This question is about thermodynamics. Consider the reaction shown: $$2 ext{Al}_2 ext{O}_3(s) + 3 ext{C}(s) \to 4 ext{Al}(s) + 3 ext{CO}(g)$$ Table 7 shows some thermodynamic data. **Table 7** | Substance | $ ext{Al}_2 ext{O}_3(s)$ | $ ext{Al}(s)$ | $ ext{C}(s)$ | $ ext{CO}(g)$ | |-------------|---------------------------|----------------|----------------|----------------| | $ ext{ΔH}^ ext{f} ext{ / } ext{kJ mol}^{-1}$ | -1669 | 0 | 0 | -394 | | $S^ ext{°} ext{ / } ext{J K}^{-1} ext{ mol}^{-1}$ | 51 | 28 | 6 | 214 | 7.1 Explain why the standard entropy value for carbon dioxide is greater than that for carbon. 7.2 State the temperature at which the standard entropy of aluminium is 0 J K⁻¹ mol⁻¹. 7.3 Use the equation and the data in Table 7 to calculate the minimum temperature, in K, at which this reaction becomes feasible.

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This question is about thermodynamics - AQA - A-Level Chemistry - Question 7 - 2021 - Paper 1

Question 7

This question is about thermodynamics. Consider the reaction shown: $$2 ext{Al}_2 ext{O}_3(s) + 3 ext{C}(s) \to 4 ext{Al}(s) + 3 ext{CO}(g)$$ Table 7 shows som... show full transcript

Worked Solution & Example Answer:This question is about thermodynamics - AQA - A-Level Chemistry - Question 7 - 2021 - Paper 1

Step 1

7.1 Explain why the standard entropy value for carbon dioxide is greater than that for carbon.

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Answer

The standard entropy value for carbon dioxide (CO₂) is greater than that for carbon (C) because CO₂ exists as a gas, which has more disorder compared to solid carbon. The gaseous state allows for more possible microstates and configurations, thus resulting in a higher entropy value.

Step 2

7.2 State the temperature at which the standard entropy of aluminium is 0 J K⁻¹ mol⁻¹.

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Answer

The standard entropy of aluminium is 0 J K⁻¹ mol⁻¹ at absolute zero, which is 0 K.

Step 3

7.3 Use the equation and the data in Table 7 to calculate the minimum temperature, in K, at which this reaction becomes feasible.

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To find the minimum temperature for feasibility, we apply the Gibbs free energy equation:

$ΔG^ ext{°} = ΔH^ ext{°} - TΔS^ ext{°}$

For the reaction to be feasible, $ΔG^ ext{°}$ must be less than or equal to 0. From the thermodynamic data:

Calculate $ΔH^ ext{°} = (4 × 0) + (3 × -394) - (2 × -1669) = -1669 + 1182 = -487 ext{ kJ mol}^{-1}$ .
Calculate $ΔS^ ext{°} = (4 × 28) + (3 × 214) - (2 × 51) - (3 × 6) = 112 + 642 - 102 - 18 = 634 ext{ J K}^{-1} ext{ mol}^{-1}$ .
Rearranging the Gibbs free energy equation gives:

$T = \frac{ΔH^ ext{°}}{ΔS^ ext{°}}$
Substituting in the values (remember to convert $ΔH^ ext{°}$ into J):

$T = \frac{-487000 ext{ J mol}^{-1}}{634 ext{ J K}^{-1} ext{ mol}^{-1}} = 766.3 K$
Since $T$ must be a positive answer, we take the minimum temperature at which the reaction is feasible as:

$T_{min} = 766.3 K$

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This question is about thermodynamics - AQA - A-Level Chemistry - Question 7 - 2021 - Paper 1

Worked Solution & Example Answer:This question is about thermodynamics - AQA - A-Level Chemistry - Question 7 - 2021 - Paper 1

7.1 Explain why the standard entropy value for carbon dioxide is greater than that for carbon.

7.2 State the temperature at which the standard entropy of aluminium is 0 J K⁻¹ mol⁻¹.

7.3 Use the equation and the data in Table 7 to calculate the minimum temperature, in K, at which this reaction becomes feasible.

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