Content Curator
Chemical Equilibrium deals with the concentration of the products and reactants. A chemical equation is said to be in equilibrium when the concentration of the reactants and the products do not change with time. It is said that at equilibrium the rate of forward reaction is same as the rate of backward reaction. For calculating the concentrations of a chemical at equilibrium we need to know its initial concentration, at the beginning of the reaction.
Read More: Concepts in Chemistry
| Table of Content |
Key Takeaways: Molarity, moles, molality, concentration, equilibrium constant, reactants, chemical reaction, products, rate of reaction, solution.
Equilibrium in Chemical Processes
[Click Here for Sample Questions]
Let’s consider a reversible equation for understanding equilibrium in chemical reactions
\(A+B \rightleftharpoons C+D\)
Here A and B are reactants while C and D are products. When the reaction begins, the concentration of A and B starts reducing while the concentration of C and D increases. A and B react to give C and D. As time goes, the reactants do not completely deplete but instead there is a decrease in forward reaction alongside an increase in reverse reaction. Later, both the reactions follow the same rate and there is no change in the concentrations of the reactants as well as of the products as they have achieved equilibrium.
In the above graph, the red line is for the reactants and the blue is for the products.
Important Read: Le Chatelier’s Principle, Henry’s Law
Equilibrium Constant Formula
[Click Here for Sample Questions]
The concentrations in an equilibrium mixture can be written in the form of an expression using equilibrium constant Formula. The concentration ratio of the products to the reactants is called the equilibrium constant formula.
For a chemical reaction with reactants as aA and bB; products cC and dD, the equilibrium constant formula can be written as,
\(K_C= \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
Where a, b, c, d denote the number of moles and A, B, C, and D denote the concentration. KC is the equilibrium constant.
Read Also:
ICE Table
[Click Here for Sample Questions]
ICE table provides an organised way of solving equilibrium constant questions. It makes it easier to place products, reactants, and the terms to be found in a tabular form. Here,
I is for ‘Initial Concentration’
C is for ‘Change in Concentration’
E is for ‘Equilibrium Concentration’
| For Reactant aA and product bB | ||
| ICE Table | A | B |
| Initial Concentration | a | b |
| Change | x | x |
| Equilibrium Concentration | a-x | b-x |
Also Read: Thermodynamics
Steps to Calculate Equilibrium Concentration
[Click Here for Sample Questions]
- The most important step will be to first write down the equation and balance it.
- Convert all the values of concentration of reactants and products into Molarity.
- Convert the given data into an ICE table, label the unknown data as ‘x’
- Apply the equilibrium constant formula \(K_C= \frac{[C]^c[D]^d}{[A]^a[B]^b}\) to get a quadratic equation
- Solve the quadratic equation to find the value of ‘x’.
Read More:
Things to Remember
- Square brackets are always used to denote the concentration of a compound.
- The equation is balanced based on the equal number of atoms/molecules participating in the reactant and its resultant product sides. The number of atoms of an element should be equal on both sides.
- At constant temperatures, the values of equilibrium constant remain the same; equilibrium constant is temperature sensitive.
- A dynamic equilibrium is achieved when the number of molecules remaining in the liquid becomes equal to the number of molecules leaving the liquid to become vapour.
- For a reverse reaction at constant temperature, the value of equilibrium constant is inverse of the equilibrium constant of the forward reaction.
- Equilibrium constant of a reaction is used to predict the direction of the reaction, the extent of a reaction and to calculate the equilibrium concentrations.
Read Also:
Sample Questions
Ques. The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [2 marks]
[N2] = 1.5 × 10–2 M,
[H2] = 3.0 ×10–2 M and
[NH3] = 1.2 ×10–2 M. Calculate equilibrium constant.
Ans.
The given equation is \(N_2+H_2 \rightarrow NH_3\)
After balancing the equation is \(N_2+3H_2 \rightarrow 2NH_3\)
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[NH_3]^2}{[N_2]^1[H_2]^3}\)
\(K_C= \frac{[1.2 \times10^{–2} M]^2}{[1.5 \times 10^{–2} M]^1[3.0 ×10^{–2} M]^3}\)
\(K_C = 0.106 \times 10^4\)
\(K_C = 1.06 \times 10^3\)
Ques. At equilibrium, the concentrations of [2 marks]
[N2] = 3.0 × 10–3 M,
[O2] = 4.2 × 10–3 M and
[NO] = 2.8 × 10–3 M in a sealed vessel at 800K. What will be KC for the reaction?
N2 (g) + O2 (g) \(\rightarrow\) 2NO (g)
Ans.
The given equation N2 (g) + O2 (g) \(\rightarrow\) 2NO (g) is a balanced equation.
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[NO]^2}{[N_2]^1[O_2]^1}\)
\(K_C= \frac{[2.8 \times 10^{–3} M]^2}{[3.0 × 10^{–3} M]^1[4.2 × 10^{–3} M]^1}\)
\(K_C = 0.622\)
Ques. Nitrogen reacts with Chlorine to produce Nitrogen Trichloride. At equilibrium, the concentration of each gas were found to be [2 marks]
[N2] = 0.15 M
[Cl2] = 0.25 M
[NCl3] = 0.50 M
Calculate the value of the equilibrium constant KC
Ans.
The given equation is N2 (g) + Cl2 (g) \(\rightarrow\) NCl3(g)
After balancing the equation is N2 (g) + 3Cl2 (g) \(\rightarrow\) 2NCl3(g)
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[NCl_3]^2}{[N_2]^1[Cl_2]^3}\)
\(K_C= \frac{[0.50\ M]^2}{[0.15\ M]^1[0.25\ M]^3}\)
\(K_C = 0.10667 \times 10^4\)
\(K_C = 1.0667 \times 10^3\)
Ques. Given the value of K for the reaction shown below, what is the value of K for the adjusted reaction? [2 marks]
2NO (g) + O2 (g) \(\rightarrow\) 2NO2 (g); K = 100
4NO (g) + 2O2 (g) \(\rightarrow\) 4NO2 (g); K′ = ?
Ans.
While observing both the reactions we can see that the second reaction is double of the first reaction thus the new value of K would be,
K′ = K2 = 1002 = 10,000
Let us see how we came to this conclusion,
K′ \(= \frac{[NO_2]^4}{[NO]^4[O_2]^2}\) as per the equilibrium constant expression for the second equation
It can also be written as,
K′ \(= \left[ \frac{[NO_2]^2}{[NO]^2[O_2]^1} \right]^2\)
As per the first equation, K \(=\frac{[NO_2]^2}{[NO]^2[O_2]^1}\) which means K′ = K2
Thus K′ can be written as K′ = K2
Ques. Given the value of K for the reaction shown below, what is the value of K for the adjusted reaction? [2 marks]
2NO (g) + O2 (g) \(\rightarrow\) 2NO2 (g); K = 100
2NO2 \(\rightarrow\) 2NO (g) + O2 (g); K′ = ?
Ans. We can see that the second equation is a reversed equation of the first equation, in such cases,
K′ \(= \frac{1}{K}=\frac{1}{100}=0.01\)
Let us see how we came to this conclusion,
K′ \(=\frac{[NO]^2[O_2]^1}{[NO_2]^2}\) as per the equilibrium constant expression for the second equation
As per the first equation, K \(=\frac{[NO_2]^2}{[NO]^2[O_2]^1}\) Thus, K′ can also be written as,
\(K' =\cfrac{1}{\cfrac{1}{\cfrac{[NO]^2[O_2]^1}{[NO_2]^2}}}\)
\(K' =\cfrac{1}{\cfrac{[NO_2]^2}{[NO]^2[O_2]^1}}\)
Thus, K′ \(= \frac{1}{K}\)
Ques. 20 mol of NOCl is placed inside an empty 4L container. At equilibrium, 8 mol of Cl2 was found to be in the container. Calculate the value of KC for this reaction. [3 marks]
2 NOCl (g) \(\rightarrow\) 2 NO (g) + Cl2 (g)
Ans.
The given equation 2 NOCl (g) \(\rightarrow\) 2 NO (g) + Cl2 (g) is a balanced equation
There are 20 moles of NOCl in 4L container thus it is 20 / 4 = 5 M of NOCl
At equilibrium, 8 mol of Cl2 was found to be in a 4L container. Thus it is 8 / 4 = 2 M
Thus the reaction direction is from left to right, thus initially the concentration of the products will be 0. The ICE table -
| ICE Table | 2 NOCl | 2 NO | Cl2 |
| Initial Concentration | 5 M | 0 | 0 |
| Change | -2x (concentration decreases) | +2x (concentration increases) | +1x (concentration increases) |
| Equilibrium Concentration | 5 – 2x | 2x | x |
We know that concentration at equilibrium of [Cl2] is 2M, thus x = 2 M
[NO] = 2x = 2 (2 M) = 4 M
[NOCl] = 5 – 2x = 5 – 4 = 1 M
Ques. The following concentrations were obtained for the formation of NH4Cl from NH3 and HCl at equilibrium at 500K. [2 marks]
[NH3] = 0.15 M
[HCl] = 0.25 M
[NH4Cl] = 0.50 M
Calculate the value of the equilibrium constant KC
Ans.
After balancing the equation is NH3 + HCl \(\rightarrow\) NH4Cl
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[NH_4Cl]^1}{[NH_3]^1[HCl]^1}\)
\(K_C= \frac{[1.2 ×10^{–2} M]^1}{[1.5 × 10^{–2} M]^1[3.0 ×10^{–2} M]^1}\)
KC = 0.2667 × 102
KC = 26.67
Ques. The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [2 marks]
[SO2] = 1.5 × 10–2 M,
[O2] = 3.0 ×10–2 M and
[2SO3] = 1.2 ×10–2 M.
2SO2 (g) + O2 (g) ? 2SO3 (g)
Calculate equilibrium constant.
Ans.
The given balanced equation is 2SO2 (g) + O2 (g) ? 2SO3 (g)
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[SO_3]^2}{[SO_2]^2[O_2]^1}\)
\(K_C= \frac{[1.2 ×10^{–2} M]^2}{[1.5 × 10^{–2} M]^2[3.0 ×10^{–2} M]^1}\)
KC = 0.2133 × 104
KC = 2.133 × 103
Ques. The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K. [2 marks]
[N2] = 1.0 × 10–2 M,
[H2] = 2.0 ×10–2 M and
[NH3] = 1.0 ×10–2 M.
Calculate equilibrium constant.
Ans.
The given equation is N2 + H2 \(\rightarrow\) NH3
After balancing the equation is N2 + 3H2 \(\rightarrow\) 2NH3
Here the question directly asks to find the equilibrium constant so we will not plot the ICE table.
\(K_C= \frac{[NH_3]^2}{[N_2]^1[H_2]^3}\)
\(K_C= \frac{[1.0 ×10^{–2} M]^2}{[1.0 × 10^{–2} M]^1[2.0 ×10^{–2} M]^3}\)
KC = 0.125 × 104
KC = 1.25 × 103
Ques. 40 mol of NH4Cl is placed inside an empty 4L container. At equilibrium, 8 mol of NH3 was found to be in the container. Calculate the value of KC for this reaction. [3 marks]
NH4Cl \(\rightarrow\) NH3 + HCl
Ans.
The given equation NH4Cl \(\rightarrow\) NH3 + HCl is a balanced equation
There are 40 moles of NH4Cl in 4L container thus it is 40 / 4 = 10 M of NH4Cl
At equilibrium, 8 mol of NH3 was found to be in a 4L container. Thus it is 8 / 4 = 2 M
Initially the concentration of the products will be 0. The ICE table -
| ICE Table | NH4Cl | NH3 | HCl |
| Initial Concentration | 10 M | 0 | 0 |
| Change | -x (concentration decreases) | +1x (concentration increases) | +1x (concentration increases) |
| Equilibrium Concentration | 10 – x | x | x |
We know that concentration at equilibrium of [HCl] is 2M, thus x = 2 M
[NH3] = x = 2 M
[NH4Cl] = 10 – x = 10 – 2 = 8 M
Read Also:
Comments