For the reaction N2 g 3H2 g 2NH3 g At a certain... - SS3 Chemistry Chemical Equilibrium Question

For the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)

At a certain temperature, the equilibrium constant (Kc) is 2.0 x 10³. If 0.20 moles of N2 and 0.40 moles of H2 are placed in a 1.0-litre container and allowed to reach equilibrium, calculate the equilibrium concentrations of N2, H2, and NH3.

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.)

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