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24/09/2021 3:02 pm
Topic starter
If bond enthalpies of N ≡ N, H –H and N –H bonds are x1, x2 and x3 respectively, ΔHfo for NH3 will be
(a) x1 + 3x2 – 6x3
(b) 1/2x1 + 3/2x2 – 3x3
(c) 3x3 – 1/2x1 – 3/2x2
(d) 6x3 – x1 – 3x2
1 Answer
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24/09/2021 3:04 pm
Correct answer: (b) 1/2x1 + 3/2x2 – 3x3
Explanation:
\(\frac{1}{2}\) N2 + \(\frac{3}{2}\)H2 → NH3
\(\frac{1}{2}\)x1 + \(\frac{3}{2}\)x2 – 3x3