Question

(i) For first order reaction, calculate the ratio between the time taken to complete \( \frac{3}{4} \)th of the reaction and the time taken to complete half of the reaction.

(ii) (I) What is meant by a complex reaction ?

(II) Give an example of a pseudo first order reaction.

Hint:

For first order, t(3/4) = (2.303/k) log 4 and t(1/2) = (2.303/k) log 2, so the ratio is 2. A complex reaction has several steps; ester hydrolysis is pseudo first order.

Answer 1

Step 1: Write the first order equation:
\[ t = \frac{2.303}{k}\log\frac{[R]_0}{[R]} \]
Take \( [R]_0 = 100 \) for convenience.

Step 2: Time for three-fourths completion:
When 3/4 is done, \( [R] = 25 \), so \( [R]_0/[R] = 4 \).
\[ t_{3/4} = \frac{2.303}{k}\log 4 = \frac{2.303}{k} \times 0.6021 = \frac{1.386}{k} \]

Step 3: Time for half completion:
When half is done, \( [R]_0/[R] = 2 \).
\[ t_{1/2} = \frac{2.303}{k}\log 2 = \frac{0.693}{k} \]

Step 4: Take the ratio:
\[ \frac{t_{3/4}}{t_{1/2}} = \frac{\log 4}{\log 2} = \frac{2\log 2}{\log 2} = 2 \]

Step 5: Complex reaction (part ii, I):
A complex reaction is one that takes place in more than one step. It is a sequence of elementary reactions.
The overall rate is fixed by the slowest step, called the rate determining step.

Step 6: Pseudo first order reaction (part ii, II):
Acid catalysed hydrolysis of ethyl acetate, with a large excess of water:
\[ \mathrm{CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH} \]
The concentration of water hardly changes, so the rate depends only on the ester. The reaction is second order by molecularity but follows first order kinetics. Inversion of cane sugar is another example.

Final Answer:
(i) \( t_{3/4} : t_{1/2} = 2 : 1 \), that is, the ratio is 2.
(ii) (I) A reaction occurring in several steps. (II) Hydrolysis of ethyl acetate.