Chemistry, asked by 24khanak, 1 year ago

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

Answers

Answered by FlameFires

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

FlameFires: Pretty sure both the questions are the same. Here, fusion is the same as formation

FlameFires: The reaction of formation is 1/2N2+3/2H2-->NH3. This will be same as fusion of one mole in this case.

FlameFires: --->

FlameFires: Reaction of formation : 1/2N2+3/2H2==NH3.

24khanak: was just confirming

24khanak: thanks again

24khanak: and good morning

FlameFires: Mornin.

Answered by dharamrajktr6

Answer:

+183.6 KJ/mol

here's the answer and hope this helps you.