Chemical Equilibrium - SS3 Chemistry Past Questions and Answers - page 1

In chemical equilibrium, the forward and reverse reactions are still occurring, but at the same rate. The concentrations of reactants and products remain constant, but they are not necessarily equal. Equilibrium is a dynamic balance where the system is continuously interconverting between reactants and products

A catalyst is a substance that speeds up the rate of a chemical reaction but does not affect the position of equilibrium. It increases the rate of both the forward and reverse reactions equally, allowing the system to reach equilibrium faster but not altering the position of equilibrium itself.

At equilibrium, the rates of the forward and reverse reactions are equal, meaning the system is in a dynamic balance. However, this does not necessarily imply that the concentrations of reactants and products are equal; rather, they remain constant but can have different values depending on the reaction's stoichiometry.

Let's denote the initial concentrations of N2 and H2 as [N2]0 and [H2]0, respectively, and the change in concentration of NH3 as x (since 2 moles of NH3 are formed for every mole of N2 reacted).

Initial concentrations:

[N2]0 = 0.20 moles / 1.0 litre = 0.20 M

[H2]0 = 0.40 moles / 1.0 litre = 0.40 M

[NH3]0 = 0 M (since no NH3 is present initially)

At equilibrium, the concentrations will be:

[N2]eq = [N2]0 - x

[H2]eq = [H2]0 - 3x

[NH3]eq = [NH3]0 + 2x

The equilibrium constant expression for the reaction is:

Kc = [NH3]eq² / ([N2]eq X [H2]eq³)

Substitute the given values into the equilibrium constant expression:

2.0 x 10³ = (0 + 2x)² / ((0.20 - x) X (0.40 - 3x)³)

Expand and rearrange the equation to solve for x:

2.0 x 10³ = (4x²) / ((0.20 - x) X (0.64 - 2.4x + 9x²))

Now, solve for x:

2.0 X 10³ = 4x² / (0.128 - 2.6x + 9x²)

Multiply both sides by (0.128 - 2.6x + 9x²):

2.0 X 10³ X (0.128 - 2.6x + 9x²) = 4x²

Distribute and rearrange:

256 - 520x + 1800x² = 4x²

Bring all terms to one side and simplify:

1800x² - 524x + 256 = 0

Use the quadratic formula to find the value of x:

x = [-(-524) ± √((-524)² - 4 X 1800 X 256)] / 2 X 1800

x ≈ [524 ± √(274576 - 230400)] / 3600

x ≈ (524 ± √442176) / 3600

x ≈ (524 ± 664.951) / 3600

x ≈ 0.447 or x ≈ -0.367

Since concentrations cannot be negative, we discard the negative value for x.

Now, calculate the equilibrium concentrations:

[N2]eq = [N2]0 - x ≈ 0.20 - 0.447 ≈ -0.247 M (discard negative value)

[H2]eq = [H2]0 - 3x ≈ 0.40 - 3 X 0.447 ≈ -0.041 M (discard negative value)

[NH3]eq = [NH3]0 + 2x ≈ 0 + 2 X 0.447 ≈ 0.894 M

Since concentrations cannot be negative, we consider the equilibrium concentrations of N2 and H2 to be 0 M, and the equilibrium concentration of NH3 to be 0.894 M.

Therefore, at equilibrium, [N2]eq ≈ 0 M, [H2]eq ≈ 0 M, and [NH3]eq ≈ 0.894 M.

(Note: The negative values for [N2]eq and [H2]eq indicate that these reactants are fully consumed at equilibrium, leaving no excess of N2 and H2.)

The equilibrium constant (Kc) expression is obtained by writing the concentrations of products (C) raised to their stoichiometric coefficients in the numerator and the concentrations of reactants (A and B) raised to their respective stoichiometric coefficients in the denominator. In this case, the stoichiometric coefficients of C, A, and B are 4, 2, and 3, respectively, leading to the expression Kc = [C]⁴ / ([A]² X [B]³).

The equilibrium constant (Kp) is expressed in terms of the partial pressures of reactants and products when the reaction involves gases. It is used when dealing with gaseous equilibria and is determined by writing the partial pressures of products raised to their stoichiometric coefficients in the numerator and the partial pressures of reactants raised to their respective stoichiometric coefficients in the denominator.

If the value of the equilibrium constant (Kc) is much greater than 1, it indicates that the concentration of products at equilibrium is significantly higher than the concentration of reactants. This suggests that the reaction favours the formation of products, and the equilibrium lies far to the right.

For an endothermic reaction (ΔH > 0), increasing the temperature will favour the endothermic direction (the forward reaction). As a result, the equilibrium position will shift to the right, favouring the formation of products.

f the value of the equilibrium constant (Kc) is less than 1, it indicates that the concentration of products at equilibrium is lower than the concentration of reactants. This suggests that the reaction favours the formation of reactants, and the equilibrium lies to the left.