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
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:
This is a useful equation to memorize when dealing with enthalpies of formation.
So
Which is
Which corresponds do answer D