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

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

At a certain temperature, the equilibrium constant (Kc) for the reaction is 0.050. If 0.20 moles of N2 and 0.40 moles of H2 are placed in a 2.0-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 / 2.0 litres = 0.10 M

[H2]0 = 0.40 moles / 2.0 litres = 0.20 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:

0.050 = (0 + 2x)² / ((0.10 - x) X (0.20 - 3x)³)

Expand and rearrange the equation to solve for x:

0.050 = (4x²) / ((0.10 - x) X (0.20 - 3x)³)

Now, solve for x:

0.050 = 4x² / (0.10 - x) X (0.008 - 0.144x + 0.648x²)

Multiply both sides by (0.10 - x) X (0.008 - 0.144x + 0.648x²):

0.050 X (0.10 - x) X (0.008 - 0.144x + 0.648x²) = 4x²

Distribute and rearrange:

0.0048 - 0.072x + 0.324x² = 4x²

Bring all terms to one side and simplify:

0.324x² - 4x² + 0.072x - 0.0048 = 0

Combine like terms:

3.676x² + 0.072x - 0.0048 = 0

Use the quadratic formula to find the values of x:

x = [-0.072 ± √(0.072² - 4 X 3.676 X (-0.0048))] / 2 X 3.676

x ≈ [-0.072 ± √(0.005184 + 0.070656)] / 7.352

x ≈ [-0.072 ± √0.07584] / 7.352

x ≈ [-0.072 ± 0.2754] / 7.352

x ≈ 0.203 or x ≈ -0.032

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

Now, calculate the equilibrium concentrations:

[N2]eq = [N2]0 - x ≈ 0.10 - 0.203 ≈ -0.103 M (discard negative value)

[H2]eq = [H2]0 - 3x ≈ 0.20 - 3 X 0.203 ≈ -0.009 M (discard negative value)

[NH3]eq = [NH3]0 + 2x ≈ 0 + 2 X 0.203 ≈ 0.406 M

Since concentrations cannot be negative, we consider the equilibrium concentration of N2 to be 0 M, the equilibrium concentration of H2 to be 0 M, and the equilibrium concentration of NH3 to be 0.406 M.

Therefore, at equilibrium, [N2]eq ≈ 0 M, [H2]eq ≈ 0 M, and [NH3]eq ≈ 0.406 M.

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