the value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)
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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
The reaction of formation is 1/2N2+3/2H2-->NH3. This will be same as fusion of one mole in this case.
--->
Reaction of formation : 1/2N2+3/2H2==NH3.
was just confirming
thanks again
and good morning
and good morning
Mornin.
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5
Answer:
+183.6 KJ/mol
here's the answer and hope this helps you.
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