6.80 g of $N H_{3}$ ; is passed over heated CuO. The standard heat enthalpIes of $N H_{3}$ , $C u O (s)$ and $H_{2} 0 (l)$ are -46.0, -155.0 and $- 2850 k J m o l^{- 1}$ respectively and the change is.
$N H_{3} + (3 / 2) C u O ⟶ \frac{1}{2} N_{2} (g) + (3 / 2) H_{2} O (l) + (3 / 2) C u (s)$ .
If the enthalpy change is $x$ kJ then $- \frac{x}{15}$ (nearest integer) is___________.

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Solution

$N H_{3} + (3 / 2) C u O ⟶ \frac{1}{2} N_{2} (g) + (3 / 2) H 2 O (l) + (3 / 2) C u (s); Δ H = ?$
$Δ H^{\circ} = Σ H_{P r o d u c t s}^{\circ} - Σ H_{R e a c t a n t s}^{\circ}$
$= [\frac{1}{2} \times H_{N_{2}}^{\circ} + (3 / 2) \times H_{H_{2} O (l)}^{\circ} + (3 / 2) H_{C u}^{\circ}] - [(3 / 2) H_{C u O}^{\circ} + H_{N H_{3}}^{\circ}]$
$= [0 + (3 / 2) \times (- 285) + (3 / 2) \times 0] - [(3 / 2) \times (- 155) + (- 46)]$
$= - 149 k J$
$[∵ H^{\circ} f o r p u r e e l e m e n t = 0 a n d H_{c o m p o u n d}^{\circ} = H_{f o r m a t i o n}^{\circ}]$
$∵ 17 g N H_{3}$ brings in change in $Δ H = - 149 k J$
$∵ 6.8 g N H_{3}$ brings in change in $Δ H = (- 149 \times 6.8) / 17 = - 59.6 k J$
$∴ Δ H f o r 6.89 g N H_{3} = - 59.6 k J = x$
$∴ \frac{- x}{15} = - 59.6 \approx 4$

The answer is 4.

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