Question

the value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)

Answer 1

Del H for formation is given. For the reverse reaction, del H changes sign as the reverse of exothermic reaction will be endothermic. So, del H for decomposition is - (-91.8)=91.8 for ONE mole. But here, two moles are decomposing, so delH=2*91.8=183.6kJ

Answer 2

Answer:

+183.6 KJ/mol

here's the answer and hope this helps you.