Question

What is standard N ≡ N bond enthalpy from following reaction,

\[\ce{N2_{(g)} + 3H2_{(g)} -> 2NH3_{(g)}; \Delta H^0 = - 83 kJ}\]

\[\ce{ΔH^0_{(H-H)}}\] = 435 kJ; \[\ce{ΔH^0_{(N-H)}}\] = 389 kJ

Options

• 435 kJ

• 1305 kJ

• 2334 kJ

• 946 kJ

Answer 1

946 kJ

Explanation:

Given,

ΔH⁰ = − 83 kJ

\[\ce{ΔH^0_{(H-H)}}\] = 435 kJ

\[\ce{ΔH^0_{(N-H)}}\] = 389 kJ

\[\ce{N2_{(g)} + 3H2_{(g)} -> 2NH3_{(g)}}\]

ΔH⁰ = `sum Δ"H"^0 ("reactant bonds") - sum Δ"H"^0 ("product bonds")` 

∴ ΔH⁰ = `[Δ"H"_(("N"≡"N"))^0 + 3Δ"H"_(("H"-"H"))^0] - [6Δ"H"_(("N"-"H"))^0]`

∴ − 83 = `Δ"H"_(("N"≡"N"))^0 + 3(435) - 6(389)`

∴ − 83 = `Δ"H"_(("N"≡"N"))^0 + 1305 - 2334`

∴ `Δ"H"_(("N"≡"N"))^0` = − 83 − 1305 + 2334

= 946 kJ