- ✓$-170$
- B$-260$
- C$-400$
- D$-450$
$\Delta \mathrm{H}$ reaction $=\Sigma \mathrm{B} . \mathrm{E} .$ (Rectant) $-\Sigma \mathrm{B} . \mathrm{E} .$ (product)
$=[\mathrm{B} \cdot \mathrm{E} \cdot(4 \mathrm{C}-\mathrm{H})+\mathrm{B} \cdot \mathrm{E} \cdot(\mathrm{C}=\mathrm{C})+\mathrm{B} \cdot \mathrm{E} \cdot(\mathrm{C}-\mathrm{C})]$
$-[\mathrm{B} \cdot \mathrm{E} \cdot 6(\mathrm{C}-\mathrm{H})+\mathrm{B} \cdot \mathrm{E} \cdot(\mathrm{C}-\mathrm{C})]$
$=\mathrm{B.E.}(\mathrm{C}=\mathrm{C})+\mathrm{B.E.}(\mathrm{H}-\mathrm{H})-2 \mathrm{B} \cdot \mathrm{E} \cdot(\mathrm{C}-\mathrm{H})-$
$\mathbf{B . E .}(C-C)$
$=600+400-2 \times 410-350\, \mathrm{kJ}$
$=-170\, \mathrm{kJ}$
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The value of $K _{ C }$ for the following reaction is :
$NH_{3}(g) \rightleftharpoons \frac{1}{2} N _{2}(g)+\frac{3}{2} H_{2}(g)$