Ammonia is produced according to the following equilibrium equation. N2(g) + 3H2(g) ⇌ 2NH3(g) There are 4.50 moles of nitrogen gas, 1.00 mole of hydrogen gas and 5.80 moles of ammonia in a 10.0 L vessel. The system is at equilibrium at 298 K. The value of K_eq at this temperature is 748. How many moles of nitrogen gas need to be added to the vessel to increase the amount of ammonia by 0.050 moles?

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Ammonia is produced according to the following equilibrium equation - HSC - SSCE Chemistry - Question 31 - 2021 - Paper 1

Question 31

Ammonia is produced according to the following equilibrium equation. N2(g) + 3H2(g) ⇌ 2NH3(g) There are 4.50 moles of nitrogen gas, 1.00 mole of hydrogen gas and 5... show full transcript

Worked Solution & Example Answer:Ammonia is produced according to the following equilibrium equation - HSC - SSCE Chemistry - Question 31 - 2021 - Paper 1

Step 1

Calculate the initial concentrations

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Answer

The moles of each gas can be converted to concentration using the formula:

$C = \frac{n}{V}$

Where

$n$ is moles, and
$V$ is volume in liters.

Concentration of $N_2$ : $C_{N_2} = \frac{4.50}{10} = 0.450 \, \text{mol L}^{-1}$
Concentration of $H_2$ : $C_{H_2} = \frac{1.00}{10} = 0.100 \, \text{mol L}^{-1}$
Concentration of $NH_3$ : $C_{NH_3} = \frac{5.80}{10} = 0.580 \, \text{mol L}^{-1}$

Step 2

Set up the changes at equilibrium

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Answer

In the equilibrium equation, let $x$ be the moles of $N_2$ added. The system will adjust as follows:

$N_2$ : $4.50 + x$ moles
$H_2$ : $1.00 - \frac{3}{2}(0.050) = 1.00 - 0.075 = 0.925$ moles
$NH_3$ : $5.80 + 0.050 = 5.85$ moles

The new concentrations become:

Concentration of $N_2$ : $C_{N_2} = \frac{4.50 + x}{10}$
Concentration of $H_2$ : $C_{H_2} = \frac{0.925}{10}$
Concentration of $NH_3$ : $C_{NH_3} = \frac{5.85}{10}$

Step 3

Apply the equilibrium constant expression

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Answer

From the equilibrium constant expression:

$K_{eq} = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Substituting the values:

$748 = \frac{\left( \frac{5.85}{10} \right)^2}{\left( \frac{4.50 + x}{10} \right)\left( \frac{0.925}{10} \right)^3}$

Step 4

Solve for the moles of nitrogen gas needed

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Answer

Rearranging and solving:

Multiply through to eliminate the denominators: $748 \left( \frac{4.50 + x}{10} \right) \left( \frac{0.925^3}{1000} \right) = \left( \frac{5.85^2}{100} \right)$
Continue simplifying and isolating $x$ to find: $x \approx 1.3$

Thus, 1.3 moles of nitrogen gas need to be added.

Ammonia is produced according to the following equilibrium equation - HSC - SSCE Chemistry - Question 31 - 2021 - Paper 1

Worked Solution & Example Answer:Ammonia is produced according to the following equilibrium equation - HSC - SSCE Chemistry - Question 31 - 2021 - Paper 1

Calculate the initial concentrations

Set up the changes at equilibrium

Apply the equilibrium constant expression

Solve for the moles of nitrogen gas needed

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