Thermodynamics — Medium JEE chemistry MCQ
Given the following thermochemical equations:
- C(s) + O₂(g) → CO₂(g); ΔH₁ = -393 kJ/mol
- CO(g) + ½O₂(g) → CO₂(g); ΔH₂ = -283 kJ/mol
The enthalpy of formation of CO(g) is:
- A. -110 kJ/mol
- B. +110 kJ/mol
- C. -676 kJ/mol
- D. +676 kJ/mol
Solution
The formation reaction for CO is:
C(s) + ½O₂(g) → CO(g); ΔH_f = ?
Using Hess's law:
Equation 1: C(s) + O₂(g) → CO₂(g); ΔH₁ = -393 kJ/mol
Reverse of Equation 2: CO₂(g) → CO(g) + ½O₂(g); -ΔH₂ = +283 kJ/mol
Adding these:
C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g)
Simplifying:
C(s) + ½O₂(g) → CO(g)
\[
\Delta H_f = \Delta H_1 + (-\Delta H_2) = -393 + 283 = -110\,\text{kJ/mol}
\]
So the correct option is A.