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. 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. Partial pressure of nitrogen __________ kPa Partial pressure of hydrogen __________ kPa Partial pressure of ammonia __________ kPa 2. Give an expression for the equilibrium constant (Kp) for this reaction. Kp = __________ 3. In a different equilibrium mixture, under different conditions, the partial pressures of the gases are shown in Table 2. | Gas | Partial pressure / kPa | |------|------------------------| | N2 | 1.20 x 10² | | H2 | 1.50 x 10² | | NH3 | 1.10 x 10² | Calculate the value of the equilibrium constant (Kp) for this reaction and give its units. Kp = __________ Units = __________ 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 Kp for this reaction. Justify your answer. Effect on Kp __________ Justification __________

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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

Question 2

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.png

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

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Answer

To find the partial pressures, we will use the total pressure and the mole fraction of NH3.

Given that:

Total pressure (P_total) = 150 kPa
Mole fraction of NH3 = 0.80

First, calculate the partial pressure of NH3:

$P_{NH3} = P_{total} imes mole ext{ }fraction_{NH3} = 150 ext{ kPa} imes 0.80 = 120 ext{ kPa}$

Next, the remaining fraction must be shared between N2 and H2. The mole fraction of N2 and H2 can be calculated as follows:

Mole fraction of NH3 = 0.80 Mole fraction of N2 + Mole fraction of H2 = 1 - 0.80 = 0.20

Assuming the mole ratio of N2 to H2 is 1:3, if x is the mole fraction of N2, then:

Mole fraction of N2 = x
Mole fraction of H2 = 3x

Thus, $x + 3x = 0.20$ $4x = 0.20$ $x = 0.05$

Therefore:

Mole fraction of N2 = 0.05
Mole fraction of H2 = 0.15

Now calculate the partial pressures:

Partial pressure of N2: $P_{N2} = P_{total} imes mole ext{ }fraction_{N2} = 150 ext{ kPa} imes 0.05 = 7.5 ext{ kPa}$

Partial pressure of H2: $P_{H2} = P_{total} imes mole ext{ }fraction_{H2} = 150 ext{ kPa} imes 0.15 = 22.5 ext{ kPa}$

Thus, the calculated values are:

Partial pressure of nitrogen: 7.5 kPa
Partial pressure of hydrogen: 22.5 kPa
Partial pressure of ammonia: 120 kPa

Step 2

2. Give an expression for the equilibrium constant (Kp) for this reaction.

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Answer

The expression for the equilibrium constant (Kp) for the reaction is:

$K_p = \frac{{(P_{NH3})^2}}{{(P_{N2})(P_{H2})^3}}$

Where:

$P_{NH3}$ is the partial pressure of ammonia
$P_{N2}$ is the partial pressure of nitrogen
$P_{H2}$ is the partial pressure of hydrogen

Step 3

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

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Answer

Using the values from Table 2:

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

Now substitute these values into the Kp expression:

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

Calculating this gives:

Find $(1.10 \times 10^2)^2 = 1.21 \times 10^4$ kPa²
Find $(1.50 \times 10^2)^3 = 3.375 \times 10^6$ kPa³
Now substitute:

$K_p = \frac{1.21 \times 10^4}{(1.20 \times 10^2) \times 3.375 \times 10^6}$

Calculate $K_p$ : This yields the value (after simplifying)...

Units for Kp are kPa $^{-1}$ due to the difference in pressures in the numerator and denominator.

Step 4

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

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Answer

Effect on Kp: Kp will decrease.

Justification: Since the reaction has a negative enthalpy change (-92 kJ mol⁻¹), it is exothermic. According to Le Chatelier's Principle, increasing the temperature would shift the equilibrium position to favor the endothermic direction (the reverse reaction in this case), thus decreasing the value of Kp.

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

2. Give an expression for the equilibrium constant (Kp) for this reaction.

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

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

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