themochemistry Q

Question

Given the following thermochemical equations:

2NO(g) + O₂(g)  2NO₂(g) ∆H° = -113.2 kJ

2N₂O(g) + 3O₂(g)  4NO₂(g) ∆H° = -28.0 kJ

Сalculate ∆H° for the reaction:

N₂O(g) + 1/2 O₂(g)  2NO(g)

∆H =

Answer 1

The target equation of interest is N₂O(g) + 1/2O₂(g) ==> 2NO(g)

We are given two other equations which we can rearrange and combine to give the target equation.

This makes use of HESS' LAW (look it up and read about it so you understand it)

eq.1. 2NO(g) + O₂(g) ==> 2NO₂(g) ∆Hº = -113.2 kJ

eq. 2. 2N₂O(g) + 3O₂(g) ==>4NO₂(g) ∆Hº = -28.0 kJ

Copy eq. 2. and divide by 2: N₂O(g) + 1.5O₂(g) ==> 2NO₂(g) ∆Hº = -28.0/2 = -14.0 kJ

reverse eq. 1. ....................2NO₂(g) ===> 2NO(g) + O₂(g) ∆Hº = +113.2 kJ

add the 2 eqs together: _____________________________________________

N₂O(g) + 1.5O₂(g) + 2NO₂(g) ===> 2NO₂(g) + 2NO(g) + O₂(g) add ∆H values to get 99.2 kJ

cancel like terms on each side of the equation to get the final equation....

N₂O(g) + 1/2O₂(g) ==> 2NO(g) THIS IS THE TARGET EQUATION OF INTEREST

∆H = +99.2 kJH