Chemistry Enthalpy

Question

Calculate ΔH^o for the process ½ N₂(g) + ½ O₂(g) → NO(g)

from the following information

N2(g) + 2 O2(g) → 2 NO2(g)	ΔHo = -128.9 kJ/mol
2 NO(g) + O2 → 2 NO2(g)	ΔHo = -639.8 kJ/mol

Answer 1

This involves Hess's law.

The final RXN is 1/2 N2 + 1/2 O2>>>>NO Find Delta H

Known RXN"s #1) N2 + 2O2>>>>>2NO2 Delta H -128.9 KJ/mol

#2) 2NO +O2>>>>> 2NO2 Delta H -639.8 KJ/mol

#2a) NO2>>>>>>>> 1/2 O2 +NO Delta H 1/2 + 639.8 KJ/mol

#1a) 1/2 N2 + O2>> NO2 Delta H 1/2 -128.9 KJ/mol

After canceling on either side of the equation you obtain

1/2 N2 + 1/2 O2>>> NO

Now Add Delta H of equation #2a + #1a to get the final Delta H=255.45 KJ/mol

METHOD:: First you reverse equation #2 and multiply the coefficients by 1/2 (note this is equation #2a) and reverse the sign and take 1/2 of the Delta H

Second add equation #1 and multiply the coefficients by 1/2 (note this is equation 1a) and multiply Delta H by 1/2

Answer 2

Enthalpy change= H products - H reactants

For first reaction as enthalpy of formation of eelements is zero.

H of NO2 formation for 2 mols

= -128.9/2

For second reaction :

-639.8= -128.9 - H NO

H NO for 2 mols= 510.90

For one mol = 255.45kj mol^-

For reaction

H=product enthalpy of formation - Reactant

H for reaction = 255.45kj mol^-