Equilibrium constant

Equilibrium constant

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

The concept of equilibrium constant was developed by the work of various scientists. Cato Maximilian Gulberg and Peter Waage as they worked on the law of mass action they are also working on the concept of the equilibrium constant in their formulation they proposed the rate of reaction is proportional to the concentration of the reactant and at equilibrium the ratio of the concentration of product to reactant is constant. Further another scientist Jacobus Henricus Vant Hoff developed the theory by relating the equilibrium constant to the temperature and predicting the impact of temperature change on the position of equilibrium.

Equilibrium constant
Equilibrium constant

Equilibrium Constant

It is the ratio of the rate of forward and backward reaction at a particular temperature or it is the ratio of active masses of the reactants to that of active masses of products at a particular temperature raised to their stoichiometric coefficients. It is denoted by Kc or Kp. The distinction between Keq and Kc is that the expression of Keq involves all the species (whether they are pure solids, pure liquids, gases, solvents or solutions) while the expression Kc involves only those species whose concentration is a variable (gases and solution). It means Kc is a devoid of pure components (like pure solids and pure liquids) and solvents.

Let us look at the definition of Equilibrium constant in terms of concentration and partial pressure

(1) Equilibrium constant in Terms of Concentration

For a reaction:

$\begin{aligned} & \mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD} \\ & \mathrm{r}_{\text {forward }} \propto[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}=\mathrm{K}_{\mathrm{f}}[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}} \\ & \mathrm{r}_{\text {backward }} \propto[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}=\mathrm{K}_{\mathrm{b}}[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}\end{aligned}$

We know that at equilibrium

$\begin{aligned} & \mathrm{r}_{\mathrm{f}}=\mathrm{r}_{\mathrm{b}} \\ & \mathrm{K}_{\mathrm{f}}[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}=\mathrm{K}_{\mathrm{b}}[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}} \\ & \frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}} \quad \text { (at constant temperature) } \\ & \frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}}=\mathrm{K}_{\mathrm{c}}\end{aligned}$

The above expression gives us the value of Kc as the activity or the active mass is expressed in terms of the concentrations (c) or the molarity

(2) Equilibrium constant in terms of Partial pressure

In this case, the equilibrium constant is known as Kp. It is applicable only for gaseous systems.
For the reaction:
$\begin{aligned} & \mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD} \\ & \mathrm{r}_{\text {forward }} \propto \mathrm{P}_{\mathrm{A}}^{\mathrm{m}} \mathrm{P}_{\mathrm{B}}^{\mathrm{n}}=\mathrm{K}_{\mathrm{f}_1} \mathrm{P}_{\mathrm{A}}^{\mathrm{m}} \mathrm{P}_{\mathrm{B}}^{\mathrm{n}} \\ & \mathrm{r}_{\text {backward }} \propto \mathrm{P}_{\mathrm{C}}^{\mathrm{p}} \mathrm{P}_{\mathrm{D}}^{\mathrm{q}}=\mathrm{K}_{\mathrm{b}_1} \mathrm{P}_{\mathrm{C}}^{\mathrm{p}} \mathrm{P}_{\mathrm{D}}^{\mathrm{q}}\end{aligned}$

At equilibrium

$\begin{aligned} & \mathrm{r}_{\mathrm{f}}=\mathrm{r}_{\mathrm{b}} \\ & \mathrm{K}_{\mathrm{f}_1} \mathrm{P}_{\mathrm{A}}^{\mathrm{m}} \mathrm{P}_{\mathrm{B}}^{\mathrm{n}}=\mathrm{K}_{\mathrm{b}_1} \mathrm{P}_{\mathrm{C}}^{\mathrm{p}} \mathrm{P}_{\mathrm{D}}^{\mathrm{q}} \\ & \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{K}_{\mathrm{f}_1}}{\mathrm{~K}_{\mathrm{b}_2}}=\frac{\mathrm{P}_{\mathrm{C}}^{\mathrm{p}} \mathrm{P}_{\mathrm{D}}^{\mathrm{q}}}{\mathrm{P}_{\mathrm{A}}^{\mathrm{m}} \mathrm{P}_{\mathrm{B}}^{\mathrm{n}}}\end{aligned}$

The above expression gives us the value of KP as the activity is expressed in terms of partial pressures.

Note: This is generally used for gaseous systems or systems where gases are in equilibrium with liquids or solids

Recommemded topic video on (Equilibrium constant)


Some Solved Examples

1.The equilibrium constant (KC) for the reaction $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature T is $4 \times 10^{-4}$. The value of KC for the given reaction at the same temperature is :

$\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$

1)0.02

2)2.5 x 10 2

3)4 x 10 -4

4) (correct)50

Solution

As we discussed in the concept

The equilibrium constant for the reverse reaction -

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

$\begin{aligned} & \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})} \rightarrow(\mathrm{i}) \\ & \mathrm{K}_{\mathrm{c}}=4 \times 10^{-4}\end{aligned}$

By multiplying the equation (i) by $\frac{1}{2}$

$\begin{aligned} & \frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{NO}_{(\mathrm{g})} \rightarrow \text { (ii) } \\ & \mathrm{K}_{\mathrm{c}}^{\prime}=\sqrt{\mathrm{K}_{\mathrm{c}}}=\sqrt{4 \times 10^{-4}}=2 \times 10^{-2}\end{aligned}$

By reversing the equation (ii), we get

$\mathrm{NO}_{(\mathrm{g})} \rightarrow \frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \mathrm{K}_{\mathrm{c}}^{\prime \prime}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^{\prime}}=\frac{1}{2 \times 10^{-2}}=50$

Hence, the answer is the option (4).

2. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is

1) (correct)1.8 atm

2)3 atm

3)0.3 atm

4)0.18 atm


Solution

Given,

$\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \leftrightharpoons 2 \mathrm{CO}_{(\mathrm{g})}$

Initial pressure 0.5 atm 0 0

Pressure at equilibrium 0.5-x 2x

Total Pressure = 0.5 - x + 2x = 0.8

x = 0.3

$\begin{aligned} & k=\frac{(p C O)^2}{\left(p C O_2\right)} \\ & k=\frac{(0.6)^2}{0.2} \\ & =1.8 \mathrm{~atm}\end{aligned}$

Hence, the answer is the option (1)

3 .Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows. $\mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}$ At equilibrium, ammonia is added such that partial pressures of NH3 now equals the original total pressure. Calculate the ratio of the total pressure now to the original total pressure.

1) (correct)$\frac{31}{27}$

2)$\frac{60}{40}$

3)$\frac{31}{9}$

4)$\frac{62}{27}$

Solution

The reaction:-

Initial $\mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons \underset{2 \mathrm{P}}{2 \mathrm{NH}_{3(\mathrm{~g})}}+\underset{\mathrm{P}}{\mathrm{CO}_{2(\mathrm{~g})}}$

$\begin{aligned} & \mathrm{K}_{\mathrm{p}}=\left(\mathrm{P}_{\mathrm{NH}_3}\right)^2\left(\mathrm{P}_{\mathrm{CO}_2}\right) \\ & \mathrm{P}_{\mathrm{T}}(\text { initial })=3 \mathrm{P} \\ & \quad \mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})} \\ & \text { Initial } \\ & \mathrm{K}_{\mathrm{P}}=(3 \mathrm{P})^2\left(\mathrm{P}^{\prime}\right) \quad \ldots \ldots \ldots \ldots \text { (ii) }\end{aligned}$

From eq (I) and (ii)

$\begin{aligned} & (2 \mathrm{P})^2(\mathrm{P})=(3 \mathrm{P})^2\left(\mathrm{P}^{\prime}\right) \\ & \mathrm{P}^{\prime}=\frac{4 \mathrm{P}}{9}\end{aligned}$

$\frac{\mathrm{P}_{\mathrm{T}}(\text { new })}{\mathrm{P}_{\mathrm{T}}(\text { old })}=\frac{3 \mathrm{P}+\mathrm{P}^{\prime}}{3 \mathrm{P}}=\frac{3 \mathrm{P}+\frac{4 \mathrm{P}}{9}}{3 \mathrm{P}}=\frac{31}{27}$

Hence, the answer is the option (1).

4. For the hypothetical reactions, the equilibrium constant (K) values are given:

$\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}_1=2 ; \quad \mathrm{B} \rightleftharpoons \mathrm{C} ; \mathrm{K}_2=4 ; \quad \mathrm{C} \rightleftharpoons \mathrm{D} ; \mathrm{K}_3=3$
The equilibrium constant (K) for the reaction $A \rightleftharpoons D$ is:

1)12

2) (correct)24

3)6

4)9

Solution

The required reaction $A \rightleftharpoons D$ can be obtained by adding all the given reactions

We know that the Equilibrium constants get multiplied when the equations are added

$\begin{aligned} & \therefore \mathrm{K}=\mathrm{K}_1 \times \mathrm{K}_2 \times \mathrm{K}_3 \\ & \Rightarrow \mathrm{K}=2 \times 4 \times 3=24\end{aligned}$

Hence, the answer is the option (2).

5 .Value of equilibrium constant depends upon:

1)Temperature

2)Method of expressing activity or active mass

3) (correct)Both 1 and 2

4)Volume

Solution

Equilibrium constants are changed if you change the temperature of the system. Kc or Kp is constant at a constant temperature, but they vary as the temperature changes.

  • The equilibrium constant K is determined by the activities of the components in the equilibrium expression.

  • The values of Kc and Kp can be different in magnitude as well as dimensions.

Hence, the answer is the option (3).

6 .In the figure shown below reactant A (represented by a square) is in equilibrium with product B (represented by a circle). The equilibrium constant is :

1)1

2)4

3)8

4) (correct)2

Solution

Now,

$A \leftrightharpoons B$

$K_c=\frac{[B]}{[A]}$

The concentration of [B] as represented by circle = 11

and the concentration of [A] as represented by square = 6

Therefore, the value of the equilibrium constant is near 2.

Hence, Option number (4) is c

7.For the following three reactions (i), (ii) and (iii), equilibrium constants are given

(i) $\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} ; \mathrm{K}_1$
(ii) $\mathrm{CH}_{4(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{~g})} ; \mathrm{K}_2$
(iii) $\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{H}_{2(\mathrm{~g})} ; \mathrm{K}_3$

Which of the following relation is correct ?

1)$\mathrm{K}_3 \cdot \mathrm{K}_2^3=\mathrm{K}_1^2$

2)$\mathrm{K}_1 \sqrt{\mathrm{K}_2}=\mathrm{K}_3$

3)$\mathrm{K}_2 \mathrm{~K}_3=\mathrm{K}_1$

4) (correct)$\mathrm{K}_3=\mathrm{K}_1 \mathrm{~K}_2$

Solution

$\begin{aligned} & \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \\ & \mathrm{K}_1=\frac{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]}{[\mathrm{CO}]\left[\mathrm{H}_2 \mathrm{O}\right]} \ldots \ldots \ldots . .(\mathrm{i}) \\ & \mathrm{CH}_{4(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{~g})} \\ & \mathrm{K}_2=\frac{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]} \ldots \ldots \ldots . .(\text { ii }) \\ & \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{H}_{2(\mathrm{~g})} \\ & \mathrm{K}_3=\frac{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]^4}{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]^2} \ldots \ldots \ldots .(\text { iii })\end{aligned}$

From equations (i), (ii) and (iii); $\mathrm{K}_3=\mathrm{K}_1 \times \mathrm{K}_2$

Hence, the answer is the option (4).

Summary

The equilibrium constant has several applications in our lives it impacts our lives by various means such as in the environment it is used to control pollution as it is used to design the system for our air and water purification by optimizing conditions to remove pollutants effectively. It is also used for drug formulation as an equilibrium constant used to design and optimize drug formulation, ensuring that medications are effective and stable. it also helps in understanding the process of how drugs interact with the body. It is also used as a preservative in the food industry. Equilibrium constants have also used process like fermentation and canning to control and optimize reactants that preserve food and enhance flavour.


Frequently Asked Questions (FAQs)

1. What is the equilibrium constant?
The equilibrium constant (K) is a numerical value that describes the relative amounts of reactants and products present at equilibrium for a specific chemical reaction at a given temperature. It represents the ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficients.
2. How does temperature affect the equilibrium constant?
Temperature affects the equilibrium constant by shifting the position of equilibrium. As temperature changes, the value of K changes. For exothermic reactions, K decreases with increasing temperature, while for endothermic reactions, K increases with increasing temperature. This relationship is described by the van 't Hoff equation.
3. Why is the equilibrium constant always written without units?
The equilibrium constant is always written without units because it is a ratio of concentrations or partial pressures, each raised to their stoichiometric coefficients. The units cancel out in this ratio, resulting in a dimensionless value for K.
4. What's the difference between Kc and Kp?
Kc and Kp are both equilibrium constants, but they differ in the units used to express concentration. Kc uses molar concentrations (mol/L), while Kp uses partial pressures of gases (atm or Pa). They are related by the equation Kp = Kc(RT)Δn, where Δn is the change in the number of moles of gas from reactants to products.
5. Can the equilibrium constant be less than 1?
Yes, the equilibrium constant can be less than 1. A value of K < 1 indicates that the equilibrium favors the reactants, meaning there are more reactants than products at equilibrium. Conversely, K > 1 indicates that the equilibrium favors the products.
6. How does the magnitude of K relate to the extent of a reaction?
The magnitude of K indicates the extent of a reaction at equilibrium. A large K (K >> 1) means the reaction strongly favors products, while a small K (K << 1) means the reaction strongly favors reactants. When K is close to 1, significant amounts of both reactants and products are present at equilibrium.
7. Why can't we use the equilibrium constant to determine reaction rates?
The equilibrium constant only provides information about the composition of the reaction mixture at equilibrium, not how fast the reaction reaches equilibrium. Reaction rates depend on the kinetics of the reaction, which are determined by factors like activation energy and collision frequency, not the equilibrium constant.
8. How is the equilibrium constant related to Gibbs free energy?
The equilibrium constant is related to Gibbs free energy through the equation ΔG° = -RT ln K, where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is temperature in Kelvin, and K is the equilibrium constant. This relationship shows that the more negative ΔG°, the larger K, and vice versa.
9. Can the equilibrium constant change if we add a catalyst?
No, adding a catalyst does not change the equilibrium constant. A catalyst speeds up both the forward and reverse reactions equally, so it affects the rate at which equilibrium is reached but not the position of equilibrium or the value of K.
10. How do we write the equilibrium constant expression for a heterogeneous reaction?
For heterogeneous reactions, we only include the concentrations of aqueous or gaseous species in the equilibrium constant expression. Pure solids and liquids are omitted because their concentrations are constant and incorporated into the value of K.
11. What happens to K when we reverse a chemical equation?
When a chemical equation is reversed, the new equilibrium constant is the reciprocal of the original K. For example, if K = 100 for the reaction A ⇌ B, then K = 1/100 = 0.01 for the reversed reaction B ⇌ A.
12. How does pressure affect the equilibrium constant for gas-phase reactions?
For gas-phase reactions, changes in pressure do not affect the equilibrium constant K. However, pressure changes can shift the position of equilibrium according to Le Chatelier's principle, which may change the concentrations or partial pressures of the species involved.
13. Can we predict the direction of a reaction using the equilibrium constant?
Yes, we can predict the direction of a reaction by comparing the reaction quotient (Q) with the equilibrium constant (K). If Q < K, the reaction will proceed forward to produce more products. If Q > K, the reaction will proceed in reverse to produce more reactants. If Q = K, the system is at equilibrium.
14. What's the relationship between Ka, Kb, and Kw for conjugate acid-base pairs?
For a conjugate acid-base pair, the product of their acid and base dissociation constants (Ka and Kb) is equal to the ion product of water (Kw). This relationship is expressed as Ka × Kb = Kw. This means that the stronger an acid, the weaker its conjugate base, and vice versa.
15. How do we calculate the equilibrium constant from experimental data?
To calculate the equilibrium constant from experimental data, we first determine the equilibrium concentrations of all species. Then, we substitute these values into the equilibrium constant expression, which is the product of the concentrations of products divided by the product of the concentrations of reactants, each raised to their stoichiometric coefficients.
16. Why is the equilibrium constant temperature-dependent?
The equilibrium constant is temperature-dependent because temperature affects the relative stability of reactants and products. Changes in temperature alter the rates of forward and reverse reactions differently, leading to a shift in the equilibrium position and thus a change in the equilibrium constant.
17. How does the equilibrium constant relate to the concept of chemical equilibrium?
The equilibrium constant quantifies the state of chemical equilibrium. At equilibrium, the rates of forward and reverse reactions are equal, resulting in no net change in concentrations. The value of K indicates the relative amounts of reactants and products present at this dynamic equilibrium state.
18. Can we use the equilibrium constant to determine if a reaction is spontaneous?
While the equilibrium constant provides information about the composition at equilibrium, it doesn't directly indicate spontaneity. However, it's related to Gibbs free energy (ΔG° = -RT ln K), which does indicate spontaneity. Generally, if K > 1, ΔG° < 0, suggesting the forward reaction is spontaneous under standard conditions.
19. How do we interpret very large or very small values of K?
Very large values of K (K >> 1) indicate that the equilibrium strongly favors the products, with nearly complete conversion of reactants. Very small values of K (K << 1) indicate that the equilibrium strongly favors the reactants, with very little product formation at equilibrium.
20. What's the difference between Kc and the reaction quotient Q?
Kc is the equilibrium constant, which represents the ratio of products to reactants at equilibrium. The reaction quotient Q has the same form as Kc but uses the concentrations of species at any point in the reaction, not necessarily at equilibrium. Q is used to determine the direction of a reaction by comparing it to K.
21. How does the stoichiometry of a reaction affect its equilibrium constant expression?
The stoichiometric coefficients in a balanced chemical equation become the exponents in the equilibrium constant expression. For example, for the reaction aA + bB ⇌ cC + dD, the expression is K = [C]^c[D]^d / [A]^a[B]^b. This means that small changes in stoichiometry can significantly affect the value of K.
22. Can we add equilibrium constants for multiple reactions?
We cannot directly add equilibrium constants. However, if we have a series of reactions that can be added to give an overall reaction, we can multiply their individual K values to get the K for the overall reaction. This is because K is related to ΔG°, and ΔG° values are additive for a series of reactions.
23. How does the concept of Le Chatelier's principle relate to the equilibrium constant?
Le Chatelier's principle describes how a system at equilibrium responds to disturbances, while the equilibrium constant quantifies the equilibrium state. Changes predicted by Le Chatelier's principle (like shifts due to concentration or pressure changes) don't alter K, but they do affect the time it takes to re-establish equilibrium.
24. Why do we use standard conditions when reporting equilibrium constants?
Standard conditions (usually 1 atm pressure, 1 M concentration for solutions, and a specific temperature, often 25°C) are used when reporting equilibrium constants to ensure consistency and comparability of data. This allows scientists to reference and use K values across different experiments and contexts.
25. How does the equilibrium constant change for the overall reaction when we multiply all coefficients in a balanced equation by a factor?
When all coefficients in a balanced equation are multiplied by a factor n, the new equilibrium constant K' is related to the original K by the equation K' = K^n. For example, if we double all coefficients, the new K will be the square of the original K.
26. What's the significance of K = 1?
When K = 1, it indicates that the concentrations of products and reactants are approximately equal at equilibrium. This means the forward and reverse reactions occur at similar rates, and neither the reactants nor the products are strongly favored in the equilibrium mixture.
27. How do we determine the units of Kc for a general reaction?
The units of Kc depend on the stoichiometry of the reaction. For a general reaction aA + bB ⇌ cC + dD, the units of Kc will be (mol/L)^(c+d-a-b). However, as mentioned earlier, we conventionally express K as a unitless quantity.
28. Can the equilibrium constant ever be zero or negative?
No, the equilibrium constant can never be zero or negative. K is always a positive number because it represents a ratio of concentrations raised to powers. A K of zero would imply no products at all, which is not possible in a reversible reaction. Negative concentrations are not physically meaningful, so K cannot be negative.
29. How does the equilibrium constant relate to the extent of reaction?
The equilibrium constant is directly related to the extent of reaction. A large K indicates a large extent of reaction (favoring products), while a small K indicates a small extent of reaction (favoring reactants). The extent of reaction can be calculated from K if initial concentrations are known.
30. Why do we need to specify the temperature when reporting an equilibrium constant?
Temperature must be specified when reporting an equilibrium constant because K is temperature-dependent. The value of K can change significantly with temperature, especially for reactions with large enthalpy changes. Without knowing the temperature, the K value lacks context and cannot be reliably used or compared.
31. How does the presence of a common ion affect the equilibrium constant?
The presence of a common ion doesn't change the equilibrium constant, as K is a fundamental property of the reaction at a given temperature. However, it can shift the equilibrium position according to Le Chatelier's principle, changing the concentrations of reactants and products while maintaining the same K value.
32. What's the relationship between ΔH°, ΔS°, and K?
The relationship between standard enthalpy change (ΔH°), standard entropy change (ΔS°), and the equilibrium constant (K) is given by the van 't Hoff equation: ln K = -ΔH°/RT + ΔS°/R. This equation shows how K depends on both the enthalpy and entropy changes of the reaction, as well as temperature.
33. How do we calculate the equilibrium constant for a multi-step reaction?
For a multi-step reaction, we can calculate the overall equilibrium constant by multiplying the equilibrium constants of each individual step. This is because the logarithm of K is proportional to ΔG°, and ΔG° values are additive for a series of reactions.
34. Can we use the equilibrium constant to calculate equilibrium concentrations?
Yes, we can use the equilibrium constant along with initial concentrations to calculate equilibrium concentrations. This often involves setting up an ICE table (Initial, Change, Equilibrium) and solving a quadratic equation. However, for very large or very small K values, approximations can sometimes be used to simplify the calculations.
35. How does the equilibrium constant relate to the solubility product (Ksp)?
The solubility product (Ksp) is a specific type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds in water. It represents the product of the concentrations of ions, each raised to the power of its stoichiometric coefficient in the dissociation equation.
36. What's the difference between the thermodynamic and concentration equilibrium constants?
The thermodynamic equilibrium constant uses activities instead of concentrations and is truly constant for a given reaction at a given temperature. The concentration equilibrium constant uses molar concentrations and can vary slightly with ionic strength. In dilute solutions, the two are approximately equal.
37. How do coupled equilibria affect the overall equilibrium constant?
In coupled equilibria, where the product of one reaction is a reactant in another, the overall equilibrium constant is the product of the individual equilibrium constants. This is because the ΔG° values for the individual reactions are additive, and K is related to ΔG° through an exponential function.
38. Can we use the equilibrium constant to predict the yield of a reaction?
While the equilibrium constant doesn't directly give the yield, it can be used to predict the maximum theoretical yield of a reaction under specific conditions. A large K suggests a high yield is possible, while a small K indicates the yield will be limited. However, kinetic factors may prevent a reaction from reaching equilibrium in practice.
39. How does the equilibrium constant relate to the concept of chemical potential?
The equilibrium constant is related to the difference in chemical potentials of products and reactants. At equilibrium, the sum of chemical potentials of products equals the sum of chemical potentials of reactants. This equality is mathematically equivalent to the definition of K in terms of activities or concentrations.
40. What's the significance of the standard state in equilibrium constant calculations?
The standard state (usually 1 M for solutes, 1 atm for gases, pure substance for liquids and solids) is crucial in equilibrium constant calculations because K is defined in terms of these standard states. This allows for consistent comparison of K values across different reactions and conditions.
41. How do we handle equilibrium constants for reactions involving very dilute solutions?
For very dilute solutions, we can often use concentrations instead of activities in equilibrium constant expressions with minimal error. However, for more accurate results, especially in non-ideal solutions, activity coefficients should be used to convert concentrations to activities.
42. Can we use the equilibrium constant to predict the pH of a solution?
Yes, for acid-base reactions, we can use the acid dissociation constant (Ka, a type of equilibrium constant) to calculate the pH of a solution. This involves setting up an ICE table and solving for the equilibrium concentration of H+ ions, then converting to pH using the equation pH = -log[H+].
43. How does the equilibrium constant relate to the concept of buffer solutions?
In buffer solutions, the equilibrium constant (usually Ka or Kb) is crucial for determining the buffer capacity and pH. The Henderson-Hasselbalch equation, which relates pH to pKa and the ratio of conjugate base to acid, is derived from the equilibrium constant expression.
44. What's the relationship between the equilibrium constant and the equilibrium law?
The equilibrium law states that for a reaction at equilibrium, the ratio of the product of the concentrations of products to the product of the concentrations of reactants, each raised to their stoichiometric coefficients, is constant. This constant ratio is the equilibrium constant, K.
45. How do we interpret equilibrium constants for competing reactions?
For competing reactions, we can compare their equilibrium constants to predict which reaction is more favorable. The reaction with the larger K will proceed to a greater extent. However, kinetic factors may also play a role in determining which reaction predominates in practice.
46. Can we use the equilibrium constant to predict the effect of a catalyst?
The equilibrium constant cannot be used to predict the effect of a catalyst. Catalysts do not change the equilibrium constant or the equilibrium composition
47. How does the concept of equilibrium constant apply to phase changes?
For phase changes, such as vaporization or sublimation, the equilibrium constant is related to the vapor pressure of the substance. For example, for the vaporization of a liquid, Kp = Pvap, where Pvap is the vapor pressure. This equilibrium

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