Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen. $$2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$$ A sample of sulfur trioxide was heated and allowed to reach equilibrium at a given temperature. The equilibrium mixture contained 6.08 g of sulfur dioxide. Calculate the mass, in g, of oxygen gas in the equilibrium mixture. --- A different mass of sulfur trioxide was heated and allowed to reach equilibrium at 1050 K. The amounts of each substance in the equilibrium mixture are shown in Table 4. | Substance | Amount at equilibrium / mol | |------------------|----------------------------| | sulfur trioxide | 0.320 | | sulfur dioxide | 1.20 | | oxygen | 0.600 | For this reaction at 1050 K the equilibrium constant, Kc = 7.62 x 10^5 Pa. Calculate the mole fraction of each substance at equilibrium. Give the expression for the equilibrium constant, Kc. Calculate the total pressure, in Pa, of this equilibrium mixture. Mole fraction $SO_3$: Mole fraction $SO_2$: Mole fraction $O_2$: Kc: Total pressure: --- For this reaction at 1050 K the equilibrium constant, Kc = 7.62 x 10^5 Pa. For this reaction at 500 K the equilibrium constant, Kc = 3.94 x 10^4 Pa. Explain how this information can be used to deduce that the forward reaction is endothermic.

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Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 1

Question 7

Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen. $$2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$$ A samp... show full transcript

Worked Solution & Example Answer:Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 1

Step 1

Calculate the mass, in g, of oxygen gas in the equilibrium mixture.

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Answer

To find the mass of oxygen gas, we first calculate the number of moles of sulfur dioxide (SO₂). Using the molar mass of sulfur dioxide (approximately 64.07 g/mol):

$n = \frac{6.08 g}{64.07 g/mol} \approx 0.0950 mol$

Since the stoichiometry of the reaction shows that 2 moles of SO₂ are produced from 2 moles of SO₃ and 1 mole of O₂, we can determine the number of moles of oxygen produced:

Given that the amount of sulfur dioxide is 1.20 mol: 1 mole of O₂ corresponds to 2 moles of SO₂ (with 1 mole of SO₂ producing half a mole of O₂), therefore,

$n_{O_2} = \frac{1.20}{2} = 0.600 mol$

The mass of oxygen gas can then be found using the molar mass of oxygen (approximately 32.00 g/mol):

$Mass = n_{O_2} \times 32.00 g/mol = 0.600 mol \times 32.00 g/mol = 19.20 g$

Step 2

Calculate the mole fraction of each substance at equilibrium.

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Answer

To calculate the mole fractions, we first sum the total number of moles:

$n_{total} = n_{SO_3} + n_{SO_2} + n_{O_2} = 0.320 + 1.20 + 0.600 = 2.120 mol$

Next, we can calculate each mole fraction:

Mole fraction of $SO_3$ : $\text{Mole fraction } SO_3 = \frac{n_{SO_3}}{n_{total}} = \frac{0.320}{2.120} \approx 0.151$
Mole fraction of $SO_2$ : $\text{Mole fraction } SO_2 = \frac{n_{SO_2}}{n_{total}} = \frac{1.20}{2.120} \approx 0.566$
Mole fraction of $O_2$ : $\text{Mole fraction } O_2 = \frac{n_{O_2}}{n_{total}} = \frac{0.600}{2.120} \approx 0.283$

Step 3

Give the expression for the equilibrium constant, Kc.

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Answer

The expression for the equilibrium constant Kc for the reaction is given by:

$K_c = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$

Step 4

Calculate the total pressure, in Pa, of this equilibrium mixture.

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Answer

The total pressure can be calculated using the ideal gas law, using the values of moles and the total volume (assuming 1 atm conditions). The total pressure is derived from:

$P = \frac{n_{total}RT}{V}$

Assuming a volume V = 1 m³ and R (ideal gas constant) = 8.314 J/(mol K), we find:

$P = \frac{(2.120 mol)(8.314 J/(mol K))(1050 K)}{1 m^3} = 17,591.88 Pa \approx 1.76 \times 10^4 Pa$

Step 5

Explain how this information can be used to deduce that the forward reaction is endothermic.

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Answer

The given values of Kc at different temperatures can be analyzed. The value of Kc at 1050 K (7.62 x 10^5 Pa) is higher than that at 500 K (3.94 x 10^4 Pa). According to Le Châtelier's principle, if the equilibrium constant increases with temperature, it indicates that the forward reaction absorbs heat, characteristic of an endothermic reaction. Thus, the forward reaction's endothermic nature is confirmed by this data.

Step 6

Use data from Question 07.3 to calculate the value of Kc at 500 K.

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Using the relation of equilibrium constants, Kc at lower temperatures (500 K) can be deduced through the change in equilibrium states. The equilibrium constants indicate that lower temperatures favor the reactants, implying a possible reversed tendency in this temperature range when considering endothermic reactions.

Step 7

Deduce the units of Kc.

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The units of Kc can be derived from its expression:

$K_c = \frac{(mol m^{-3})^2(mol m^{-3})}{(mol m^{-3})^2} = mol m^{-3}$

Thus, the units for Kc can be expressed as mol/m³.

Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 1

Worked Solution & Example Answer:Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 1

Calculate the mass, in g, of oxygen gas in the equilibrium mixture.

Calculate the mole fraction of each substance at equilibrium.

Give the expression for the equilibrium constant, Kc.

Calculate the total pressure, in Pa, of this equilibrium mixture.

Explain how this information can be used to deduce that the forward reaction is endothermic.

Use data from Question 07.3 to calculate the value of Kc at 500 K.

Deduce the units of Kc.

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