MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Rate of Reaction

The following...

Question

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.

Open in App

Solution

$\begin{matrix} S O_{2} C l_{2} (g) \to & S O_{2} (g) & + & C l_{2} (g) I n i t i a l l y & p_{i} & 0 & 0 A f t e r t i m e t & p_{i} - p & p & p \end{matrix}$
Total pressure after time t i.e.,
$p_{t} = p_{i} - p + p + p = p_{i} + p$
So, $a = p_{i}$
$a - x = p_{i} - (p_{t} - p_{i}) = p_{i} - p_{t} + p_{i} = 2 p_{i} - p_{t}$
Substitutions of the value of a and (a-x) gives
$k = \frac{2.303}{t} l o g \frac{p_{i}}{(2 p_{i} - p_{t})}$

(a) Calculation of rate constant k
Given $p_{i} = 0.5 a t m; p_{t} = 0.6 a t m k = \frac{2.303}{t} l o g \frac{p_{i}}{(2 p_{i} - p_{t})} = \frac{2.303}{(100 s)} l o g \frac{0.5 a t m}{0.4 a t m} = \frac{2.303}{(100 s)} l o g 1.25 = \frac{2.303}{(100 s)} \times 0.0969 = 2.23 \times 10^{- 3} s^{- 1}$

(b) Calculation of reaction rate when total pressure is 0.65 atm
$p_{S O_{2} C l_{2}} = 0.5 - (0.65 - 0.50) = (1 - 0.65) = 0.35 a t m k = 2.23 \times 10^{- 3} s^{- 1}$
Rate $= k \times p_{S O_{2} C l_{2}} = (2.23 \times 10^{- 3} s^{- 1}) \times (0.35 a t m)$
Rate $= 7.8 \times 10^{- 4} a t m s^{- 1}$

Suggest Corrections

thumbs-up

1

Similar questions

Q. 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)

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

Calculate the rate of the reaction when total pressure is

0.65

Q. The following data were obtained during the first order thermal composition 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)}

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

What is the rate of reaction when total pressure is

0.65 a t m

?

The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

Experiment	Time/s⁻¹	Total pressure/atm
1	0	0.5
2	100	0.6

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

Q. The following data were obtained during the first order thermal decomposition of

S O_{2} C l_{2}

at constant volume

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

Expt	Time / $S^{- 1}$	Total pressure/ atm
1	0	0.5
2	100	0.6

when the total pressure is 0.65 atm then the rate constant of reaction is ..............
if a = pi, and a - x = 2pi - Pt

Q. 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)

Experiment	Time in $s^{- 1}$	Total pressure in atm
1	0	0.4
2	100	0.7

Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Rate of Reaction

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Rate of Reaction

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon