Question

The enthalpy of formation of ammonia
-46.0 kJ mol-1. The enthalpy for the reaction
2N2(g) + 6H2(g) → 4NH3(g) is
1) -46 kJ
(2) 46 kJ
3) 184 kJ
(4) –184 kJ
​

Answer 1

Answer:

4)-184kj

Explanation:

∆H=∆f Hproduct -∆f Hreactant

W.K.T.

enthalpy of N2,H2=0

::0+0--->4*-46

=-184kj.