Relation between Kp and Kc

The equilibrium constants ( K_p ) and ( K_c ) are fundamental concepts in chemical equilibrium, describing the relationship between the concentrations or partial pressures of reactants and products in a chemical reaction at equilibrium. In chemical thermodynamics, the equilibrium constants ( K_p ) and ( K_c ) provide crucial information about the position of equilibrium for a reaction.

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This Story also Contains

1. Relation Between Kp And Kc
2. Some Solved Examples
3. Summary

K_c is the equilibrium constant expressed in terms of the concentrations of reactants and products, while ( K_p ) is expressed in terms of the partial pressures of gases. The relationship between ( Kp ) and ( Kc ) is particularly important for understanding how changes in pressure and temperature affect chemical equilibria.

The specific relationship between ( Kp ) and ( Kc ) was derived later as the field of chemical thermodynamics evolved. The equilibrium constants ( K_p ) and ( K_c ) are extremely helpful in various ways. They provide insight into the extent to which a reaction will proceed to form products or remain as reactants. A large value of ( K_c ) or ( K_p ) indicates a reaction that favors the formation of products, while a small value suggests a reaction that favors the reactants.

Relation Between K_p And K_c

Let us suppose we react:
$
\mathrm{n}_1 \mathrm{~A}(\mathrm{~g})+\mathrm{n}_2 \mathrm{~B}(\mathrm{~g}) \leftrightharpoons \mathrm{n}_3 \mathrm{C}(\mathrm{g})+\mathrm{n}_4 \mathrm{D}(\mathrm{g})
$

The equilibrium constant $\mathrm{K}_{\mathrm{c}}$ for this reaction is given as:
$
\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{C}]^{\mathrm{n}_3}[\mathrm{D}]^{\mathrm{n}_4}}{[\mathrm{~A}]^{\mathrm{n}_1}[\mathrm{~B}]^{\mathrm{n}_2}}
$

For the reaction:
$
\mathrm{n}_1 \mathrm{~A}(\mathrm{~g})+\mathrm{n}_2 \mathrm{~B}(\mathrm{~g}) \leftrightharpoons \mathrm{n}_3 \mathrm{C}(\mathrm{g})+\mathrm{n}_4 \mathrm{D}(\mathrm{g})
$

The equilibrium constant $\mathrm{K}_{\mathrm{p}}$ is given as:
$
K_p=\frac{\left(P_C\right)^{n_3}\left(P_D\right)^{n_4}}{\left(P_A\right)^{n_1}\left(P_B\right)^{n_2}}
$

Now, from the Ideal gas Equation
$
\mathrm{PV}=\mathrm{nRT}
$

$
\begin{aligned}
& \mathrm{P}=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT} \\
& \mathrm{P}=\mathrm{CRT}
\end{aligned}
$

Putting the value of $P$ in terms of $C$ in the expression for $K_P$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=\frac{[\mathrm{C}]^{\mathrm{n}_3}(\mathrm{RT})^{\mathrm{n}_3}[\mathrm{D}]^{\mathrm{n}_4}(\mathrm{RT})^{\mathrm{n}_4}}{[\mathrm{~A}]^{\mathrm{n}_1}(\mathrm{RT})^{\mathrm{n}_1}[\mathrm{~B}]^{\mathrm{n}_2}(\mathrm{RT})^{\mathrm{n}_2}} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{[\mathrm{C}]^{\mathrm{n}_3}[\mathrm{D}]^{\mathrm{n}_4}}{[\mathrm{~A}]^{\mathrm{n}_1}[\mathrm{~B}]^{\mathrm{n}_2}}[\mathrm{RT}]^{\left.\mathrm{n}_3+\mathrm{n}_4\right)-\left(\mathrm{n}_1+\mathrm{n}_2\right)}
\end{aligned}
$

Putting $\Delta \mathrm{n}_{\mathrm{g}}=\left(\mathrm{n}_3+\mathrm{n}_4\right)-\left(\mathrm{n}_1+\mathrm{n}_2\right)$, we have
$
\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_5}
$

It can be seen that:
- When $\Delta \mathrm{ng}_{\mathrm{g}}=0$, then $\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}$
. When $\Delta \mathrm{ng}_{\mathrm{g}}>0$, then $\mathrm{K}_{\mathrm{P}}>\mathrm{K}_{\mathrm{Q}}$
When $\Delta \mathrm{n}_{\mathrm{g}}<0$, then $\mathrm{K}_{\mathrm{P}}<\mathrm{K}_{\mathrm{C}}$

Recommended topic video on(Relation between Kp and Kc)

Some Solved Examples

Example 1. An amount of solid NH₄HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH₃ and H₂S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.85 atm. The equilibrium constant forNH₄HS decomposition at this temperature is

1)0.30

2)0.18

3) 0.17

4)0.11 (correct)

\begin{aligned}
&\text { Solution }\\
&\begin{array}{lccc}
& \mathrm{NH}_4 \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \\
\text { Initial pressure } & 0 & 0.5 & 0 \\
\text { At eq. } & 0 & 0.5+\mathrm{x} & \mathrm{x}
\end{array}
\end{aligned}

Total pressure = 0.5 + 2x = 0.84

x = 0.17 atm

Now,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{p}}=\mathrm{pNH}_3 \times \mathrm{pH}_2 \mathrm{~S} \\
& \mathrm{~K}_{\mathrm{p}}=(0.5+0.17)(0.17)=0.11 \mathrm{~atm}^2
\end{aligned}
$

Example 2. Two solids dissociate as follows

$\begin{aligned} & A(s) \rightleftharpoons B(g)+C(g) ; K_{p 1}=x a t m^2 \\ & D(s) \rightleftharpoons C(g)+E(g) ; K_{p 2}=y a t m^2\end{aligned}$

The total pressure when both the solids dissociate simultaneously is :

1) $(x+y) \operatorname{atm} 2)$
2) $\sqrt{x+y} \mathrm{~atm}$
3) $x^2+y^2 a t m$
4) $($ correct) $2(\sqrt{x+y}) \mathrm{atm}$

Solution

Relation between pressure and concentration -
$
P V=n R T
$
or $P=\frac{n}{V} R T$
or $P=C R T$
$R=0.0831$ bar inter $/ \mathrm{mol} K$

- wherein

P is pressure in Pa. C is concentration in mol/liter. T is the temperature in kelvin

As we have learned in total pressure at equilibrium
$
\begin{aligned}
& A_{(s)} \rightleftharpoons B_{(g)}+C_{(g)} K_p=\text { xatm }^2 \\
& D_{(s)} \rightleftharpoons C_{(g)}+E_{(g)} K p_2=\text { yatm }^2 \\
& p_1+p_2 \quad p_2 \\
& K p_1=p_1\left(p_1+p_2\right) \quad \text { Kp } p_2=p_2\left(p_1+p_2\right) \\
& K p_2=p_2\left(p_1+p_2\right)^2 \\
& x+y=\left(p_{1+} p_2\right)^2 \\
& \left(p_{1+} p_2\right)=\sqrt{x+y} \\
& P_{\text {total }}=P_b+P_c+P_e \\
& =2\left(p_{1+} p_2\right)=2 \sqrt{x+y}
\end{aligned}
$

Hence, the answer is the option (4).

Example 3. Consider the reaction $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$. The equilibrium constant of the above reaction is $K_p$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by ( Assume that $P_{N H_3}<<p_{\text {total }}$ at equilibrium )

$\begin{aligned} & \text { 1) (correct) } \frac{3^{\frac{3}{2}} K_p^{\frac{1}{2}} P^2}{16} \\ & \text { 2) } \frac{3^{\frac{3}{2}} K_p^{\frac{1}{2}} P^2}{4} \\ & \text { 3) } \frac{K_p^{\frac{1}{2}} P^2}{4} \\ & \text { 4) } \frac{K_P^{\frac{1}{2}} P^2}{16}\end{aligned}$

Solution
$
\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \quad \text { Equilibrium Constant }=K_p
$

Now,
$
\begin{aligned}
& 2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2+3 \mathrm{H}_2 ; K_p^{\prime}=\frac{1}{K_p} \\
& P_{\text {total }}=P=P_{\mathrm{N}_2}+P_{\mathrm{H}_2}+P_{\mathrm{NH}_3}
\end{aligned}
$

Due to $P_{\mathrm{NH}_3}<<P_{\text {Total }}$
$
\mathrm{P}_{\text {total }} \approx \mathrm{P}_{\mathrm{N}_2}+\mathrm{P}_{\mathrm{H}_2}
$

Total moles are 4,1 of $\mathrm{NH}_3$, and 3 of $\mathrm{H}_2$.
Partial Pressure of $\mathrm{A}=\frac{\text { moles of } \mathrm{A}}{\text { Total moles }} \times$ Total Pressure

$\begin{aligned} & \mathrm{P}_{\mathrm{N}_2}=\frac{1}{4} \times \mathrm{P} \text { and } \mathrm{P}_{\mathrm{H}_2}=\frac{3}{4} \times \mathrm{P} \\ & \frac{1}{K_P}=\frac{P_{N_2}\left(P_{H_3}\right)^3}{\left(P_{N_3}\right)^2}=\frac{\left(\frac{P}{4}\right)\left(\frac{3 P}{4}\right)^3}{\left(P_{N H_3}\right)^2} \\ & \left(P_{N H_3}\right)^2=\frac{3^3 P^4}{4^4} K_p\end{aligned}$

Example 4. For the reaction, $\mathrm{CO}_{(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{COCl}_{2(\mathrm{~g})}$ the value of $\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{c}}}$ is equal to

1) $\left(\right.$ correct) $\frac{1}{R T}$
2)RT
3) $\sqrt{\mathrm{RT}}$
4) 1.0

Solution

Relation between $K p$ and $K_c-K_p=K_c(R T)^{\triangle n}$
Now, for the given reaction, $\mathrm{CO}_{(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{COCl}_{2(\mathrm{~g})}$
$
\begin{aligned}
& \Delta \mathrm{n}=-1 \\
& \therefore \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}
\end{aligned}
$

Thus $\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{c}}}=\frac{1}{\mathrm{RT}}$

Hence, the answer is the option (1).

Example 5. For the reaction,
$\mathrm{SO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{3(\mathrm{~g})}$ if $\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\mathrm{x}}$ where the symbols have the usual meaning then the value of x is

1) -1
2) (correct) $-\frac{1}{2}$
3) $\frac{1}{2}$
4) 1

Solution

We know that

Relation between Kp and Kc -
$
\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}}
$

According to the data given in the question
$
\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\mathrm{x}}
$

For the given reaction
$
\begin{aligned}
& \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g}) \\
& \Delta \mathrm{n}=1-\frac{3}{2}=-\frac{1}{2}
\end{aligned}
$

Hence, the value of $x$ is $-\frac{1}{2}$

Hence, the answer is an option (2).

Summary

The relationship between ( K_c ) (the equilibrium constant for concentration) and ( K_p ) (the equilibrium constant for partial pressures) is crucial in understanding chemical equilibria, especially for reactions involving gases. This relation has various benefits such as predicting reaction behavior: Knowing the values of ( Kc ) and ( Kp ) helps to predict the direction and extent of a reaction under varying conditions.