What is ΔH in kJ for the reaction below?

Given the following enthalpies of formation, ΔHfo:

N2H4(l) = 50.4 kJ/mol;

H2O(l) = -286 kJ/mol;

N2O(g) = 81.5 kJ/mol;

NO2(g) = 33.8 kJ/mol;

what is ΔH in kJ for the reaction below?

N2H4(l) + 2NO2(g) → 2N2O(g) + 2H2O(l)

A) 81.5 - 286 + 50.4 + 33.8
B) 2x81.5 - 2x286 + 50.4 + 2x33.8
C) -2x81.5 + 2x286 - 50.4 - 2x33.8
D) 2x81.5 - 2x286 - 50.4 - 2x33.8
E) -2x81.5 - 2x286 - 50.4 - 2x33.8

1 Answer
Apr 4, 2018

Option D.

Explanation:

#DeltaH# (reaction) = #DeltaH#(products) - #DeltaH# (reactants)

This is a useful equation to memorize when dealing with enthalpies of formation.

So #DeltaH# should be:

#(2(81.5)+2(-286))-(2(33.8)+50.4)#
Which is

#2(81.5)-2(286)-2(33.8)-50.4#

Which corresponds do answer D