What volume of $N H_{3} (g)$ at $27^{o} C$ and 1 atm pressure will be obtained by thermal decomposition of $26.25 g N H_{4} C l$ ?

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Solution

Ammonium chloride undergoes decomposition as:
$N H_{4} C l (s) 1 m o l 53.5 g \to N H_{3} (g) 1 m o l 1 m o l + H C l (g)$
$∵ 53.5 g N H_{4} C l$ give 1 mol $N H_{3}$

$∴ 26.25 g N H_{4} C l$ will give $\frac{1}{53.5} \times 26.25 m o l e N H_{3}$ $= 0.5 m o l e$

$P V = n R T$
$1 \times V = 0.5 \times 0.0821 \times 300$
$V = 12.315 l i t r e a m p l e$

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S O_{2} C l_{2}

S O_{2} C l_{2 (g)} \to S O_{2 (g)} + C l_{2 (g)}

The following data were obtained during the first order thermal decomposition of $S O_{2} C l_{2}$ at a constant volume.

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$\begin{matrix} E x p e r i m e n t & T i m e / s^{- 1} & T o t a l p r e s s u r e / a t m 1 & 0 & 0.5 2 & 100 & 0.6 \end{matrix}$

Calculate the rate of the reaction when total pressure is 0.65 atm.

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S O_{2} C l_{2} (g) \to S O_{2} (g) + C l_{2} (g)

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1.0

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2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)

Time (s)	0	100
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