Chemical Equilibrium: Law of mass action and equilibrium constant expression, explanation and derivation

Chemical Equilibrium and Equilibrium constant expression

Question: What is law of mass action? Derive the equilibrium constant expression for a general reaction given as follows.

Answer.

Law of mass action:

This law states that the rate at which a substance reacts is directly proportional to its active mass and the rate at which reaction takes place is directly proportional to product of active masses of all reactants.

Consider a reaction in which a moles of a reactant react with b moles of reactant B to produce a c moles of product c and d moles of product d.

According to law of mass action:

\[{\color{DarkGreen} Rate \: of \, forward\: reaction \propto \left [ A \right ]^{a}\left [ B \right ]^{b}}\]

\[{\color{DarkGreen} Rate \: of \, forward\: reaction = {K}_{f}\left [ A \right ]^{a}\left [ B \right ]^{b}}\cdot \cdot \cdot (i)\]

\[ {\color{DarkGreen} Rate \: of \, reverse\: reaction = {K}_{r}\left [ C \right ]^{c}\left [ D \right ]^{d}}\cdot \cdot \cdot (ii)\]

\[Where \: K_{f}\: and\: K_{r}\: are \: rate \: constants\: for \: forward\: and \: reverse\: reaction\:,\: respectively.\]

At equilibrium state,

\[{\color{DarkGreen} Rate \: of \: forward \: reaction = Rate \: of\: reverse\: reaction }\]

\[{\color{DarkGreen} K_{f}\left [ A \right ]^{a}\left [ B \right ]^{b} = K_{r}\left [ C \right ]^{c}\left [ D \right ]^{d}}\]

\[{\color{DarkGreen} {}\frac{K_{f}}{K_{r}} = \frac{\left [ C \right ]^{c}\left [ D \right ]^{d}}{\left [ A \right ]^{a}\left [ B \right ]^{b}}}\]

\[{\color{DarkGreen} K_{c} = \frac{\left [ C \right ]^{c}\left [ D \right ]^{d}}{\left [ A \right ]^{a}\left [ B \right ]^{b}}}\]

\[Where \: K_{c}\: is\: called\: equilibrium\: constant\]and above equation is called equilibrium constant expression. Where square brackets indicate the equilibrium concentrations of substances in mole/dm3

Example: 9.2: Write equilibrium constant expression for following reactions.

Solution:

Self- Assessment Exercise 9.2

Following reaction can occur during lightening storms. Derive equilibrium constant expression for this reaction.

Solution:-

According to law of mass action:

\[{\color{DarkGreen} Rate \: of \, forward\: reaction \propto \left [ {O}_{2} \right ]^{3}}\]

\[{\color{DarkGreen} Rate \: of \, forward\: reaction = {K}_{f}\left [ {O}_{2} \right ]^{3}}\cdot \cdot \cdot (i)\]

\[{\color{DarkGreen} Rate \: of \, reverse\: reaction \propto \left [ {O}_{3} \right ]^{2}}\]

\[ {\color{DarkGreen} Rate \: of \, reverse\: reaction = {K}_{r}\left [ {O}_{3} \right ]^{2}}\cdot \cdot \cdot (ii)\]

\[Where \: K_{f}\: and\: K_{r}\: are \: rate \: constants\: for \: forward\: and \: reverse\: reaction\:,\: respectively.\]

At equilibrium state,

\[{\color{DarkGreen} Rate \: of \: forward \: reaction = Rate \: of\: reverse\: reaction }\]

\[{\color{DarkGreen} K_{f}\left [ {O}_{3} \right ]^{2} = K_{r}\left [ {O}_{2} \right ]^{3}}\]

\[{\color{DarkGreen} {}\frac{K_{f}}{K_{r}} = \frac{\left [ {O}_{2} \right ]^{3}}{\left [ {O}_{3} \right ]^{2}}}\]

\[{\color{DarkGreen} K_{c} = \frac{\left [ {O}_{2} \right ]^{3}}{\left [ {O}_{3} \right ]^{2}}}\]

\[Where \: K_{c}\: is\: called\: equilibrium\: constant\] for above reaction.

Self Assessment Exercise 9.2

Q.2 Write equilibrium constant expression for the following reactions.

$4HCl_{(g)} + O_{2(g)} \rightleftharpoons 2H_{2}O_{(g)} +2Cl_{2(g)}$

$CH_{3}COOH_{(l)} + C_{2}H_{5}OH_{(l)} \rightleftharpoons CH_{3}COOC_{2}H_{5(l)} + H_{2}O_{(l)}$

$2HF_{(g)} \rightleftharpoons H_{2(g)} + F_{2(g)}$

$2NO_{2(g)} \rightleftharpoons 2N_{2}O_{4(g)}$

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