Experiment Time $(s^{- 1})$ Total pressure (atm)
1 0 0.4
2 100 0.7
The following data were obtained during the first order thermal decomposition of $S O_{2} {C l}_{2}$ at a constant volume.
$S O_{2} C l (g) ⟶ S O_{2} (g) + {C l}_{2} (g)$
Calculate the rate constant. (Given: $log 4 = 0.6021, log 2 = 0.3010$ )

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Solution

The thermal decomposition of $S O_{2} C l_{2}$ at a constant volume is represented by the following equation.
$\begin{matrix} S O_{2} C l_{2 (g)} & ⟶ & S O_{2 (g)} & + & C l_{2 (g)} A t t = 0 & P_{0} & 0 & 0 A t t = t & P_{0} - p & p & p \end{matrix}$
After time t, total pressure is given as,
$P_{t} = (P_{0} - p) + p + p$
$P_{t} = P_{0} + p$
which on rearrangement gives:
$p = P_{t} - P_{0}$
$∴ P_{0} - p = P_{0} - (P_{t} - P_{0}) = 2 P_{0} - P_{t}$
For a first-order reaction,
$k = \frac{2.303}{t} l o g \frac{P_{0}}{P_{0} - p}$
$= \frac{2.303}{r} l o g \frac{P_{0}}{2 P_{0} - P_{t}}$
$t = 100 s$
$k = 2.303100 L o g 0.42 \times 0.4 - 0.7 = 1.386 \times 10^{- 2} s^{- 1}$

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Similar questions

S O_{2} {C l}_{2}

S O_{2} {C l}_{2} (g) ⟶ S O_{2} (g) + {C l}_{2} (g)

log 4 = 0.6021, log 2 = 0.3010

S O_{2} C l_{2}

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

0.65

S O_{2} C l_{2}

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

S O_{2} C l_{2}

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

Experiment	$T i m e / s^{- 1}$	Total pressure (atm)
$1$ $2$	$0$ $100$	$0.5$ $0.6$

0.65 a t m

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

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

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

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

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