Question

Given the following reactions N2 (g) + O2 (g) → 2NO (g) ΔH = +180.7 kJ 2NO( g) + O2 (g) → 2NO2 (g) ΔH = -113.1 kJ the enthalpy for the decomposition of nitrogen dioxide into molecular nitrogen and oxygen 2NO2 (g) → N2 (g) + 2O2 (g) is ________ kJ.

Answer 1

Given, Equations are

N₂(g) + O₂(g) 2NO (g) (a).  H = +180.7 KJ (1)

2NO(g) + O₂ (g) 2NO₂ (g).H = -113.1 KJ (2)

Now, add Eq.1 + Eq.2

N₂ (g) + O₂ (g)+ O₂ (g) + 2NO (g)   2NO (g) + 2NO₂ (g)

Cancelling same species on both side.

N₂ + 2O₂ 2NO₂ (3)

Now, according to Hess law total heat change in multiple steps = total heat change in a single step.

So, H in Eq.3 = H of Eq.1 + H of Eq.2

= +180.7 +(-113.1)

= + 67.6 KJ.

Now , the target equation decomposition of Nitrogen dioxide is reverse of Eq.3

2NO₂ (g) N₂ (g) + 2O₂ (g) .(4)

By Lavoisier-Laplace's law, heat change ( Enthalpy change) in any process is equal and opposite to heat change in reverse process.

Therefore, Enthalpy change in Eq.4 = - Enthalpy change in Eq.3

= - 67.6 KJ.