Sales Toll Free No: 1-855-666-7446

Equilibrium Constant Calculator

Top
Equilibrium constant is the value which shows when the chemical reaction reaches the equilibrium. Consider a chemical reaction, aA + bB $\Leftrightarrow$ cC + dD

At equilibrium, the equilibrium constant $K_{c}$ is given as,
$K_{c}$=$\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^b}$

where,
[A],[B],[C],[D] = molar concentration of A, B, C, D.
a, b, c, d = coefficients in the chemical equation.
 

Steps

Back to Top
Step 1: Note down the concentration of both reactants and products of the given reaction.

Step 2: Plug in the concentrations [A],[B],[C] and [D] into the equilibrium constant $K_{c}$ formula given as,
$K_{c}$=$\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^b}$
and get the $K_{c}$ value.

Problems

Back to Top
For your reference, few problems based on equilibrium constant given below.

Solved Examples

Question 1: Consider the reaction
$H_{2}(g)+I_{2}(g)\leftrightarrow 2HI(g)$

Calculate the equilibrium constant $K_c$, if the concentration of [$H_2$] is 0.106 M, [$I_2$] is 0.035 M and [HI] is 1.29 M.
Solution:
 
Step 1: Given : $H_{2}(g)+I_{2}(g)\leftrightarrow 2HI(g)$
[$H_2$] = 0.106 M,
[$I_2$] = 0.035 M,
[HI] is 1.29 M.

Step 2: Using the above formula,
$K_{c}$=$\frac{[C]^{c}}{[A]^{a}[B]^b}$
(Here in this equation, there is only one product. So, we left $[D]^{d}$)

$K_{c}$=$\frac{[1.29 M]^{2}}{[0.106 M][0.035 M]}$

$K_{c}$=$4.49 \times 10^2$

Therefore, equilibrium constant $K_c$ = $4.49 \times 10^2$.
 

Question 2: Consider the reaction
$CO_{2}(g)+H_{2}(g)\leftrightarrow CO(g) + H_{2}O$

Calculate the equilibrium constant $K_c$, if the concentration of [$CO_{2}$] is 0.0954 M, [$H_{2}$] is 0.0454 M, [CO] is 0.004 M and [$H_2$] is 0.0046 M.

Solution:
 
Step 1: Given : $CO_{2}(g)+H_{2}(g)\leftrightarrow CO(g) + H_{2}O$
[$CO_{2}$] = 0.0954 M,
[$H_{2}$] = 0.0454 M,
[CO] = 0.004 M, 
[$H_2$] = 0.0046 M.

Step 2: Using the above formula,
$K_{c}$=$\frac{[C]^{c}[D]^{D}}{[A]^{a}[B]^b}$

$K_{c}$=$\frac{[0.0046 M][0.0046 M]}{[0.0954 M][0.0454 M]}$

$K_{c}$=$5 \times 10^{-3}$

Therefore, equilibrium constant $K_c$ = $5 \times 10^{-3}$.