Question

the enthalpy of formation of ammonia is -46 Kilo joule per mole the enthalpy change for the reaction to NH3 gives and 2 + 3 H2 is​

Answer 1

enthalpy of formation of any compound is change in enthalpy for the formation of one mole of compound from its elements.

so, enthalpy of formation of ammonia is change in enthalpy for the formation of one mole ammonia from its elements.

i.e., (1/2)N2 + (3/2)H2 → NH3 , ∆Hf = -46 KJ/mole

⇒ NH3 → (1/2)N2 + (3/2)H2 , ∆H =-(∆Hf) = -(-46) = 46 kJ/mole

⇒2NH3 → N2 + 3H2, ∆H' = 2 × ∆H = 2 × 46 = 92 KJ/mole

hence, enthalpy change for the chemical reaction to NH3 gives N2 and 3H2 is 92 KJ/mole .

Answer 2

Answer:

answer is -184k/J

Explanation:

-46.0×4 as 4 Nh3