AP Chemistry Determine the value of the enthalpy change, ΔHrxn, for the following reaction using the information below: N2O(g) + NO2(g) → 3NO(g)

Question

Determine the value of the enthalpy change, ΔHrxn, for the following reaction using the information below:

N2O(g) + NO2(g) → 3NO(g)

2NO(g) → N2(g) + O2(g) ΔHrxn = -181.0 kJ/mol

2N2O(g) → 2N2(g) + O2(g) ΔHrxn = -163.0 kJ/mol

NO(g) + ½O2(g) → NO2(g) ΔHrxn = -57.0 kJ/mol

Answer 1

Hess's Law

N₂O(g) + NO₂(g) ==> 3NO(g) TARGET EQUATION

Given:

(1). 2NO(g) → N2(g) + O2(g) ΔHrxn = -181.0 kJ/mol

(2). 2N2O(g) → 2N2(g) + O2(g) ΔHrxn = -163.0 kJ/mol

(3). NO(g) + ½O2(g) → NO2(g) ΔHrxn = -57.0 kJ/mol

Procedure:

Divide eq.(2) by 2: N2O(g) → N2(g) + 1/2 O2(g) ΔHrxn = -81.5 kJ/mol

Reverse eq.(3): NO2(g) --> NO(g) + 1/2 O2(g) ∆H = +57.0 kJ/mol

Reverse eq.(1): N2(g) + O2(g) --> 2NO(g) ∆H = +181.0 kJ/mol

Add all three equations to get:

N2O(g) + NO2(g) + N2(g) + O2(g) ==> N2(g) + 1/2 O2(g) + NO(g) + 1/2 O2(g) + 2NO(g)

Combine and/or cancel like terms to obtain the final equation:

N2O(g) + NO2(g) ==> 3NO(g) TARGET EQUATION

∆H_rxn = -81.5 + 57.0 + 181.0 = 156.5 kJ