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 - VCE - SSCE Chemistry - Question 7 - 2008 - Paper 1

For the second reaction:

2CuO(s) + 1/2O2(g) → Cu2O(s)

The enthalpy change is

ΔH = -170 kJ mol⁻¹.

This indicates that this reaction is also exothermic but less so than the first.

To find the total enthalpy change for the overall reaction, we can sum the enthalpy changes of the individual reactions:

Total ΔH = ΔH1 + ΔH2 Total ΔH = -312 kJ mol⁻¹ + (-170 kJ mol⁻¹) Total ΔH = -482 kJ mol⁻¹

This indicates that the overall reaction is highly exothermic.