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The equation for the reaction between ammonia and oxygen is shown - AQA - A-Level Chemistry - Question 3 - 2018 - Paper 1

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The equation for the reaction between ammonia and oxygen is shown. 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) ΔH = -905 kJ mol⁻¹ Some standard entropies are given in Tabl... show full transcript

Worked Solution & Example Answer:The equation for the reaction between ammonia and oxygen is shown - AQA - A-Level Chemistry - Question 3 - 2018 - Paper 1

Step 1

Calculate the entropy change for the reaction between ammonia and oxygen

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Answer

To calculate the entropy change (ΔS) for the reaction, we use the formula:

ΔS=ΣS°productsΣS°reactantsΔS = Σ S°_{products} - Σ S°_{reactants}

For the products:

4NO(g) + 6H₂O(g) →

=4(211)+6(189)=844+1134=1978extJK1extmol1= 4(211) + 6(189) = 844 + 1134 = 1978 ext{ J K}^{-1} ext{ mol}^{-1}

For the reactants:

4NH₃(g) + 5O₂(g) →

=4(193)+5(205)=772+1025=1797extJK1extmol1= 4(193) + 5(205) = 772 + 1025 = 1797 ext{ J K}^{-1} ext{ mol}^{-1}

Thus, the entropy change is:

ΔS=19781797=181extJK1extmol1ΔS = 1978 - 1797 = 181 ext{ J K}^{-1} ext{ mol}^{-1}

Step 2

Calculate a value for the Gibbs free-energy change (ΔG), in kJ mol⁻¹, for the reaction between ammonia and oxygen at 600 °C

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Answer

To calculate ΔG, we use the Gibbs free energy equation:

ΔG=ΔHTΔSΔG = ΔH - TΔS

Where:

  • ΔH = -905 kJ mol⁻¹ (from the reaction)
  • T = 600 °C = 873 K (convert to Kelvin)
  • ΔS = 181 J K⁻¹ mol⁻¹ = 0.181 kJ K⁻¹ mol⁻¹ (convert J to kJ)

Now calculate the value:

ΔG=905(873)(0.181)ΔG = -905 - (873)(0.181) ΔG=905158.553=1063.553extkJmol1ΔG = -905 - 158.553 = -1063.553 ext{ kJ mol}^{-1}

Therefore,

ΔGextat600°Cextisapproximately1063.6extkJmol1ΔG ext{ at } 600 °C ext{ is approximately } -1063.6 ext{ kJ mol}^{-1}

Step 3

Explain how this change affects the value of ΔG for the reaction

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Answer

When the reaction between ammonia and oxygen is conducted at a higher temperature, the Gibbs free energy change (ΔG) becomes less negative. This is due to the equation:

ΔG=ΔHTΔSΔG = ΔH - TΔS

Having a larger T results in a higher value subtracted from ΔH, which decreases the overall negativity of ΔG. Consequently, if the temperature increases, the reaction becomes less spontaneous.

Step 4

Describe the stages of this alternative route

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Answer

  1. Adsorption: Reactants (NH₃ and O₂) are adsorbed onto the platinum surface, creating a bond with the surface atoms, which lowers the activation energy.

  2. Bond breaking: The existing bonds in the NH₃ and O₂ molecules are weakened as they interact with the platinum surface, making it easier for the bonds to break.

  3. Transition state formation: An alternative reaction pathway is formed due to the presence of the catalyst, stabilizing the transition state.

  4. Reaction: The bonds in the reactants are reformed to produce products (NO and H₂O) once the transition state is reached.

  5. Desorption: The produced products are released from the platinum surface back into the reaction mixture.

Step 5

Deduce the change in oxidation state of nitrogen, when NH₃ is oxidised to NO

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Answer

In ammonia (NH₃), nitrogen has an oxidation state of -3. When oxidised to nitric oxide (NO), nitrogen has an oxidation state of +2. Thus, the change in oxidation state is:

+2(3)=+5ext(anincrease)+2 - (-3) = +5 ext{ (an increase)}

Step 6

Give an equation for this reaction

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Answer

The equation for the reaction where ammonia reacts with oxygen to produce nitrous oxide is:

2NH3+O2ightarrowN2O+3H2O2NH₃ + O₂ ightarrow N₂O + 3H₂O

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