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Relation between Kp and Kc

Relation between Kp and Kc

Edited By Shivani Poonia | Updated on Jul 02, 2025 06:01 PM IST

The equilibrium constants ( Kp ) and ( Kc ) 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 ( Kp ) and ( Kc ) provide crucial information about the position of equilibrium for a reaction.

This Story also Contains
  1. Relation Between Kp And Kc
  2. Some Solved Examples
  3. Summary
Relation between Kp and Kc
Relation Between Kp and Kc

Kc is the equilibrium constant expressed in terms of the concentrations of reactants and products, while ( Kp ) 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 ( Kp ) and ( Kc ) 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 ( Kc ) or ( Kp ) indicates a reaction that favors the formation of products, while a small value suggests a reaction that favors the reactants.

Relation Between Kp And Kc

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

\therefore 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 ( Kc ) (the equilibrium constant for concentration) and ( Kp ) (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.

Frequently Asked Questions (FAQs)

1. What is the relationship between Kp and Kc?
Kp and Kc are equilibrium constants related to gas-phase reactions. Kp is the equilibrium constant in terms of partial pressures, while Kc is in terms of concentrations. They are related by the equation Kp = Kc(RT)^Δn, where R is the gas constant, T is temperature in Kelvin, and Δn is the change in the number of moles of gas from reactants to products.
2. Why do we need both Kp and Kc?
We use both Kp and Kc because some reactions are more conveniently expressed in terms of pressure (Kp), while others are better described using concentration (Kc). Kp is often used for gas-phase reactions, while Kc is more common for reactions in solution or involving solids and liquids.
3. When are Kp and Kc equal?
Kp and Kc are equal when there is no change in the number of moles of gas from reactants to products (Δn = 0). This occurs in reactions where the number of gaseous molecules on both sides of the equation is the same.
4. How does temperature affect the relationship between Kp and Kc?
Temperature directly affects the relationship between Kp and Kc through the equation Kp = Kc(RT)^Δn. As temperature increases, the value of RT increases, potentially changing the relationship between Kp and Kc if Δn is not zero.
5. What does Δn represent in the Kp-Kc relationship?
Δn represents the change in the number of moles of gas from reactants to products. It is calculated by subtracting the number of moles of gaseous reactants from the number of moles of gaseous products in the balanced chemical equation.
6. Can Kp be less than Kc?
Yes, Kp can be less than Kc when Δn is negative, meaning there are fewer moles of gas on the product side than on the reactant side. In this case, (RT)^Δn will be less than 1, making Kp smaller than Kc.
7. How do you convert Kp to Kc?
To convert Kp to Kc, use the equation Kc = Kp / (RT)^Δn. You need to know the temperature (T) and the change in the number of moles of gas (Δn) for the reaction.
8. Why is the unit of R important in the Kp-Kc relationship?
The unit of R is crucial because it must be consistent with the units used for pressure and concentration. Typically, R is expressed in L⋅atm/(mol⋅K) when working with standard conditions, ensuring dimensional consistency in the equation Kp = Kc(RT)^Δn.
9. How does the Kp-Kc relationship change for reactions involving only solids and liquids?
For reactions involving only solids and liquids, Δn = 0 because there are no gaseous species. In this case, Kp = Kc, regardless of temperature or pressure.
10. What happens to the Kp-Kc relationship in very high-pressure systems?
In very high-pressure systems, the ideal gas law assumptions break down, and the simple Kp-Kc relationship may no longer hold. More complex equations of state are needed to accurately describe the system behavior.
11. How does the presence of a catalyst affect the Kp-Kc relationship?
A catalyst does not affect the Kp-Kc relationship. Catalysts speed up the rate at which equilibrium is reached but do not change the equilibrium position or the values of Kp and Kc.
12. Can you have a Kp value but no corresponding Kc value?
Yes, this can occur in reactions where all species are gases and concentrations are difficult to measure or define. In such cases, it's more practical to work with partial pressures and Kp.
13. How does the magnitude of Kp compare to Kc when Δn is positive?
When Δn is positive (more moles of gas in products than reactants), Kp will be greater than Kc. This is because (RT)^Δn will be greater than 1, making Kp larger than Kc in the equation Kp = Kc(RT)^Δn.
14. Why is it important to specify standard conditions when discussing Kp and Kc?
Standard conditions (typically 1 atm pressure and 25°C) are important because equilibrium constants depend on temperature. Specifying standard conditions ensures that values can be compared across different sources and experiments.
15. How does the Kp-Kc relationship help in understanding Le Chatelier's principle?
The Kp-Kc relationship helps illustrate how pressure changes affect equilibrium. For reactions where Δn ≠ 0, changing pressure will shift the equilibrium, which is reflected in the different responses of Kp and Kc to these changes.
16. Can Kp and Kc have different units?
Yes, Kp and Kc often have different units. Kp is typically unitless (assuming pressures are in atm), while Kc usually has units related to concentration (e.g., mol/L) raised to some power, depending on the reaction stoichiometry.
17. How does the ideal gas law relate to the Kp-Kc relationship?
The ideal gas law (PV = nRT) is fundamental to deriving the Kp-Kc relationship. It allows us to relate pressure to concentration through the equation P = CRT, which is the basis for the Kp = Kc(RT)^Δn relationship.
18. What role does the gas constant R play in the Kp-Kc equation?
The gas constant R serves as a proportionality factor in the Kp-Kc relationship. It ensures that the units are consistent and allows for the conversion between pressure and concentration-based equilibrium constants.
19. How would you experimentally determine both Kp and Kc for a reaction?
To determine Kp, you would measure the partial pressures of gases at equilibrium. For Kc, you'd measure equilibrium concentrations. You could then use the Kp-Kc relationship to verify your results or calculate one constant from the other.
20. Why is Δn always an integer in the Kp-Kc relationship?
Δn is always an integer because it represents the change in the number of moles of gas molecules, which must be a whole number in a balanced chemical equation.
21. How does the Kp-Kc relationship apply to heterogeneous equilibria?
In heterogeneous equilibria, only gaseous species are considered in the Kp-Kc relationship. Solids and pure liquids do not contribute to Δn or appear in the equilibrium constant expressions.
22. Can you use the Kp-Kc relationship for non-ideal gases?
The Kp-Kc relationship as typically presented assumes ideal gas behavior. For non-ideal gases, you would need to use fugacity instead of pressure and activity coefficients instead of concentrations, leading to a more complex relationship.
23. How does the Kp-Kc relationship change with altitude?
The Kp-Kc relationship itself doesn't change with altitude, but the values of Kp and Kc might change due to different atmospheric pressures and temperatures at different altitudes, affecting the equilibrium position.
24. Why is it sometimes easier to measure Kp than Kc for gas-phase reactions?
For gas-phase reactions, measuring pressure is often simpler and more direct than measuring concentration, especially at high temperatures or in closed systems. This makes Kp more convenient to determine experimentally in many cases.
25. How does the Kp-Kc relationship help in understanding the effect of adding an inert gas to an equilibrium system?
The Kp-Kc relationship helps explain why adding an inert gas at constant volume doesn't affect the equilibrium position. While total pressure increases, partial pressures (and thus Kp) remain constant, consistent with no change in Kc.
26. Can the Kp-Kc relationship be used for reactions in solution?
The Kp-Kc relationship is primarily used for gas-phase reactions. For reactions in solution, we typically work with Kc or other concentration-based equilibrium constants, as pressure is less relevant in liquid systems.
27. How does the Kp-Kc relationship relate to the concept of fugacity in non-ideal gases?
For non-ideal gases, fugacity (f) replaces pressure (P) in equilibrium calculations. The relationship becomes Kf = Kc(RT)^Δn, where Kf is the equilibrium constant in terms of fugacities. This allows the relationship to be extended to non-ideal conditions.
28. Why doesn't the Kp-Kc relationship depend on the stoichiometric coefficients of a reaction?
The Kp-Kc relationship depends only on Δn, which is the change in total moles of gas. Stoichiometric coefficients affect the form of Kp and Kc individually, but their ratio (and thus the relationship between them) depends only on the net change in gas moles.
29. How does the Kp-Kc relationship help in understanding the effect of pressure on equilibrium yield?
The Kp-Kc relationship shows that for reactions where Δn ≠ 0, changing pressure will affect the equilibrium position. If Δn > 0, increasing pressure will favor the reverse reaction (decreasing yield), while if Δn < 0, increasing pressure will favor the forward reaction (increasing yield).
30. Can you use the Kp-Kc relationship to predict the direction of a reaction?
The Kp-Kc relationship itself doesn't predict reaction direction, but knowing both Kp and Kc can help. If the reaction quotient Q is less than K, the reaction will proceed forward; if Q is greater than K, it will proceed in reverse.
31. How does the Kp-Kc relationship apply to reactions with fractional stoichiometric coefficients?
The Kp-Kc relationship applies the same way for reactions with fractional coefficients. Δn is still calculated as the difference between the sum of gaseous product coefficients and the sum of gaseous reactant coefficients, which may result in a fractional Δn.
32. Why is it important to use the same temperature when comparing Kp and Kc values?
Temperature affects both Kp and Kc individually and their relationship through the RT term. Using the same temperature ensures that you're comparing equilibrium constants under the same conditions, allowing for meaningful comparisons and calculations.
33. How does the Kp-Kc relationship help in understanding the concept of partial pressure?
The Kp-Kc relationship reinforces the concept of partial pressure by showing how the pressure-based equilibrium constant (Kp) relates to the concentration-based constant (Kc). It illustrates that partial pressures, like concentrations, are measures of the amount of a gas in a mixture.
34. Can the Kp-Kc relationship be used to calculate the equilibrium composition of a gas mixture?
Yes, if you know either Kp or Kc and the initial composition of the gas mixture, you can use the Kp-Kc relationship along with the equilibrium constant expression to calculate the equilibrium composition of the gas mixture.
35. How does the Kp-Kc relationship relate to the van 't Hoff equation?
While the Kp-Kc relationship relates pressure and concentration-based equilibrium constants, the van 't Hoff equation describes how these constants change with temperature. Both are important for understanding equilibrium behavior under different conditions.
36. Why is it necessary to specify the phase of each species when using the Kp-Kc relationship?
Specifying the phase is crucial because only gaseous species contribute to Δn in the Kp-Kc relationship. Solids and liquids are typically omitted from Kp and Kc expressions, so knowing the phase helps in correctly applying the relationship.
37. How does the Kp-Kc relationship help in understanding the concept of standard state in thermodynamics?
The Kp-Kc relationship uses standard conditions (often 1 atm and 25°C) as a reference point. This reinforces the concept of standard state in thermodynamics, where properties are measured or calculated under specific, defined conditions for consistency and comparison.
38. Can the Kp-Kc relationship be used for reactions at extremely high temperatures?
The basic Kp-Kc relationship can be used at high temperatures, but caution is needed. At extremely high temperatures, gases may deviate significantly from ideal behavior, and other factors like dissociation or ionization may come into play, potentially complicating the relationship.
39. How does the Kp-Kc relationship relate to the concept of chemical potential?
While not directly related, both the Kp-Kc relationship and chemical potential are important in understanding equilibrium. The relationship between Kp and Kc reflects how the system responds to changes in pressure and concentration, which are related to changes in chemical potential.
40. Why is it important to consider significant figures when using the Kp-Kc relationship?
Significant figures are important in the Kp-Kc relationship to maintain the precision of your calculations. The number of significant figures in your result should reflect the precision of your input values for Kp or Kc, temperature, and Δn.

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