Question

For the first order thermal decomposition reaction, the following data was obtained:

C₂H₅Cl(g) &rarr; C₂H₄(g) + HCl(g)<table border="1"><tr><th>S. No.</th><th>Time (s)</th><th>Total Pressure (atm)</th></tr><tr><td>1</td><td>0</td><td>0.30</td></tr><tr><td>2</td><td>30</td><td>0.50</td></tr></table>Calculate the rate constant. [Given: log 3 = 0.48]

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Hint:

Let the initial pressure of C 2 H 5 Cl be p 0 = 0.30 atm. If x atm of C 2 H 5 Cl decomposes, it produces x atm of C 2 H 4 and x atm of HCl. So the pressures become: C 2 H 5 Cl = (0.30 - x), C 2 H 4 = x, HCl = x.

Answer 1

Let the initial pressure of C₂H₅Cl be p₀ = 0.30 atm.

If x atm of C₂H₅Cl decomposes, it produces x atm of C₂H₄ and x atm of HCl. So the pressures become: C₂H₅Cl = (0.30 - x), C₂H₄ = x, HCl = x.

Total pressure = (0.30 - x) + x + x = 0.30 + x.

At t = 30 s, total pressure = 0.50 atm, so 0.30 + x = 0.50, giving x = 0.20 atm.

Pressure of C₂H₅Cl remaining = 0.30 - 0.20 = 0.10 atm.

For a first order reaction:

k = (2.303 / t) &times; log (p₀ / p)

k = (2.303 / 30) &times; log (0.30 / 0.10)

k = (2.303 / 30) &times; log 3

k = (2.303 / 30) &times; 0.48

k = (2.303 &times; 0.48) / 30 = 1.105 / 30

k = 0.0368 s^-1 (approximately 3.68 &times; 10^-2 s^-1).