Question

Kinetic data for hydrolysis of a 0.3 M solution of ethyl acetate in presence of HCl as catalyst is given in the following table. A constant volume of reaction mixture is taken at different time and titrated against standard alkali solution at different time.

It the rate law is given by rate = k[Ester]¹[H⁺]¹, where H⁺ is a catalyst, find the rate (in M/min) of ester hydrolysis in a solution which is 0.1 M each in ester and acid HCl (M/min)

Use: (ln 2 = 0.7, ln5 = 1.6)

If your answer is x × 10⁻y then fill 'x' in OMR sheet.

Answer 1

4

Answer 2

The hydrolysis of ethyl acetate in the presence of HCl catalyst is given by the reaction:

$CH₃COOC₂H₅ + H₂O \xrightarrow{H^+} CH₃COOH + C₂H₅OH$

The rate law is given as rate = k[Ester]¹[H⁺]¹. Since H⁺ is a catalyst and its concentration remains constant throughout the reaction, the reaction is pseudo-first-order with respect to the ester concentration. Rate = k'[Ester], where k' = k[H⁺].

The given data is from the titration of the reaction mixture with a standard alkali solution. The alkali neutralizes the acid present in the mixture. At time t=0, only the catalyst acid (HCl) is present. At time t>0, both the catalyst acid and the product acetic acid are present. At time t=$\infty$, the reaction is complete, and the maximum amount of acetic acid is present along with the catalyst acid.

Let $V_t$ be the volume of alkali used at time t. $V_0$ is proportional to the amount of HCl catalyst. $V_\infty$ is proportional to the total amount of acid at completion (HCl + acetic acid from complete hydrolysis of ester). $V_\infty - V_0$ is proportional to the amount of acetic acid produced from the complete hydrolysis of the initial amount of ester, which is proportional to the initial concentration of ester, $[Ester]_0$. $V_t - V_0$ is proportional to the amount of acetic acid produced at time t, which is proportional to the amount of ester hydrolyzed at time t. The amount of ester remaining at time t is proportional to the initial amount of ester minus the amount hydrolyzed. $[Ester]_t \propto (V_\infty - V_0) - (V_t - V_0) = V_\infty - V_t$.

For a pseudo-first-order reaction, the integrated rate law is:

$\ln([Ester]_0/[Ester]_t) = k't$

$\ln((V_\infty - V_0)/(V_\infty - V_t)) = k't$

From the table, at t=0, $V_0 = 25.00$ ml. At t=10 min, $V_{10} = 28.00$ ml. At t=$\infty$, $V_\infty = 40.00$ ml.

$V_\infty - V_0 = 40.00 - 25.00 = 15.00$ ml.

$V_\infty - V_{10} = 40.00 - 28.00 = 12.00$ ml.

Substitute these values into the integrated rate law at t=10 min:

$\ln(15.00/12.00) = k' \times 10$

$\ln(5/4) = 10k'$

$\ln 5 - \ln 4 = 10k'$

$\ln 5 - 2\ln 2 = 10k'$

Using the given values $\ln 2 = 0.7$ and $\ln 5 = 1.6$:

$1.6 - 2(0.7) = 10k'$

$1.6 - 1.4 = 10k'$

$0.2 = 10k'$

$k' = 0.02$ min⁻¹.

The pseudo-first-order rate constant $k'$ is given by $k' = k[H⁺]$. In the experiment described in the table, the initial concentration of ethyl acetate was 0.3 M. Let the concentration of HCl catalyst be $[HCl]_1$. The volume of alkali required is proportional to the amount of acid. Let the constant of proportionality be C. Amount of HCl $\propto V_0$. So, $[HCl]_1 \propto V_0$. Amount of initial ester $\propto (V_\infty - V_0)$. So, $[Ester]_0 \propto (V_\infty - V_0)$. We are given $[Ester]_0 = 0.3$ M, which corresponds to $(V_\infty - V_0) = 15.00$ ml. The concentration of HCl in this experiment, $[HCl]_1$, corresponds to $V_0 = 25.00$ ml. Since the volume of alkali is proportional to the concentration of the acid it neutralizes (for a constant volume of reaction mixture and standard alkali), the ratio of concentrations is equal to the ratio of volumes:

$[HCl]_1 / [Ester]_0 = V_0 / (V_\infty - V_0)$

$[HCl]_1 / 0.3 = 25.00 / 15.00 = 5/3$

$[HCl]_1 = (5/3) \times 0.3 = 0.5$ M. Since HCl is a strong acid, $[H⁺]_1 = [HCl]_1 = 0.5$ M.

Now we can find the actual rate constant k using $k' = k[H⁺]_1$:

$0.02$ min⁻¹ $= k \times 0.5$ M

$k = 0.02 / 0.5 = 0.04$ M⁻¹min⁻¹.

We need to find the rate of ester hydrolysis in a solution which is 0.1 M each in ester and acid HCl. Let the new concentrations be $[Ester]_2 = 0.1$ M and $[HCl]_2 = 0.1$ M. The concentration of the catalyst is $[H⁺]_2 = [HCl]_2 = 0.1$ M.

The rate under these new conditions is given by the rate law:

Rate₂ = k$[Ester]_2$[H⁺]₂

Rate₂ = $(0.04$ M⁻¹min⁻¹) $\times (0.1$ M$) \times (0.1$ M$)$

Rate₂ = $0.04 \times 0.1 \times 0.1$ M/min

Rate₂ = $0.0004$ M/min.

The answer should be in the form $x \times 10^{-y}$.

$0.0004 = 4 \times 10^{-4}$ M/min.

So, $x = 4$ and $y = 4$.

The question asks to fill 'x' in the OMR sheet. So, the answer is 4.