Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K. The equation for the reaction between nitrogen and hydrogen is shown. N2(g) + 3H2(g) ⇌ 2NH3(g) 1.1 When equilibrium was reached, the total pressure in the flask was 150 kPa and the mole fraction of NH3(g) in the mixture was 0.80 Calculate the partial pressure of each gas in this equilibrium mixture. 1.2 Give an expression for the equilibrium constant (Kc) for this reaction. 1.3 In a different equilibrium mixture, under different conditions, the partial pressures of the gases are shown in Table 2. Table 2 Gas Partial pressure / kPa N2 1.20 × 10² H2 1.50 × 10² NH3 1.10 × 10² Calculate the value of the equilibrium constant (Kc) for this reaction and give its units. 1.4 The enthalpy change for the reaction is -92 kJ mol⁻¹. State the effect, if any, of an increase in temperature on the value of Kc for this reaction. Justify your answer.

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Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K - AQA - A-Level Chemistry - Question 2 - 2018 - Paper 1

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Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K. The equation for the reaction between nitrog... show full transcript

Worked Solution & Example Answer:Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K - AQA - A-Level Chemistry - Question 2 - 2018 - Paper 1

Step 1

1.1 Calculate the partial pressure of each gas in this equilibrium mixture.

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Answer

To find the partial pressures, we start with the total pressure (150 kPa) and the mole fraction of NH3.

Mole fraction of NH3, $ext{X}_{NH3} = 0.80$ , thus:
- Total moles = 1 (as a representative sample)
- Moles of NH3 = $0.80 \times 1 = 0.80$ moles.
Using the remaining moles:
- Moles of N2 = $0.20 \text{ moles}$ (1 - 0.80 = 0.20)
- Moles of H2 = $\frac{3}{2}\text{moles of N2} = 3 \times 0.20 = 0.60 \text{moles}$ .
The mole fraction of nitrogen ( $X_{N2}$ ) and hydrogen ( $X_{H2}$ ) can be calculated:
- $X_{N2} = \frac{0.20}{0.20 + 0.60 + 0.80} = 0.20/1.60 = 0.125$
- $X_{H2} = \frac{0.60}{1.60} = 0.375$
Now, use the mole fractions to find the partial pressures:
- $pp_{N2} = X_{N2} \times P_{total} = 0.125 \times 150 = 18.75 \text{ kPa}$
- $pp_{H2} = X_{H2} \times P_{total} = 0.375 \times 150 = 56.25 \text{ kPa}$
- $pp_{NH3} = X_{NH3} \times P_{total} = 0.80 \times 150 = 120 \text{ kPa}$ .

Step 2

1.2 Give an expression for the equilibrium constant (Kc) for this reaction.

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Answer

The expression for the equilibrium constant ( $K_c$ ) can be derived from the balanced reaction:

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

where $[NH_3]$ , $[N_2]$ , and $[H_2]$ are the molar concentrations of ammonia, nitrogen, and hydrogen respectively.

Step 3

1.3 Calculate the value of the equilibrium constant (Kc) for this reaction and give its units.

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Answer

Given the partial pressures:

$pp_{N2} = 1.20 \times 10^2 \text{ kPa}$
$pp_{H2} = 1.50 \times 10^2 \text{ kPa}$
$pp_{NH3} = 1.10 \times 10^2 \text{ kPa}$

Using these values, we can substitute them into the equilibrium constant expression:

$K_c = \frac{(1.10 \times 10^2)^2}{(1.20 \times 10^2) \times (1.50 \times 10^2)^3}$

Calculating this gives: $K_c = \frac{(1.21 \times 10^4)}{(1.20 \times 10^2) \times (3.375 \times 10^6)} = \frac{1.21 \times 10^4}{4.05 \times 10^8} = 2.99 \times 10^{-5}$

The units for the equilibrium constant (assuming pressures in kPa) are:

$K_c: \text{kPa}^{-1}$

Step 4

1.4 State the effect, if any, of an increase in temperature on the value of Kc for this reaction. Justify your answer.

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Answer

Effect on Kc: Decrease / Smaller.

Justification: The given reaction is exothermic (negative enthalpy change). According to Le Chatelier's principle, increasing the temperature of an exothermic reaction shifts the equilibrium position to the left, favoring the reactants and decreasing the concentration of products. Thus, the value of $K_c$ will decrease.

Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K - AQA - A-Level Chemistry - Question 2 - 2018 - Paper 1

Worked Solution & Example Answer:Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K - AQA - A-Level Chemistry - Question 2 - 2018 - Paper 1

1.1 Calculate the partial pressure of each gas in this equilibrium mixture.

1.2 Give an expression for the equilibrium constant (Kc) for this reaction.

1.3 Calculate the value of the equilibrium constant (Kc) for this reaction and give its units.

1.4 State the effect, if any, of an increase in temperature on the value of Kc for this reaction. Justify your answer.

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