# Calculating Equilibrium Concentration

## Equilibrium Concentration

When we talk about a balanced chemical reaction, we mean that each element has an equal number of atoms on both sides of the equation. These balanced chemical reactions form the basis for the concept of equilibrium concentration. We say that a chemical is in an equilibrium concentration when the products and reactants do not change as time moves on. In other words, chemical equilibrium or equilibrium concentration is a state when the rate of forward reaction in a chemical reaction becomes equal to the rate of backward reaction. At the same time, there is no change in the products and reactants, and it seems that the reaction has stopped.

It can be understood from the graph above, that initially, the concentration of the product is zero. Still, with time, the concentration of the product increases and the concentration of the reactant decreases as it is getting consumed. After some time, the concentration does not change any further. Then it is said that the reaction is in equilibrium concentration. We will now see how to calculate equilibrium concentration using the equilibrium concentration equation.

### Equilibrium Constant Kc

To understand how to calculate equilibrium concentration using the equilibrium concentration equation, you need to know the formula for equilibrium constant Kc. When the chemical is in equilibrium, the ratio of the products to the reactants is called the equilibrium constant.

Consider a chemical reaction,

aA +bB cC + dD

For this equation, the equilibrium constant is defined as:

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

### The ICE Table

The simplest way of finding the equilibrium concentration equation is by adopting the ICE table. It is an organised table to identify what quantity of products and reactants are given and what quantity needs to be found. The equilibrium constant and table will be very beneficial when we look at how to calculate equilibrium concentration.

• ‘I’ stands for initial concentration.

• ‘C’ stands for the change in concentrations.

• ‘E’ stands for equilibrium concentration.

### Steps to Calculate Equilibrium Concentration

There are a few steps that need to be carried out to find the equilibrium concentration of a chemical reaction. The steps are as below.

1. The first step is to write down the balanced equation of the chemical reaction. aA +bB cC + dD

2. The second step is to convert the concentration of the products and the reactants in terms of their Molarity.

3. The third step is to form the ICE table and identify what quantities are given and what all needs to be found.

## ICE Table

 ICE A B C D Initial a b c d Change x x x X Equilibrium a-x b-x c-x d-x

1. Now using the formula for equilibrium constant, we will obtain an equation in terms of the unknown variable ‘x’.

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

1. The last step is to solve the quadratic equation to find the value of ‘x’.

Now that you know how to calculate equilibrium concentration let’s look at some solved problems for better understanding.

### Solved Problems

Question 1) Find the equilibrium concentration of 6 moles of PCl5 is kept in a 1L vessel at 300K temperature. Assume Kc to be equal to 1.

Answer 1) the first step is to write the chemical reactions

PCl5PCl3+Cl2

PCl5 = 6 moles

## Concentration of PCl5 = 6 moles / 1L = 6 M

 ICE PCl5 Cl3 Cl2 Initial 6 0 0 Change x -x -x Equilibrium 6-x x X

Using Kc formula, we get,

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

$K_{c}$ = $\frac{[PCI_{3}][Cl_{2}]}{[PCI_{5}]}$

1 = $\frac{x * x}{(6 - x)}$

$x^{2}$ + x - 6 = 0

Upon solving the quadratic equation, we get, x = 2, and x = -3.

X cannot be a negative number, therefore x = 2.

Substituting the value of x we get,

[PCl5] = 6 – x = 6 – 2 = 4M

[PCl3] = [Cl2] = x = 2M

Question 2) Find the concentration for each substance in the following reaction.

C2H4 + H4 C2H6

 ICE C2H4 H2 C2H6 Initial 0.33 0.53 0 Change -x -x x Equilibrium 0.33-x 0.53-x x

Using Kc formula, we get,

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

$K_{c}$  = $\frac{[C_{2} H_{6}]}{[C_{2}H_{4}][H_{2}]}$

0.98 = $\frac{x}{(0.33 - x)(0.53-x)}$

0.98 = $\frac{x}{x^{2} - 0.86x\: +\: 0.1749}$

0.98($x^{2}$ - 0.86x + 0.1749) = x

0.98$x^{2}$ - 1.8428x + 0.1714 = 0

Upon solving the quadratic equation, we get, x = 1.78, and x = 0.098.

Let x = 1.78,

[C2H4] = 0.33 – 1.78 = -1.45,

The concentration cannot be negative; hence we discard x = 1.78.

Let x = 0.098,

[C2H4] = 0.33 – 0.098= 0.23,

Therefore, we get the following equilibrium concentration,

[C2H4] = 0.23M

[H2] = 0.43M

[C2H6] = 0.098M.