Question

Given the reaction,
N₂ (g) + 3H₂ (g) → 2NH₃ (g),
ΔH = -92.6 kJ
The enthalpy of formation of NH₃ is
(a) -92.6 kJ
(b) 92.6 kJ mol⁻¹
(c) -46.3 kJ mol⁻¹
(d) -185.2 kJ mol⁻¹

Answer 1

@HEY MATE!!

Enthalpy formation of a compound is the change in enthalpy that take place during the formation of 1 mole of a substance in standard form .

Re- write the given equation for 1 mole of NH3(g),

1/2 N2(g) +3/2 H2(g) ➡NH3(g)

Enthalpy formation of NH3

= 1/2 delta H

= 1/2(-92.6 kj mol-1)

= -46.3 kj mol-1

Thanks!!

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