# Law of Chemical Equilibrium and Equilibrium Constant

During a chemical process, **chemical equilibrium** refers to the state in which the concentrations of both reactants and products have no tendency to fluctuate over time. When the forward and reverse reaction rates are equal, a chemical reaction is said to be in chemical equilibrium. The state is known as a **dynamic equilibrium**, and the rate constant is known as the **equilibrium constant**, because the rates are equal and there is no net change in the concentrations of the reactants and products. Let’s look into it more.

Consider the universal reversible reaction: **A + B ⇌ C + D**.

At the beginning (i.e., at time t = 0), the absorption of A and B are consummate and the absorption of C and D are minimal ( equal to zero because no C and D are ultimately formed). As the reaction proceeds, the attention of A and B are reduced with time whereas the attention of C and D are added. Thus, the rate of advancing reaction is dwindling while the rate of backward reaction goes on adding.

Eventually, a stage comes, when the rate of encouraging response becomes equal to the rate of backward reaction. The reaction is also said to be in a state of chemical equilibrium. The interpretation of the reaction rates with time and eventually the accomplishment of chemical equilibrium may be represented diagrammatically.

The succeeding exemplifications illustrate how the equilibrium is achieved:

- Decomposition of Calcium carbonate in a closed vessel,
- Decomposition of N
_{2}O_{4}in a closed vessel,- Combination of H
_{2 }and I_{2 }to form HI, and- Reaction between ferric nitrate and potassium thiocyanate solution.

### Law of Chemical Equilibrium

The Law of Chemical Equilibrium is a result obtained by applying the Law of Mass Action to a reversible reaction in equilibrium. For example, consider the general reversible reaction,

** A + B ⇌ C + D**

At equilibrium, assume the active masses of A, B, C and D are represented as [A], [B], [C], and [D] respectively. Applying the Law of Mass Action,

The rate at which A and B react together, i.e.

**Rate of the forward reaction ∝ [A] [B] = k _{f}[A] [B] **

- where k
_{f}could be a constant of proportionality and is termed velocity constant for the forward reaction.

Similarly, the rate at which C and D react together, i.e.,

**Rate of the backward reaction ∝ [C] [D]= k _{b}[C] [D]**

- where k
_{b}represents the velocity constant for the backward reaction.

At equilibrium,

Rate of forward reaction = Rate of backward reaction

Hence,

**k _{f}[A][B] = k_{b}[C][D] or ([C][D])/([A][B]) = k_{f}/k_{b} = K**

At constant temperature, as k_{f} and k_{b} are constant, therefore, k_{f}/ k_{b} = K is also constant at constant temperature and is called** Equilibrium constant.**

Again, consider the furthermore universal reversible reaction.

productions are written in the numerator and those

**aA + bB +…… **⇌ ** xX+yY + …**

Applying the Law of Mass Action, as before, we get

**[X] ^{x} [Y]^{y}/[A]^{a}[B]^{b} = K or K_{c}**

- where K is the equilibrium constant. It’s constant at stationary temperature.

### Equilibrium Constant

The aforementioned mathematical equation is called the Law of Chemical Equilibrium. stated in words, it may be outlined as follows-

The product of the molar concentrations of the products, each raised to the power adequate to its stoichiometric coefficient divided by the product of the molar concentrations of the reactants, each raised to the power adequate to its stoichiometric coefficient is constant at constant temperature and is termed as

Equilibrium constant.

It’s conventional to use K. for equilibrium constant raised in terms of concentrations. Where there’s no distrustfulness that K is in terms of concentration, c is neglected.

For gas-phase reactions, (that is when the reactants and products are gaseous), the equilibrium constant can be expressed either as concentrations in moles per litre or in terms of the partial pressures of the reactants and products. if stated in terms of partial pressures. However, it is denoted by K_{p} so, if A, B, X, and Y are gaseous, it can be-

K_{P}= (P_{X}^{x}. P_{Y}^{y})/(P_{A}^{a}. P_{B}^{b})Where P

_{A }P_{B}, P_{X,}and P_{Y}are the partial pressures of A, B, X, and Y independently in the reaction amalgam at equilibrium. It may be observed that the pressures in the aforementioned equation are grasped in atmospheres or bars or pascals (in SI units).

### Characteristics of Equilibrium Constant

Some of the most consequential characteristics of the Equilibrium constant are as follows:

- The value of the Equilibrium constant for an individual reaction is always constant hinging exclusively upon the temperature of the reaction and is self-subsistent of the concentrations of the reactions with which we start or the direction from the Equilibrium is came up.
- If the value of the Equilibrium constant is inversed If the reaction is switched.
- Nevertheless, the Equilibrium constant for the substitute equation is the quadrate root of K(that is √K), If the equation is divided by 2.
- The Equilibrium constant for the new equation is the square of K(that is K
^{2}) If the equation is multiplied by 2. - If the equation is scratched in two steps also
**K**._{1}× K_{2}= K - The value of the Equilibrium constant isn’t affected by the extension of the catalyst to the reaction.

### Effect of Temperature on Equilibrium Constant

The numerical value of the equilibrium constant for an individual reaction is stable as long as its temperature is commemorated stable. It’s a well-understood reality that the rate of a chemical reaction increases with an increase in temperature.

Still, the termination of this rate increase depends upon the energy of activation of the reaction. anymore, since the energy of activation for the forward and backward reactions are disparate, so a presented increment in temperature will increase the rate of the forward and backward reactions to different extents. Indistinguishable terms, the values of the velocity constants for forwarding and backward responses. i.e, k_{b} and will change else with a given rise or fall in temperature.

Further, since K = k_{f}/k_{b} thus, the value of the equilibrium constant (K) will change, i.e. the state of equilibrium is altered. Therefore, we conclude that the equilibrium constant for individual reaction fluctuations with temperature. Further, it has been found that the value of the equilibrium constant of an endothermic reaction increases (k_{f} increases more than k_{b}) and that of an exothermic reaction diminishments (k_{b} increases additional than k_{f}) with elevation in temperature. For reactions having zero heat of reaction, the temperature has no aftereffect on the value of K.

**Factors Affecting Equilibrium**

- Change of concentration of any product or reactant
- Change of temperature of the system
- Change the pressure of the system
- Addition of catalyst
- Addition of inert gas.

### Sample Problems

**Problem 1: At 773 K, the equilibrium constant K _{c} for the reaction, N₂ (g) + 3 H₂ (g) **⇌

**2 NH**

_{3}(g) is 6.02 x 10^{-2}L² mol^{–}². Calculate the value of K_{p}at the same temperature.**Solution:**

△n

_{g}= 2-4=-2K

_{p }= K_{c}(RT)^{△n}= 6.02 × 10

^{-2 }L² mol^{–}²× (0.0821 L atm K^{–}¹ mol^{–}¹ × 773 K)^{-2}=

1.5 × 10.5 atm^{-2}

**Problem 2: For the equilibrium, 2 NOCI (g) ⇌ 2 NO(g) + Cl₂ (g), the value of the equilibrium constant, K _{c} is 3.75 x 10^{-6} at 1069 K. Calculate K_{p} for the reaction at this temperature.**

**Solution: **

△n=(2+1)-2= 1

K

_{p}= K_{c}(RT)^{△n }= (3.75 × 10^{-6 })(0.0831 × 1069)=

3.33 × 10^{-4}

**Problem 3: The following concentrations were obtained for the formation of NH _{3} from N_{2} and H_{2} at equilibrium at 500 K: [N_{2}]=1.5× 10^{-2} M, [H_{2}] = 3.0 x 10^{-2 }× M and [NH_{3}] = 1.2 × 10^{-2} M.**

**Solution: **

The equilibrium reaction is :N

_{2}(g)+3H_{2}(g) ⇌ 2NH_{3}(g)K

_{c}= [NH_{3}]^{2}/[N_{2}][H_{2}]^{3}= (1.2 × 10

^{-2})^{2}/(1.5 × 10^{-2})(3.0 × 10^{-2})^{3}=

3.55 × 10^{2}

**Problem 4: For an equilibrium reaction, the rate constants for the forward and the backward reaction are 2.38 x10 ^{-4} and 8.15 x 10^{-5} respectively. Calculate the equilibrium constant for the reaction.**

**Solution: **

Equilibrium Constant K= k

_{f}/k_{b}= (2.38 x 10^{-4})/(8.15×10^{-8})=2.92

**Problem 5: The value of K, for the reaction, 2 AB + C is 2.0 × 10³ At a given time, the composition of the reaction mixture is [A] = [B] = [C] = 3 × 10.4 M In which direction, the reaction will proceed?**

**Solution: **

For the given reaction. Q

_{c}= [B][C]/[A]^{2}=

^{ }(3 x 10^{-4})(3 x 10^{-4})/ (3 x 10^{-4})^{2 }=1Thus, Q

_{c}> K_{c}Hence, the reaction will proceed in the backward direction.

**Problem 6: In the equilibrium, CaCO _{3} (s) **⇌

**CaO (s) + CO₂ (g), at 1073 K, the pressure of CO₂ is found to be 2.5 x 10 Pa. What is the equilibrium constant of this reaction at 1073 K?**

**Solution: **

With reference to the standard state pressure of 1 bar, i.e., 10

^{5}Pa,K

_{p }= p_{CO₂ }= 2.5 ×10^{4}Pa/p^{0}= (2.5 × 10

^{4}Pa)/10^{5}Pa=

0.25

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