Question

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

\(C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)\)

[structure - image pending]

Calculate rate constant.
[Given: log 3 = 0.48]

Hint:

For a gaseous first order reaction where one mole of reactant gives two moles of products, the increase in total pressure is used to find how much reactant has decomposed, and the first order integrated rate law gives th

Answer 1

Concept:
For a gaseous first order reaction where one mole of reactant gives two moles of products, the increase in total pressure is used to find how much reactant has decomposed, and the first order integrated rate law gives the rate constant.

Step 1:
Let initial pressure of \(C_2H_5Cl\) be \(P_0 = 0.30\) atm. If \(x\) atm decomposes, then \(C_2H_4\) formed = \(x\) and \(HCl\) formed = \(x\), so total pressure \(P_t = (P_0 - x) + x + x = P_0 + x\).

Step 2:
At t = 30 s, \(P_t = 0.50\) atm, so \(x = P_t - P_0 = 0.50 - 0.30 = 0.20\) atm. Pressure of reactant left = \(P_0 - x = 0.30 - 0.20 = 0.10\) atm.

Step 3:
First order rate constant: \(k = \frac{2.303}{t}\log\frac{P_0}{P_0 - x} = \frac{2.303}{30}\log\frac{0.30}{0.10} = \frac{2.303}{30}\log 3\).

Answer: \(k = \frac{2.303}{30}\times 0.48 = \frac{1.105}{30} = 3.68\times10^{-2}\ s^{-1}\) (approximately \(0.0368\ s^{-1}\)).