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## Chemistry library

### Course: Chemistry library>Unit 12

Lesson 1: Equilibrium constant

# Calculating equilibrium constant Kp using partial pressures

Definition of equilibrium constant Kp for gas phase reactions, and how to calculate Kp from Kc.

## Key points

• The equilibrium constant, 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, 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, start subscript, start text, p, end text, end subscript is
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, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, K, start subscript, start text, c, end text, end subscript, by the equation below:
K, 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, start text, n, end text is
delta, 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

## Introduction: a short review of equilibrium and $K_\text c$K, start subscript, start text, c, end text, end subscript

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.
Why penguins, you ask? Keep reading!! Photo credit: Wikimedia Commons, CC BY-SA 3.0.
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 or K, start subscript, start text, c, end text, end subscript to represent equilibrium constants. When we use the symbol 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 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.

## $K_\text p$K, start subscript, start text, p, end text, end subscript vs. $K_\text c$K, start subscript, start text, c, end text, end subscript: using partial pressure instead of concentration

When a reaction component is a gas, we can also express the amount of that chemical at equilibrium in terms of its partial pressure. When the equilibrium constant is written with the gases in terms of partial pressure, the equilibrium constant is written as the symbol K, start subscript, start text, p, end text, end subscript. The subscript p stands for penguins.
For example, let's say we have the generic balanced gas-phase reaction below:
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, start text, a, end text moles of reactant start text, A, end text react with start text, b, end text moles of reactant start text, B, end text to make start text, c, end text moles of product start text, C, end text and start text, d, end text moles of product start text, D, end text.
If we know the partial pressures for each component at equilibrium, where the partial pressure of start text, A, end text, left parenthesis, g, right parenthesis is abbreviated as start text, P, end text, start subscript, start text, A, end text, end subscript, then the expression for K, start subscript, start text, p, end text, end subscript for this reaction is
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, 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 in the equilibrium expression. This is the same as when calculating K, start subscript, start text, c, end text, end subscript.
• K, start subscript, start text, p, end text, end subscript is often written without units. Since the value of 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, start subscript, start text, p, end text, end subscript problem.
• All the partial pressures used for calculating K, start subscript, start text, p, end text, end subscript should have the same units.
• We can write 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.

## Converting between gas concentration and partial pressure

Close up of someone pouring light brown soda from a can into a glass. The bubbles coming out of the soda create a thick layer of froth at the top of the glass.
Soda is pressurized with carbon dioxide, which is slightly soluble in the soda liquid. When the can is opened, the gas partial pressure above the liquid surface decreases, which causes the dissolved carbon dioxide to go from the aqueous to the gas phase. Therefore, bubbles! Photo credit: Marnav Sharma, CC BY 2.0
We can convert between gas concentration—in units of start text, M, end text or 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 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 start text, P, end text and 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}
We can use this relationship to derive an equation to convert directly between K, start subscript, start text, c, end text, end subscript and K, start subscript, start text, p, end text, end subscript at temperature start text, T, end text, where start text, R, end text is the gas constant:
K, 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, 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, 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!

## Example 1: finding $K_\text p$K, start subscript, start text, p, end text, end subscript from partial pressures

Let's try finding K, start subscript, start text, p, end text, end subscript for the following gas-phase reaction:
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 start text, T, end text:
\begin{aligned} \text P_{\text N_2 \text O_5} &= 2.00\,\text{atm}\\ \\ \text P_{\text O_2} &= 0.296\,\text{atm}\\ \\ \text P_{\text{NO}_2} &= 1.70\,\text{atm}\end{aligned}
At temperature start text, T, end text, what is K, start subscript, start text, p, end text, end subscript for this reaction?
First we can write the K, start subscript, start text, p, end text, end subscript expression for our balanced equation:
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, start subscript, start text, p, end text, end subscript by plugging in the equilibrium partial pressures in the equilibrium expression:
K, 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

## Example 2: finding $K_\text p$K, start subscript, start text, p, end text, end subscript from $K_\text c$K, start subscript, start text, c, end text, end subscript

Now let's look at a different reversible reaction:
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, start subscript, start text, c, end text, end subscript for this reaction is 4, point, 5, times, 10, start superscript, 4, end superscript at 400, start text, K, end text, what is the equilibrium constant, K, start subscript, start text, p, end text, end subscript, at the same temperature?
Use the gas constant that will give 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, 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, 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}
We can now substitute in our values for K, start subscript, start text, c, end text, end subscript, start text, T, end text, and delta, start text, n, end text to find K, start subscript, start text, p, end text, end subscript. We will want to keep track of the units of the gas constant start text, R, end text in our equation since that will determine if we are calculating K, start subscript, start text, p, end text, end subscript for partial pressures of bar or atm. Since we want to calculate K, start subscript, start text, p, end text, end subscript for when partial pressure has units of bar, we will use 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.
\begin{aligned}K_\text p &= K_\text c(\text{RT})^{\Delta \text n} \\ \\ &= (4.5\times 10^4)(\text R \cdot400)^{-2} \\ \\ &= (4.5\times 10^4)(0.08314 \cdot400)^{-2} \\ \\ &= 41\end{aligned}
Note that if we had used a gas constant defined in terms of atm, we would have gotten a different value for K, start subscript, start text, p, end text, end subscript.

## Example 3: find $K_\text p$K, start subscript, start text, p, end text, end subscript from total pressure

Finally, let's consider the equilibrium reaction for the decomposition of water:
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, start subscript, start text, p, end text, end subscript for the reaction?
To do this problem, it might be helpful to visualize our partial pressures using an ICE table.
Note that we don't include pure liquids in our calculations for 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.
Equation2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons2, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesisstart text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis
InitialN/A0, start text, a, t, m, end text0, start text, a, t, m, end text
ChangeN/Aplus, 2, xplus, x
EquilibriumN/A2, xx
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 start text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript increases by x, the change for start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript will be twice that much, 2, 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. We know from Dalton's Law that the total pressure of a system, 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:
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}
Using our observed total pressure of 2.10atm, we can solve for x:
\begin{aligned}\text P_\text{total} &= 2.10\,\text{atm}=3x\\ \\ x&=0.70\,\text{atm}\end{aligned}
By substituting in 0.70atm for x in the last row of our ICE table, we can now find the equilibrium partial pressures for the two gases:
start 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
start 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, start subscript, start text, p, end text, end subscript:
\begin{aligned}K_\text p &= ({\text P_{\text{H}_2}})^2 \cdot \text P_{\text{O}_2}\\ \\ &=(1.40) ^2 \cdot (0.70)\\ \\ & = 1.37\end{aligned}

## Summary

• The equilibrium constant 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 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, start subscript, start text, p, end text, end subscript is
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, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, K, start subscript, start text, c, end text, end subscript, by the equation below
K, 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, start text, n, end text is
delta, 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

## Want to join the conversation?

• Why do we not include liquids and solids in equilibrium expressions? What do you mean by "their concentration is equal to 1"?
• The concentration of the solids never changes, as its density remains the same. Regarding liquids, adding or removing liquids has an insignificant effect on the concentration of the system, as the system is in an aqueous solution. Yes, the liquids do change the concentration, but not by a measurable amount. It's like adding another fish to the sea, making pretty much no difference to the concentration of fish in the ocean
• In example 2, how do you know that you are finding the constant in terms of bar and not atm?
• It is because the Universal Gas Constant R=8.314 is substituted and pressure calculated by using this will give you the answer in terms of barr
If you want to calculate in terms of atm then use R=0.0820
• Can anyone explain why Pressure does not change the equilibrium constant for a reaction but temperature does?
• The equilibrium constant is a constant.
If you change a pressure, all the pressures change so that their new values will give the same value for the equilibrium constant.
• For the equation Kp = Kc(RT)^(delta N), shouldn't there be two instances in which Kp = Kc? First, when delta N = 0 (mols of product gas = mols of reactant gas); second when temperature T is the exact reciprocal of constant R or when R*T = 1 (if R = 0.08206 L*atm*mol^(-1)*K^(-1), T = 1/0.08206 K)? What is the significance of Kp = Kc?
• For the ICE Table, how do we know that the Change has +2x and +x? I mean, why did we choose plus instead of minus?
• With the ICE table, you initially assume that the reaction is entirely on the left, with water but no hydrogen or oxygen gases. Then, as the reaction proceeds, hydrogen and oxygen are formed, which are written as +2x and +x atm in the "change" line.

If you used minus signs, that would indicate that the gas concentrations were decreasing as the reaction moved towards equilibrium, which is not the case.
• in the formula Kc = Kp * (RT)^n
if we use double the reactants then the difference b/w the number of moles will also get double Kc will change
• but you cant arbitrarily double the moles of reactants or products in a particular BALANCED reaction.
all values for the moles should be calculated using a balanced equation simplified to its lowest form.
increasing quantity of reactants will not change the value of kc either, kc always is a constant for a given temperature.
but it has no effect on difference in moles for a balanced reaction.
(correct me if i'm wrong)
• When the do the PV=nRT, shouldn't the final equation be P=nRT/V, not P=(n/V)RT?
(1 vote)
• The two expressions P=nRT/V and P=(n/V)RT are the same. The second one is written specifically to separate out the expression n/V, which is equal to the concentration. This helps emphasize the relationship between concentration and the ideal gas equation.
• As in example 3, why are we taking the partial pressure of water in liquid form as 1, why not any other value (say less than 1)?