Law of Mass Action- Law, Equilibrium Constant, Characteristics, Practice Problems and FAQs

Let us assume a reaction between gaseous reactants in a closed container at a specific temperature to form again gaseous products. With increasing time, you will observe that the concentrations of the reactants to decrease and products increase. After some time, there is then no longer any change in the concentrations of the reactants or products. Do you know why is this happening?

A dynamic equilibrium between the reactants and product gasses is created because the container is closed and the generated product is unable to escape.

This constancy of concentration of reactants and products after some time due to the simultaneous existence of opposing changes is called equilibrium. Then you may ask what will be the concentration of them? How it is fixed?

The answer is given by the Law of Mass Action. Come let us learn about it.

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Table of Content

Law of Mass Action
Mathematical Expression of Law of mass action
Characteristics of Equilibrium Constant
Practice Problems
Frequently Asked Questions

Law of Mass Action

Guldberg and Waage made the suggestion that the rate of a chemical reaction, at a given temperature, is exactly proportional to the product of the active masses of the reactants raised to the powers of their stoichiometric coefficients.

Active mass: The mass of the actually reacting molecules is called the active mass. It is different form the total mass of all the molecules(including that are non reacting). Although the phrase "active mass" is more correct, for practical purposes the concentration of the reactive species is used as the active mass and the partial pressure of gaseous species as active mass. It is an indivisible amount.

Law of Mass Action says that the higher the active mass of the reactants larger the number of molecules reacting to form the product.

“The rate of a chemical reaction is proportional to the product of the active masses of reactants raised to the powers of their stoichiometric coefficients at a given temperature, according to this law.”

Mathematical Expression of Law of mass action

Consider a chemical reaction

$a A + b B \to p r o d u c t$

According to the law of mass action

$r α {[A]}^{a} {[B]}^{b}$

Where [A] and [B] are the concentrations of reactants A and B, respectively, and r is the reaction rate.

When the proportionality sign is removed,

$r = K {[A]}^{a} {[B]}^{b}$

Where K = Rate Constant

Consider a chemical reaction

$a A + b B ⇌ c C + d D$

According to the law of mass action,

For forward reaction; $r_{f} = K_{f} {[A]}^{a} {[B]}^{b}$

Where $K_{f}$ = rate constant for forward reaction

For backward reaction; $r_{b} = K_{b} {[C]}^{c} {[D]}^{d}$

Where K_b = rate constant for backward reaction

At equilibrium, the rate of the forward reaction = the rate of backward reaction

$r_{f} = {r_{b}}_{}$

$K_{f} {[A]}^{a} {[B]}^{b} = K_{b} {[C]}^{c} {[D]}^{d}$

$\frac{K_{f}}{K_{b}} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$

$K_{c} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ ……………..(1)

K_c is the equilibrium constant.

Note: A pure solid or pure liquid has a constant concentration.

Characteristics of Equilibrium Constant for a related equilibrium

Consider a reaction: $a A + b B ⇌ c C + d D w i t h; K_{C}$

When a reaction is reversible, the new equilibrium constant K_Cn' is the reciprocal of the old one.	$c C + d D ⇌ a A + b B; K_{C}^{'}$ $K_{C n}^{'} = \frac{1}{K_{C}}$
The value of the K_Cn'' equals the KC raised to the power n when a reaction is multiplied by a factor 'n'.	$n a A + n b B ⇌ n c C + n d D; K_{C}^{'}$ $K_{C n}^{'} = {(K_{C})}^{n}$
The K_Cn' is the product of the KC when two or more equilibrium reactions are added	$a A ⇌ c C; K_{C_{1}}$ , $b B ⇌ d D; K_{C_{2}}$ $a A + b B ⇌ c C + d D; K_{C}$ $K_{C n} = K_{C_{1}} \times K_{C_{2}}$
The K_Cn' is the ratio of the K_C when two or more equilibrium reactions are subtracted.	$2 a A + b B ⇌ c C + 2 d D; K_{C_{1}}$ , $a A ⇌ d D; K_{C_{2}}$ $a A + b B ⇌ c C + d D; K_{C}$ $K_{C n} = \frac{K_{C_{1}}}{K_{C_{2}}}$

Practice Problems

Q1. Why is the concentration of a pure solid or pure liquid assumed to be constant?

Consider density ( $ρ$ ) (in gmL^-1) as the pure solid and liquid density. Then,

$D e n s i t y (ρ) = \frac{M a s s}{V o l u m e} = \frac{m}{V}$

$M a s s = m = ρ V$

The number of moles in a pure solid or liquid is calculated as follows:

$m o l e s = \frac{M a s s}{M o l a r M a s s} = \frac{ρ V}{M}$

Volume of the solution = volume of the solvent (for pure liquid).

The molarity of the solution is calculated as

$M o l a r i t y = \frac{N u m b e r o f m o l e s}{V o l u m e (L)} = \frac{ρ V}{M V} = \frac{ρ}{M}$

The molar mass of any pure solid or liquid is always constant.
The density of any pure solid or liquid is always constant.

As a result, for any pure solid or liquid, the molarity or concentration is always constant.

Q2. In a one-litre tank at 250°C, the reaction $A (g) + B (g) ⇌ C (g) + D (g)$ is investigated. The starting concentrations of A and B were 3n and n, respectively. The equilibrium concentration of C was found to be identical to the equilibrium concentration of B when equilibrium was reached. What is the equilibrium concentration of D?

A. $\frac{n}{4}$
B. $\frac{n}{2}$
C. $\frac{n}{6}$
D. $\frac{n}{3}$

Solution: $A (g) + B (g) ⇌ C (g) + D (g)$

At initial 3n n 0 0

At equilibrium 3n-x n-x x x

Equilibrium concentration of C was determined to be equal to B's equilibrium concentration, as given.

[B]=[C]

$\frac{n - x}{V} = \frac{x}{V}$

$n - x = x$

$n = 2 x$

$x = \frac{n}{2}$

$[C] = x = \frac{n}{2}$

Equilibrium concentration of $[C] = [D] = \frac{n}{2}$

Hence, option (B) is correct answer.

Q3. The rate constant for the following forward reaction is $2.5 \times 10^{- 4} s^{- 1}$ at 300 K.

P(g) ⇌ 2Q(g)

If 10^-7 mol of P and 10 mol of Q are present in a 10-litre vessel at equilibrium, then the rate constant of backward reaction at this temperature is $X \times 10^{- 11} M^{- 1} s^{- 1}$ . Find the value of 4X.

A. 5|
B. 0.5
C. 3
D. 1

Solution: Given: $K_{f} = 2.5 \times 10^{- 4} s^{- 1}$

Mole of P at equilibrium= $10^{- 7}$ mol

Mole of Q at equilibrium= 10 mol

[P] = $\frac{10^{- 7} m o l}{10 L} = 10^{- 8} m o l L^{- 1} = 10^{- 8} M$

[Q] = $\frac{10 m o l}{10 L} = 1 m o l L^{- 1} = 1 M$

P(g) ⇌ 2Q(g)

$K_{e q} = \frac{{[Q]}^{2}}{[P]} = \frac{{(1)}^{2}}{10^{- 8}} = 10^{8} M$

As $K_{e q} = \frac{K_{f}}{K_{b}} = \frac{2.5 \times 10^{- 4} s^{- 1}}{K_{b}}$

$10^{8} M = \frac{2.5 \times 10^{- 4} s^{- 1}}{K_{b}}$

$K_{b} = \frac{2.5 \times 10^{- 4} s^{- 1}}{10^{8} M} = 2.5 \times 10^{- 12} M^{- 1} s^{- 1}$

$X \times 10^{- 11} M^{- 1} s^{- 1} = 2.5 \times 10^{- 12} M^{- 1} s^{- 1} = 0.25 \times 10^{- 11} M^{- 1} s^{- 1}$

X=0.25

$4 X = 4 (0.25) = 1$

Hence, correct answer should be option (D).

Q4. For a reaction A (g) ⇌ 3 B(g), if $3.2 \times 10^{- 4}$ mol of A and $2 \times 10^{- 3}$ mol of B are present in a 1-litre vessel at equilibrium, then the rate constant is

A. $2.5 \times 10^{- 4} M$
B. $2.5 \times 10^{- 5} M$
C. $2.5 \times 10^{- 3} M$
D. $2.5 \times 10^{- 6} M$

Solution:

Mole of A at equilibrium= $3.2 \times 10^{- 4}$ mol

Mole of B at equilibrium= $2 \times 10^{- 3}$ mol

[A] = $\frac{3.2 \times 10^{- 4} m o l}{1 L} = 3.2 \times 10^{- 4} m o l L^{- 1} = 3.2 \times 10^{- 4} M$

[B] = $\frac{2 \times 10^{- 3} m o l}{1 L} = 2 \times 10^{- 3} m o l L^{- 1} = 2 \times 10^{- 3} M$

A(g) ⇌ 3B (g)

$K_{e q} = \frac{{[B]}^{3}}{[A]} = \frac{{(2 \times 10^{- 3} M)}^{3}}{3.2 \times 10^{- 4} M} = 2.5 \times 10^{- 5} M$

Hence, correct answer should be option (B).

Frequently Asked Questions

Q1. What is an effective concentration in chemistry?
Answer: The effective concentration is the level of focus that actively contributes to the reaction. For instance, if the reactant initially contained 8 moles and only 3 moles remained at equilibrium. In relation to three moles, then, is the effective concentration.

Q2. Why do we usually enclose the concentration at equilibrium in square brackets?
Answer: The amount of a reactant that is actually taking part in the reaction—its effective concentration or active mass—is indicated by the square bracket. The square brackets are used for this reason.

Q3. What uses does the Law of Mass Action have in Science?
Answer: This equation holds true for semiconductors as well, and it has a number of significant ramifications for the fields of semiconductor physics and electronics. The law of mass action establishes a connection between the concentrations of electron holes and free electrons when the semiconductor system is in thermal equilibrium.

Q4. Is it possible for the equilibrium constant to have a negative value?
Answer: No, it is not feasible for the equilibrium constant to have a negative value because it is equal to the ratio of the product and reactant concentrations. The ratio is always positive since the concentration of the product and reaction cannot be negative, which foretells a positive value for the equilibrium constant.