If the concentrations of reactants and products are same, one can call this reaction as equilibrium reaction. If A and B are the reactants, C and D are products, then an equilibrium reaction can be written as, A+B → C+D. From this reaction equation, we can calculate the equilibrium constant, $K_{c}$.

The**formula for the equilibrium constant** is,

The

where A, B, C and D are the molar concentrations

a, b, c and d are the respective coefficients

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Problems related to equilibrium constant are given below:

### Solved Examples

**Question 1: **Calculate the equilibrium constant of the following equation.

Na_{2}O+H_{2}O→2NaOH; the concentration of each reactants are NaOH = 0.23M, Na_{2}O = 0.015M and H_{2}O = 0.1M

** Solution: **

Given equation is,

Na_{2}O+H_{2}O→2NaOH

The molar concentration is,

NaOH = 0.23M, Na_{2}O = 0.015M and H_{2}O = 0.1M

The formula for equilibrium constant is given by,

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

$K_{c}$ = $\frac{[0.23]^{2}}{[0.015]^{1}[0.1]^{1}}$

$K_{c}$ = 35.266

**Question 2: **Calculate the equilibrium constant for the following equation

2HNO_{3} → N_{2}O_{5}+H_{2}O; the molar concentration is HNO_{3} = 0.05, N_{2}O_{5 }= 0.3 and H_{2}O = 0.1

** Solution: **

Given equation is,

2HNO_{3} → N_{2}O_{5}+H_{2}O

The molar concentration is HNO_{3} = 0.05, N_{2}O_{5 }= 0.3 and H_{2}O = 0.1

The formula for equilibrium constant is given by,

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

$K_{c}$ = $\frac{[0.3]^{1}[0.1]^{1}}{[0.05]^{2}}$

$K_{c}$ = 12

Na

Given equation is,

Na

The molar concentration is,

NaOH = 0.23M, Na

The formula for equilibrium constant is given by,

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

$K_{c}$ = $\frac{[0.23]^{2}}{[0.015]^{1}[0.1]^{1}}$

$K_{c}$ = 35.266

2HNO

Given equation is,

2HNO

The molar concentration is HNO

The formula for equilibrium constant is given by,

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

$K_{c}$ = $\frac{[0.3]^{1}[0.1]^{1}}{[0.05]^{2}}$

$K_{c}$ = 12