Question

State and explain Hess's Law of constant heat summation with a suitable example.

Answer 1

According to this law, "If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature."

Δ_rH = Δ_rH₁ + Δ_rH₂ + Δ_rH₃ + ⋯

It can be represented as:

Example:

C(graphite, s) + \(\frac{1}{2}\) O₂ (g) → CO(g); Δ_rH^Θ = ?

Steps:

(i) C(graphite, s) + O₂ (g) → CO₂ (g); ∆_rH^Θ = −393.5 kJ mol⁻¹

(ii) CO(g) + \(\frac{1}{2}\)O₂ (g) → CO₂ (g); ∆rH^Θ = −283.0 kJ mol⁻¹

To get one mole of CO(g) on the right, we reverse equation (i). In this, heat is absorbed instead of being released, so we change sign of ∆_rH^Θ value.

(iii) CO₂ (g) → CO(g) + \(\frac{1}{2}\)O₂ (g) ; ∆_rH_x^Θ = +283.0 kJ mol⁻¹

Adding equation (i) and (iii) to get the desired equation

C(graphite, s) + \(\frac{1}{2}\)O₂ (g) → CO(g)

For which, ∆_rH^Θ = (−395.5 + 283.0)

= −110.5 kJ mol⁻¹