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

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.

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.

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)

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)?

adding or removing liquid makes a very negligible difference in the pressure of the system. you can assume it as 1 without changing the answer. at least for calculation purposes besides taking any value apart from 1 will unnecessarily increase the difficulty of your calculations.

In example 3, what does N/A in the ICE table indicate?

N/A is a common abbreviation for "Not applicable".

When the do the PV=nRT, shouldn't the final equation be P=nRT/V, not P=(n/V)RT?

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.

Main content

Course: AP®︎/College Chemistry > Unit 11

Lesson 1: Equilibrium constant

Calculating equilibrium constant Kp using partial pressures

Google Classroom

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

Key points

The equilibrium constant, $K_{p}$ ‍, describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
For a gas-phase reaction, $aA (g) + bB (g) ⇌ cC (g) + dD (g)$ ‍, the expression for $K_{p}$ ‍ is

$K_{p} = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}}$ ‍

$K_{p}$ ‍ is related to the equilibrium constant in terms of molar concentration, $K_{c}$ ‍, by the equation below:

$K_{p} = K_{c} (RT)^{Δ n}$ ‍

where $Δ n$ ‍ is

$Δ n = mol of product gas - mol of reactant gas$ ‍

Introduction: a short review of equilibrium and $K_{c}$ ‍

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

K_{c}

to represent equilibrium constants. When we use the symbol

K_{c}

, the subscript c means that all concentrations are being expressed in terms of molar concentration, or

\frac{mol solute}{L of solution}

$K_{p}$ ‍ vs. $K_{c}$ ‍: 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_{p}

. The subscript p stands for penguins.

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

$aA (g) + bB (g) ⇌ cC (g) + dD (g)$ ‍

In this equation,

a

moles of reactant

A

react with

b

moles of reactant

B

to make

c

moles of product

C

and

d

moles of product

D

If we know the partial pressures for each component at equilibrium, where the partial pressure of

A (g)

is abbreviated as

P_{A}

, then the expression for

K_{p}

for this reaction is

$K_{p} = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}}$ ‍

Remember the following important points when calculating

K_{p}

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 $K_{c}$ ‍.
$K_{p}$ ‍ is often written without units. Since the value of $K_{p}$ ‍ depends on the units used for the partial pressure, you will need to check the pressure units used in your textbook when solving a $K_{p}$ ‍ problem.
All the partial pressures used for calculating $K_{p}$ ‍ should have the same units.
We can write $K_{p}$ ‍ for reactions that include solids and pure liquids since they do not appear in the equilibrium expression.

Converting between gas concentration and partial pressure

We can convert between gas concentration—in units of

M

\frac{mol}{L}

—and partial pressure using the ideal gas equation. Since molar concentration is the number of moles of gas per volume, or

\frac{n}{V}

, we can rearrange the ideal gas equation to get the relationship between

P

and

\frac{n}{V}

as follows:

$\begin{aligned} PV & = nRT Divide both sides by V. \\ P & = (\frac{n}{V}) RT \end{aligned}$ ‍

We can use this relationship to derive an equation to convert directly between

K_{c}

and

K_{p}

at temperature

T

, where

R

is the gas constant:

$K_{p} = K_{c} (RT)^{Δ n}$ ‍

The symbol

Δ n

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:

$Δ n = mol of product gas - mol of reactant gas$ ‍

The derivation for this equation is based on the ideal gas law and our previously stated definitions of

K_{p}

and partial pressure, so no crazy math skills required!

Recall that for some chemical

A

, the partial pressure

P_{A}

can be written in terms of the molar concentration

[A]

as follows:

\begin{aligned} P_{A} & = (\frac{moles A}{V}) RT \\ = [A] RT \end{aligned}

If we plug in the partial pressure expression in terms of concentration into the

K_{p}

equation, we get:

\begin{aligned} K_{p} & = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}} \\ = \frac{([C] RT)^{c} ([D] RT)^{d}}{([A] RT)^{a} ([B] RT)^{b}} \\ = \frac{[C]^{c} (RT)^{c} [D]^{d} (RT)^{d}}{[A]^{a} (RT)^{a} [B]^{b} (RT)^{b}} \end{aligned}

We can rearrange the equation to separate the molar concentration terms from the

RT

terms:

K_{p} = \underset{⏟}{\frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}} \cdot \frac{(RT)^{c} (RT)^{d}}{(RT)^{b} (RT)^{a}}

K_{c}

Hey, the first part looks pretty familiar! We can now simplify the giant mess of

RT

terms by combining the exponents:

\begin{aligned} K_{p} & = K_{c} (RT)^{(c + d) - (a + b)} \\ = K_{c} (RT)^{mol product gas - mol reactant gas} \\ = K_{c} (RT)^{Δ n} \end{aligned}

Where we have defined

Δ n

as follows:

Δ n = mol of product gas - mol of reactant gas

Let's practice using these equations in some examples!

Example 1: finding $K_{p}$ ‍ from partial pressures

Let's try finding

K_{p}

for the following gas-phase reaction:

$2 N_{2} O_{5} (g) ⇌ O_{2} (g) + 4 {NO}_{2} (g)$ ‍

We know the partial pressures for each component at equilibrium for some temperature

T

\begin{aligned} P_{N_{2} O_{5}} & = 2.00 atm \\ P_{O_{2}} & = 0.296 atm \\ P_{{NO}_{2}} & = 1.70 atm \end{aligned}

At temperature $T$ ‍, what is $K_{p}$ ‍ for this reaction?

First we can write the

K_{p}

expression for our balanced equation:

$K_{p} = \frac{(P_{O_{2}}) (P_{{NO}_{2}})^{4}}{(P_{N_{2} O_{5}})^{2}}$ ‍

We can now solve for

K_{p}

by plugging in the equilibrium partial pressures in the equilibrium expression:

$K_{p} = \frac{(0.296) (1.70)^{4}}{(2.00)^{2}} = 0.618$ ‍

Example 2: finding $K_{p}$ ‍ from $K_{c}$ ‍

Now let's look at a different reversible reaction:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$ ‍

If $K_{c}$ ‍ for this reaction is $4.5 \times 10^{4}$ ‍ at $400 K$ ‍, what is the equilibrium constant, $K_{p}$ ‍, at the same temperature?

Use the gas constant that will give

K_{p}

for partial pressure units of bar.

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

$K_{p} = K_{c} (RT)^{Δ n}$ ‍

To find

Δ n

, we compare the moles of gas from the product side of the reaction with the moles of gas on the reactant side:

$\begin{aligned} Δ n & = mol of product gas - mol of reactant gas \\ = 2 {mol NH}_{3} - (1 {mol N}_{2} + 3 {mol H}_{2}) \\ = - 2 mol gas \end{aligned}$ ‍

We can now substitute in our values for

K_{c}

T

, and

Δ n

to find

K_{p}

. We will want to keep track of the units of the gas constant

R

in our equation since that will determine if we are calculating

K_{p}

for partial pressures of bar or atm. Since we want to calculate

K_{p}

for when partial pressure has units of bar, we will use

R = 0.08314 \frac{L \cdot bar}{K \cdot mol}

\begin{aligned} K_{p} & = K_{c} (RT)^{Δ n} \\ = (4.5 \times 10^{4}) (R \cdot 400)^{- 2} \\ = (4.5 \times 10^{4}) (0.08314 \cdot 400)^{- 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_{p}

Example 3: find $K_{p}$ ‍ from total pressure

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

$2 H_{2} O (l) ⇌ 2 H_{2} (g) + O_{2} (g)$ ‍

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 $K_{p}$ ‍ for the reaction?

To do this problem, it might be helpful to visualize our partial pressures using an ICE table.

An ICE table is just a way of organizing and visualizing the concentration information in an equilibrium problem. ICE stands for: Initial concentration-Change in concentration-Final concentration.

We can fill out our ICE table using the known concentrations, and unknown concentrations are represented using a variable such as

x

. By using an ICE table and the equilibrium expression, it is possible to solve for

x

and calculate the initial, non-equilibrium concentrations and/or the final concentrations once the reaction reaches equilibrium.

Note that we don't include pure liquids in our calculations for

K_{p}

; 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.

Equation	$2 H_{2} O (l) ⇌$ ‍	$2 H_{2} (g)$ ‍	$O_{2} (g)$ ‍
Initial	N/A	$0 atm$ ‍	$0 atm$ ‍
Change	N/A	$+ 2 x$ ‍	$+ x$ ‍
Equilibrium	N/A	$2 x$ ‍	$x$ ‍

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

P_{O_{2}}

increases by

x

, the change for

P_{H_{2}}

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,

P_{total}

, is equal to the sum of the partial pressures for each of the components in the system:

$P_{total} = P_{A} + P_{B} + P_{C} + …$ ‍

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

$\begin{aligned} P_{total} & = P_{H_{2}} + P_{O_{2}} \\ = 2 x + x \\ = 3 x \end{aligned}$ ‍

Using our observed total pressure of 2.10atm, we can solve for

x

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

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:

$P_{H_{2}} = 2 x = 1.40 atm$ ‍

$P_{O_{2}} = x = 0.70 atm$ ‍

Now we can set up an equilibrium expression for the reaction and solve for

K_{p}

\begin{aligned} K_{p} & = (P_{H_{2}})^{2} \cdot P_{O_{2}} \\ = (1.40)^{2} \cdot (0.70) \\ = 1.37 \end{aligned}

Summary

The equilibrium constant $K_{p}$ ‍ describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
For a gas-phase reaction $aA (g) + bB (g) ⇌ cC (g) + dD (g)$ ‍, the expression for $K_{p}$ ‍ is

$K_{p} = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}}$ ‍

$K_{p}$ ‍ is related to the equilibrium constant in terms of molar concentration, $K_{c}$ ‍, by the equation below

$K_{p} = K_{c} (RT)^{Δ n}$ ‍

where $Δ n$ ‍ is

$Δ n = mol of product gas - mol of reactant gas$ ‍

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gizem.cetin.usa
Posted 8 years ago. Direct link to gizem.cetin.usa's post “Why do we not include liq...”
Why do we not include liquids and solids in equilibrium expressions? What do you mean by "their concentration is equal to 1"?
Button navigates to signup pageComment on gizem.cetin.usa's post “Why do we not include liq...”
(19 votes)
Answer
- Alex Wrede
  Posted 8 years ago. Direct link to Alex Wrede's post “The concentration of the ...”
  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
  Comment on Alex Wrede's post “The concentration of the ...”
  (35 votes)
Mushahid Osman
Posted 8 years ago. Direct link to Mushahid Osman's post “In example 2, how do you ...”
In example 2, how do you know that you are finding the constant in terms of bar and not atm?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- kausik sivakumar
  Posted 8 years ago. Direct link to kausik sivakumar's post “It is because the Univers...”
  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
  Button navigates to signup page
  (6 votes)
caroxxwu
Posted 6 years ago. Direct link to caroxxwu's post “For the equation Kp = Kc(...”
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?
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(7 votes)
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Clara I-W
Posted 8 years ago. Direct link to Clara I-W's post “Can anyone explain why Pr...”
Can anyone explain why Pressure does not change the equilibrium constant for a reaction but temperature does?
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(3 votes)
Answer
- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “The equilibrium constant ...”
  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.
  Button navigates to signup page
  (7 votes)
AJ
Posted 7 years ago. Direct link to AJ's post “For the ICE Table, how do...”
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?
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(4 votes)
Answer
- RogerP
  Posted 7 years ago. Direct link to RogerP's post “With the ICE table, you i...”
  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.
  Button navigates to signup page
  (2 votes)
Adit Khokar
Posted 8 years ago. Direct link to Adit Khokar's post “in the formula Kc = Kp * ...”
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
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(2 votes)
Answer
- ŞΣΔşђ ĐŗДġΦЇđ
  Posted 7 years ago. Direct link to ŞΣΔşђ ĐŗДġΦЇđ's post “but you cant arbitrarily ...”
  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)
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  (5 votes)
Lucian Rex
Posted 7 years ago. Direct link to Lucian Rex's post “When the do the PV=nRT, s...”
When the do the PV=nRT, shouldn't the final equation be P=nRT/V, not P=(n/V)RT?
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(1 vote)
Answer
- yuki
  Posted 7 years ago. Direct link to yuki's post “The two expressions P=nRT...”
  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.
  Button navigates to signup page
  (5 votes)
apndolby297
Posted 8 years ago. Direct link to apndolby297's post “As in example 3, why are ...”
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)?
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(2 votes)
Answer
- ŞΣΔşђ ĐŗДġΦЇđ
  Posted 7 years ago. Direct link to ŞΣΔşђ ĐŗДġΦЇđ's post “adding or removing liquid...”
  adding or removing liquid makes a very negligible difference in the pressure of the system.
  you can assume it as 1 without changing the answer.
  at least for calculation purposes
  besides taking any value apart from 1 will unnecessarily increase the difficulty of your calculations.
  Button navigates to signup page
  (3 votes)
Sneha Shrestha
Posted 7 years ago. Direct link to Sneha Shrestha's post “why it is difficult to fi...”
why it is difficult to find equilibrium constant at room temperature
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(3 votes)
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- Anshari Hasanbasri
  Posted 6 years ago. Direct link to Anshari Hasanbasri's post “Why is it difficult? You ...”
  Why is it difficult? You just need to plug in the room temperature for T in Kp=Kc(RT)^(delta n) (assuming you were to find Kp from Kc of the molecules)
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  (1 vote)
Swarna
Posted 7 years ago. Direct link to Swarna's post “In example 3, what does N...”
In example 3, what does N/A in the ICE table indicate?
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(2 votes)
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- RogerP
  Posted 7 years ago. Direct link to RogerP's post “N/A is a common abbreviat...”
  N/A is a common abbreviation for "Not applicable".
  Comment on RogerP's post “N/A is a common abbreviat...”
  (2 votes)

AP®︎/College Chemistry

Course: AP®︎/College Chemistry > Unit 11

Key points

Introduction: a short review of equilibrium and Kc‍

Kp‍ vs. Kc‍ : using partial pressure instead of concentration

Converting between gas concentration and partial pressure

Example 1: finding Kp‍ from partial pressures

Example 2: finding Kp‍ from Kc‍

Example 3: find Kp‍ from total pressure

Summary

Want to join the conversation?

Introduction: a short review of equilibrium and $K_{c}$ ‍

$K_{p}$ ‍ vs. $K_{c}$ ‍: using partial pressure instead of concentration

Example 1: finding $K_{p}$ ‍ from partial pressures

Example 2: finding $K_{p}$ ‍ from $K_{c}$ ‍

Example 3: find $K_{p}$ ‍ from total pressure