The enthalpy change for the reaction is 1215.75 kJ/mol.
The term enthalpy change is the energy that accompanies a reaction. Now we have the reaction; 4HNO3(g) + 2H2O(g) → 2N2H4(g) + 7O2(g)
ΔHf HNO3(g) = -134.31 kJ/mol
ΔHf H2O(g)= -241.83 kJ/mol
ΔHf N2H4(g) = 97.42 kJ/mol
ΔHf O2(g) = 0 kJ/mol
ΔHreaction = [(2 * 97.42) + ( 7 * 0)] - [ (4 * -134.31 ) + (2 * -241.83 )
= 194.84 - (-537.25 - -483.66)
=1215.75 kJ/mol
The enthalpy change for the reaction is therefore 1215.75 kJ/mol.
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