sum of the partial pressures for each of the components in the system:

$\text P_\text{total}=\text P_\text{A}+\text P_\text{B}+\text P_\text{C}+...$Ptotal​=PA​+PB​+PC​+...start text, P, end text, start subscript, start text, t, o, t, a, l, end text, end subscript, equals, start text, P, end text, start subscript, start text, A, end text, end subscript, plus, start text, P, end text, start subscript, start text, B, end text, end subscript, plus, start text, P, end text, start subscript, start text, C, end text, end subscript, plus, point, point, point

Using our equilibrium values, we can express the total pressure for our reaction as follows:

$\begin{aligned}\text P_\text{total}&=\text P_{\text{H}_2}+\text P_{\text{O}_2}\\\\&=2x+x\\\\&=3x\end{aligned}$Ptotal​​=PH2​​+PO2​​=2x+x=3x​

Using our observed total pressure of 2.10atm, we can solve for $x$xx:

$\begin{aligned}\text P_\text{total} &= 2.10\,\text{atm}=3x\\\\x&=0.70\,\text{atm}\end{aligned}$Ptotal​x​=2.10atm=3x=0.70atm​

By substituting in 0.70atm for $x$xx in the last row of our ICE table, we can now find the equilibrium partial pressures for the two gases:

$\text P_{\text{H}_2} = 2x = 1.40\,\text{atm}$PH2​​=2x=1.40atmstart text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, equals, 2, x, equals, 1, point, 40, start text, a, t, m, end text

$\text P_{\text{O}_2} = x = 0.70\,\text{atm}$PO2​​=x=0.70atmstart text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, equals, x, equals, 0, point, 70, start text, a, t, m, end text

Now we can set up an equilibrium expression for the reaction and solve for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript:

• The equilibrium constant $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.

• For a gas-phase reaction $\text{aA}(g)+\text{bB}(g) \rightleftharpoons \text{cC}(g)+\text{dD}(g)$aA(g)+bB(g)⇌cC(g)+dD(g)start text, a, A, end text, left parenthesis, g, right parenthesis, plus, start text, b, B, end text, left parenthesis, g, right parenthesis, \rightleftharpoons, start text, c, C, end text, left parenthesis, g, right parenthesis, plus, start text, d, D, end text, left parenthesis, g, right parenthesis, the expression for $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript is

$K_\text p =\dfrac{(\text P_{\text C})^c (\text P_{\text D})^d}{(\text P_{\text A})^a (\text P_{\text B})^b}$Kp​=(PA​)a(PB​)b(PC​)c(PD​)d​K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, start text, P, end text, start subscript, start text, C, end text, end subscript, right parenthesis, start superscript, c, end superscript, left parenthesis, start text, P, end text, start subscript, start text, D, end text, end subscript, right parenthesis, start superscript, d, end superscript, divided by, left parenthesis, start text, P, end text, start subscript, start text, A, end text, end subscript, right parenthesis, start superscript, a, end superscript, left parenthesis, start text, P, end text, start subscript, start text, B, end text, end subscript, right parenthesis, start superscript, b, end superscript, end fraction

• $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, $K_\text c$Kc​K, start subscript, start text, c, end text, end subscript, by the equation below

$K_\text p = K_\text c(\text{RT})^{\Delta \text n}$Kp​=Kc​(RT)ΔnK, start subscript, start text, p, end text, end subscript, equals, K, start subscript, start text, c, end text, end subscript, left parenthesis, start text, R, T, end text, right parenthesis, start superscript, delta, start text, n, end text, end superscript

where $\Delta \text n$Δndelta, start text, n, end text is

$\Delta \text n=\text{mol of product gas}-\text{mol of reactant gas}$Δn=molofproductgas−molofreactantgasdelta, start text, n, end text, equals, start text, m, o, l, space, o, f, space, p, r, o, d, u, c, t, space, g, a, s, end text, minus, start text, m, o, l, space, o, f, space, r, e, a, c, t, a, n, t, space, g, a, s, end text