Sample 1Medium
An enthalpy diagram shows C(graphite) + O2(g) at the top. One arrow goes straight down to CO2(g), labelled ΔH = −393.5 kJ mol⁻¹. A second path goes first down to CO(g) + ½O2(g), with an unknown ΔH₁, and then from there down to CO2(g), labelled ΔH = −283.0 kJ mol⁻¹. What is ΔH₁, the standard enthalpy of formation of CO(g)?
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The answer is B.
By Hess's law the enthalpy change is the same by either path: ΔH₁ + (−283.0) = −393.5 kJ mol⁻¹. So ΔH₁ = −393.5 + 283.0 = −110.5 kJ mol⁻¹.
- A.
−221.0 kJ would be for forming 2 mol of CO; the diagram shows 1 mol.
- C.
Adding the two given values is wrong; by Hess's law ΔH₁ + (−283.0) = −393.5.
- D.
Forming CO from carbon and oxygen releases heat, so ΔH₁ is negative.
NCERT: Class 11 Chemistry, Chapter 5