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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) + 1/2O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy - 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)-+-1/2O2(g)-→-Cu2O(s)--ΔH-=--170-kJ-mol⁻¹--Enthalpy-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) + 1/2O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy

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) + 1/2O2(g) → Cu2O(s) ΔH = -170 kJ mol⁻¹ Enthalpy - 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 presented are:

  1. The formation of copper(II) oxide:

    2Cu(s)+O2(g)→2CuO(s)(ΔH=−312 kJ mol−1)2Cu(s) + O_2(g) → 2CuO(s) \quad (\Delta H = -312 \text{ kJ mol}^{-1})

  2. The conversion of copper(II) oxide to copper(I) oxide:

    2CuO(s)+12O2(g)→Cu2O(s)(ΔH=−170 kJ mol−1)2CuO(s) + \frac{1}{2}O_2(g) → Cu_2O(s) \quad (\Delta H = -170 \text{ kJ mol}^{-1})

Step 2

Calculate the overall enthalpy change for the reaction

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Answer

To find the overall enthalpy change for the transformation from 2Cu(s) and O2(g) to Cu2O(s), we can combine the two reactions. The total enthalpy change, ( \Delta H_{total} ), is calculated as follows:

ΔHtotal=ΔHreaction1+ΔHreaction2\Delta H_{total} = \Delta H_{reaction 1} + \Delta H_{reaction 2}

Substituting in the values:

ΔHtotal=(−312)+(−170)\Delta H_{total} = (-312) + (-170)

Therefore:

ΔHtotal=−482 kJ mol−1\Delta H_{total} = -482 \text{ kJ mol}^{-1}

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