Calculate the Arrhenius factor for the hydrolysis of methyl acetate in acidic medium at \(298\, \ K\). This reaction was studied by titrating the liberated acetic acid against sodium hydroxide at different time interval. The concentration of ester at different time interval is:
| t/min | 0 | 30 | 60 | 90 |
| c/mol/L | 0.55 | 0.35 | 0.22 | 0.14 |
\(E_a\) for the reaction \(= 18.2 kJ/mol\)
[log 1.57 = 0.196; log 2.5 = 0.3979; log 3.93 = 0.5942; \(e^{-7.3459} = 6.45 \times 10^{-4}\); \(e^{-0.00734} = 0.9926\)]
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 23.25.
To calculate the Arrhenius factor (\(A\)) for the hydrolysis of methyl acetate in an acidic medium, we will use the Arrhenius equation:
\(k = A e^{-E_a/RT}\)
where:
\(k\) is the rate constant,
\(A\) is the Arrhenius factor,
\(E_a\) is the activation energy,
\(R\) is the gas constant (\(8.314 \, \text{J/mol·K}\)),
\(T\) is the temperature in Kelvin.
The rate constant \(k\) for a first-order reaction can be calculated using the formula:
\(k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\)
From the given data, when \(t = 30\, \ min\), then
\([A]_0 = 0.55\, \ molL^{-1}\), \([A] = 0.35\, \ molL^{-1}\)
\(k = \frac{2.303}{30}log\frac{0.55}{0.35}\)
\(⇒ k = \frac{2.303}{30}log(1.57)\)
\(⇒ k = \frac{2.303 \times 0.196}{30}\)
\(⇒ k = 0.0152\)
We know, from Arrhenius equation
\(A = \frac{k}{e^{-E_a/RT}}\)
\(⇒ A = \frac{0.0152}{e^{-18200/8.314 \times 298}}\)
\(⇒ A= \frac{0.0152}{e^{-7.3459}}\)
\(⇒ A = \frac{0.0152}{6.45 \times {-4}}\)
\(⇒ A \approx 23.25\)