\(100\, \ cm^3\) of \(1\, \ M\) \(CH_3COOH\) and \(100\, \ cm^3\) of \(2\, \ M\) \(CH_3OH\) were mixed to form an ester which follow \(1^{st}\) order kinetic with respect to both \(CH_3COOH\) and \(CH_3OH\). The change in their initial rate, if each solution is mixed with equal number of volume of water would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.25 times
The esterification is first order in each reactant, so the rate law is
\( \text{Rate} = k\,[\ce{CH3COOH}][\ce{CH3OH}] \)
Mixing 100 cm3 of each solution doubles the volume, so each concentration is halved:
\( [\ce{CH3COOH}] = 0.5\,\mathrm{M}, \quad [\ce{CH3OH}] = 1\,\mathrm{M} \)
\( \text{Rate}_1 = k(0.5)(1) = 0.5k \)
If each solution is first mixed with an equal volume of water, each concentration is halved once more:
\( [\ce{CH3COOH}] = 0.25\,\mathrm{M}, \quad [\ce{CH3OH}] = 0.5\,\mathrm{M} \)
\( \text{Rate}_2 = k(0.25)(0.5) = 0.125k \)
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.125k}{0.5k} = 0.25 \)
So the initial rate becomes one-fourth (0.25 times).
Shortcut: The reaction is second order in all and every concentration is halved, so the rate changes by \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).