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Chemical Kinetics — Chemistry STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics3 Marks
Question
For the first order thermal decomposition reaction, the following data were obtained:
$\text{C}_{2}\text{H}_{5}\text{Cl}\text{(g)}\rightarrow \text{C}_{2}\text{H}_{4}\text{(g)} + \text{HCl}\text{(g)} $
Time/sec Total pressure/atm
0 0.30
300 0.50
Calculate the rate constant
(Given: log 2 = 0.301, log 3=0.4771, log 4 =0.6021)
✓

Answer

Given: Initial pressure, Po = 0.30 atm
Pt = 0.50 atm
t = 300 s
Rate constant, $\text{k} = \frac{2.303}{t}\text{log}\frac{\text{p}_{o}}{2\text{p}_{o} - \text{p}_{t}}$
$ = \frac{2.303}{300\text{ s }}\text{log}\frac{0.30}{2\times0.30-0.50}$
$ = \frac{2.303}{300\text{ s }}\text{log}\frac{0.30}{0.60 - 0.50}$
$ = \frac{2.303}{300\text{ s}}\text{log}\frac{0.30}{0.10}$
$ = \frac{2.303}{300\text{ s}}\text{log }3$
$=\frac{2.303}{300\text{ s }}\times0.4771$
$ = \frac{1.099}{300\text{ s }}$
$ = 0.0036\text{ s}^{-1} / 3.66\times10^{-3}\text{ s}^{-1}.$

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