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Q.

Determine enthalpy change for the following polymerization reaction per mole of $N_{2} (g)$ consumed $n N_{2} (g) + n H_{2} (g) ⟶ (NH - NH)_{n} (g)$
Given:
Bond enthalpy:
$\begin{aligned} N \equiv N = 942 kJ / mole \\ N - N = 163 kJ / mole \\ H - H = 436 kJ / mole \\ N - H = 390 kJ / mole \end{aligned}$

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a

-400 kJ/mol

b

140 kJ/mol

c

-110 kJ/mol

d

272 kJ/mol

answer is A.

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Detailed Solution

Enthalpy change for the polymerization reaction per mole,

$\begin{array}{l} [n \times 942 + n \times 436 - n \times (163 \times 2 + 2 \times 390)] \times \frac{1}{n} \\ = 272 kJ / mole \end{array}$

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Determine enthalpy change for the following polymerization reaction per mole of N2(g) consumed nN2(g)+nH2(g)⟶(NH−NH)n(g)Given: Bond enthalpy:N≡N=942kJ/moleN−N=163kJ/moleH−H=436kJ/moleN−H=390kJ/mole