Calculate the standard reaction enthalpy for the reaction.
N2H4(l) + H2(g) 2NH3(g)
Given:
N2H4(l) + O2(g) N2(g) + 2H2O(g)
H° = -543 kJ.mol-1
2H2(g) + O2(g) 2H2O(g)
H° = -484 kJ.mol-1
N2(g) + 3H2(g) 2NH3(g)
H° = -92.2 kJ.mol-1
A) (-935 kJ.mol-1 )
B) (-1119 kJ.mol-1 )
C) (-151 kJ.mol-1 )
D) (-59 kJ.mol-1 )
E) (-243 kJ.mol-1 )
Correct Answer:
Verified
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