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The following energy profile relates to the two reactions 2Cu(s) + O2(g) → 2CuO(s) ΔH = -312 kJ mol⁻¹ 2CuO(s) + 1/2 O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy 2Cu(s) + O2(g) ΔH = -170 kJ mol⁻¹ Cu2O(s) + 1/2 O2(g) → 2CuO(s) ΔH = -312 kJ mol⁻¹ - VCE - SSCE Chemistry - Question 7 - 2008 - Paper 1

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Question 7

The-following-energy-profile-relates-to-the-two-reactions--2Cu(s)-+-O2(g)-→-2CuO(s)-ΔH-=--312-kJ-mol⁻¹-2CuO(s)-+-1/2-O2(g)-→-Cu2O(s)-ΔH-=--170-kJ-mol⁻¹--Enthalpy--2Cu(s)-+-O2(g)-ΔH-=--170-kJ-mol⁻¹-Cu2O(s)-+-1/2-O2(g)-→-2CuO(s)-ΔH-=--312-kJ-mol⁻¹-VCE-SSCE Chemistry-Question 7-2008-Paper 1.png

The following energy profile relates to the two reactions 2Cu(s) + O2(g) → 2CuO(s) ΔH = -312 kJ mol⁻¹ 2CuO(s) + 1/2 O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy 2C... show full transcript

Worked Solution & Example Answer:The following energy profile relates to the two reactions 2Cu(s) + O2(g) → 2CuO(s) ΔH = -312 kJ mol⁻¹ 2CuO(s) + 1/2 O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy 2Cu(s) + O2(g) ΔH = -170 kJ mol⁻¹ Cu2O(s) + 1/2 O2(g) → 2CuO(s) ΔH = -312 kJ mol⁻¹ - VCE - SSCE Chemistry - Question 7 - 2008 - Paper 1

Step 1

Identify the reactions and their enthalpy changes

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Answer

The two reactions can be identified along with their respective enthalpy changes:

  1. Formation of Copper(I) Oxide:

    2Cu(s)+O2(g)2CuO(s)(ΔH=312 kJ mol1)2Cu(s) + O2(g) \rightarrow 2CuO(s) \quad (\Delta H = -312 \text{ kJ mol}^{-1})

    This reaction shows the formation of copper(II) oxide from elemental copper and oxygen, with an enthalpy change of -312 kJ mol⁻¹, indicating that it is exothermic.

  2. Formation of Copper(I) Oxide:

    2CuO(s)+12O2(g)Cu2O(s)(ΔH=170 kJ mol1)2CuO(s) + \frac{1}{2}O2(g) \rightarrow Cu2O(s) \quad (\Delta H = -170 \text{ kJ mol}^{-1})

    This reaction indicates the conversion of copper(II) oxide to copper(I) oxide, which also releases energy, shown by a negative enthalpy change of -170 kJ mol⁻¹.

Step 2

Determine the overall enthalpy change

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Answer

To find the overall enthalpy change of the complete reaction from 2Cu(s) + O2(g) to Cu2O(s), we must combine the enthalpy changes from both reactions:

Starting with reaction 1, we have:

  1. Formation of Copper (II) Oxide: 2Cu(s)+O2(g)2CuO(s)(ΔH=312 kJ mol1)2Cu(s) + O2(g) \rightarrow 2CuO(s) \quad (\Delta H = -312 \text{ kJ mol}^{-1})

Next, using the reverse of reaction 2 to connect it with the first:

  1. Conversion of Copper (II) Oxide to Copper (I) Oxide: Cu2O(s)2CuO(s)+12O2(g)(ΔH=+170 kJ mol1)Cu2O(s) \rightarrow 2CuO(s) + \frac{1}{2}O2(g) \quad (\Delta H = +170 \text{ kJ mol}^{-1})

Overall enthalpy change is then calculated as follows:

ΔHtotal=(312)+(+170)=142 kJ mol1\Delta H_{total} = (-312) + (+170) = -142 \text{ kJ mol}^{-1}

Thus, the overall enthalpy change for the reaction converting 2Cu and O2 to Cu2O is -142 kJ mol⁻¹.

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