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

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

## Key points

• The equilibrium constant, $K_\text p$KpK, 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 $K_\text p$KpK, 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)dK, 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

• $K_\text p$KpK, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, $K_\text c$KcK, 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=molofproductgasmolofreactantgasdelta, 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 $K_\text c$Kc​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 $K$KK or $K_\text{c}$KcK, start subscript, start text, c, end text, end subscript to represent equilibrium constants. When we use the symbol $K_\text{c}$KcK, 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 $\dfrac{\text{mol solute}}{\text {L of solution}}$Lofsolutionmolsolutestart 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.

## $K_\text p$Kp​K, start subscript, start text, p, end text, end subscript vs. $K_\text c$Kc​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 $K_\text p$KpK, start subscript, start text, p, end text, end subscript. The subscript p stands for penguins.

[Seriously? Penguins??]

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 $\text P_\text A$PAstart text, P, end text, start subscript, start text, A, end text, end subscript, then the expression for $K_\text p$KpK, 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)dK, 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 $K_\text p$KpK, 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 $K_\text c$KcK, start subscript, start text, c, end text, end subscript.

• $K_\text p$KpK, start subscript, start text, p, end text, end subscript is often written without units. Since the value of $K_\text p$KpK, 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 $K_\text p$KpK, start subscript, start text, p, end text, end subscript problem.

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

• We can write $K_\text p$KpK, 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 $\text M$Mstart text, M, end text or $\dfrac{\text {mol}}{\text L}$Lmolstart 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 $\dfrac{\text n}{\text V}$Vnstart 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 $\dfrac{\text n}{\text V}$Vnstart 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 $K_\text c$KcK, start subscript, start text, c, end text, end subscript and $K_\text p$KpK, 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=molofproductgasmolofreactantgasdelta, 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

[How did we get this equation?]

Let's practice using these equations in some examples!

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

Let's try finding $K_\text p$KpK, 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)$2N2O5(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:

\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}PN2O5PO2PNO2=2.00atm=0.296atm=1.70atm

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

First we can write the $K_\text p$KpK, 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=(PN2O5)2(PO2)(PNO2)4K, 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 $K_\text p$KpK, 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

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

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 $K_\text c$KcK, 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, $K_\text p$KpK, start subscript, start text, p, end text, end subscript, at the same temperature?

Use the gas constant that will give $K_\text p$KpK, 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=molofproductgasmolofreactantgas=2molNH3(1molN2+3molH2)=2molgas

We can now substitute in our values for $K_\text c$KcK, 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 $K_\text p$KpK, 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 $K_\text p$KpK, start subscript, start text, p, end text, end subscript for partial pressures of bar or atm. Since we want to calculate $K_\text p$KpK, 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=0.08314KmolLbarstart 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}Kp=Kc(RT)Δn=(4.5×104)(R400)2=(4.5×104)(0.08314400)2=41

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

## Example 3: find $K_\text p$Kp​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\text H_2 \text O(l) \rightleftharpoons 2 \text H_2(g)+\text O_2(g)$2H2O(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 $K_\text p$KpK, 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.

[What is an ICE table??]

Note that we don't include pure liquids in our calculations for $K_\text p$KpK, 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.

Equation$2\text H_2 \text O(l) \rightleftharpoons$2H2O(l)2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons$2 \text H_2(g)$2H2(g)2, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis$\text O_2(g)$O2(g)start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis
InitialN/A$0\,\text {atm}$0atm0, start text, a, t, m, end text$0\,\text {atm}$0atm0, start text, a, t, m, end text
ChangeN/A$+2x$+2xplus, 2, x$+x$+xplus, x
EquilibriumN/A$2x$2x2, x$x$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 $\text P_{\text{O}_2}$PO2start 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 $\text P_{\text{H}_2}$PH2start 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, $\text P_\text{total}$Ptotalstart 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}Ptotalx=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 $K_\text p$KpK, 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}Kp=(PH2)2PO2=(1.40)2(0.70)=1.37

## Summary

• The equilibrium constant $K_\text p$KpK, 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 $K_\text p$KpK, 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)dK, 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

• $K_\text p$KpK, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, $K_\text c$KcK, 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=molofproductgasmolofreactantgasdelta, 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

## FAQs

### How do you find the equilibrium constant of KP? ›

The general expression: Kp = Kc(RT) ∆n can be derived where ∆n = moles of gaseous products - moles of gaseous reactants.

How to write equilibrium constant expression in terms of partial pressure? ›

First, write Keq (equilibrium constant expression) in terms of activities.
1. K=(aNH3)2(aN2)(aH2)3.
2. Kp=(PNH3)2(PN2)(PH2)3.
3. Kp=(0.003)2(0.094)(0.039)3=1.61.
Jan 29, 2023

What is the equilibrium constant with respect to partial pressure KP? ›

Equilibrium constant expression in terms of partial pressure is designated as Kp. Equilibrium constant Kp is equal to the partial pressure of products divided by partial pressure of reactants and the partial pressure are raised with some power which is equal to the coefficient of the substance in balanced equation.

Does KP use partial pressures? ›

Writing an expression for Kp

Kp has exactly the same format as Kc, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.

Does KP depend on partial pressure? ›

Pressure does not affect the value of Kp, just as concentration does not affect the value of Kc. An increase in pressure causes equilibrium to shift in favor of the direction with the fewer moles so that the pressure decreases. The partial pressure ratio of reactant to products stays the same so Kp does not change.

What is the formula of equilibrium constant in pressure? ›

∆ n = mol e s of product gas - mol e s of reactant gas.

What is partial pressure formula? ›

Using diving terms, partial pressure is calculated as: partial pressure = (total absolute pressure) × (volume fraction of gas component)

What is the equilibrium expression Kp? ›

Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures.

Are Kp and Kp the equilibrium constant? ›

For a given exothermic reaction , Kp and k'p are the equilibrium constants at temperatures T1 and T2 respectively. Assuming that heat of reaction is constant in temperature range reaction is constant in temperature range between T1andT2 , it is readily observed that.

What is the relation between KP and pressure? ›

Kp And Kc are the equilibrium constant of an ideal gaseous mixture. Kp is equilibrium constant used when equilibrium concentrations are expressed in atmospheric pressure and Kc is equilibrium constant used when equilibrium concentrations are expressed in molarity.

### Is KP and KC are in equilibrium constant in terms of partial pressure and concentration? ›

Kc and Kp are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

What is the relation between KP and total pressure? ›

An increase in pressure will lead to an increase in Kx to maintain a constant value of Kp. So the reaction will shift to form more of the products C and D.

What happens to the equilibrium constant KP when pressure is increased? ›

Equilibrium constants are not changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature.

Why KP does not depend on pressure? ›

Solution: The equilibrium constant depends on the f(temperature), so all the reaction Kp does not depend upon the initial pressure of reactants.

How does partial pressure affect equilibrium? ›

1) When the partial pressure of any of the gaseous reactants or of the products is increased, the position of equilibrium is shifted so as to decrease its partial pressure. This is usually achieved by favoring the reaction in which there is decrease in the number of moles of gaseous components.

What is the rule of partial pressure? ›

According to Dalton's law of partial pressures, the total pressure by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases. The partial pressure is defined as the pressure each gas would exert if it alone occupied the volume of the mixture at the same temperature.

What is partial pressure for dummies? ›

In a mixture of gases, each gas contributes to the total pressure of the mixture. This contribution is the partial pressure. The partial pressure is the pressure the gas if the gas were in the same volume and temperature by itself.

What is the formula of KP and KC in equilibrium? ›

Relationship between Kp and Kc is Kp = Kc(RT)^Δn .

How is KP related to pressure? ›

Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unitless number, although it relates the pressures.

What is the formula with partial pressure? ›

Using diving terms, partial pressure is calculated as: partial pressure = (total absolute pressure) × (volume fraction of gas component) For the component gas "i": pi = P × F.

### What is KP in equilibrium? ›

Kc = Equilibrium constant measured in moles per liter. Kp = Equilibrium constant calculated from the partial pressures.

How do we solve mathematically in law of partial pressure? ›

3.
1. Find the number of moles of oxygen and nitrogen using PV=nRT which is n=PV/RT. oxygen: ((1 atm)(12L))/(0.08206 atm L mol-1 K-1)(273 K)=0.536 moles oxygen. ...
2. Use PV=nRT or P=(nRT)/V to find the total pressure. ...
3. PA/Ptot=nA/Ntot can be rearranged to PA=(Ptot)(nA/Ntot) to find the partial pressures.
Jan 29, 2023

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