Problem 24 - Chemistry The Molecular Science

Question

The ester methyl acetate, \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\), reacts with base to break one of the \(\mathrm{C}-\mathrm{O}\) bonds. CC(=O)OCC(O)C(C)O CC(=O)OCCOCCC(=O)O The rate law is rate \(=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{OH}^{-}\right]\) where $$ k=0.14 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} \text {at } 25^{\circ} \mathrm{C} $$ (a) Calculate the initial rate at which the methyl acetate is converted to products when both reactants, \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\) and \(\mathrm{OH}^{-}\), have a concentration of \(0.025 \mathrm{M}\). (b) Calculate the rate at which methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), initially appears in the solution.

Step-by-Step Solution

Answer

The initial rate of reaction is \(8.75 \times 10^{-5}\,\text{M}\,\text{s}^{-1}\), and the rate of methyl alcohol appearance is the same.

Step 1: Identify the Rate Law

The rate law for this reaction is given by: rate \( = k[\text{CH}_3\text{COOCH}_3][\text{OH}^-] \). This tells us that the rate depends on the concentrations of both \(\text{CH}_3\text{COOCH}_3\) and \(\text{OH}^-\).

Step 2: Input Known Values for Initial Rate

We are given that \(k = 0.14\,\text{L}\,\text{mol}^{-1}\,\text{s}^{-1}\) and the initial concentrations for both \([\text{CH}_3\text{COOCH}_3]\) and \([\text{OH}^-]\) are \(0.025\,\text{M}\). Substitute these values into the rate law equation to find the initial rate of the reaction.

Step 3: Calculate Initial Rate

Using the rate law:\[\text{Initial Rate} = 0.14 \times 0.025 \times 0.025 = 8.75 \times 10^{-5}\,\text{M}\,\text{s}^{-1}\]This is the rate at which methyl acetate is converted into products initially.

Step 4: Relate Initial Rate to Methyl Alcohol Production

In the reaction, every molecule of methyl acetate that reacts will produce one molecule of methyl alcohol, \(\text{CH}_3\text{OH}\). Therefore, the initial rate of appearance of \(\text{CH}_3\text{OH}\) is the same as the rate calculated in Step 3.

Step 5: Calculate Rate of Methyl Alcohol Appearance

Thus, the initial rate at which methyl alcohol appears in the solution is:\[8.75 \times 10^{-5}\,\text{M}\,\text{s}^{-1}\]

Key Concepts

Rate Law

In chemical kinetics, the rate law is a mathematical expression that describes how the rate of a chemical reaction depends on the concentration of the reactants. For the ester hydrolysis reaction involving methyl acetate (\(\mathrm{CH}_3\mathrm{COOCH}_3\)) and hydroxide ions (\(\mathrm{OH}^-\)}, the rate law is given as:rate = \(k[\mathrm{CH}_3\mathrm{COOCH}_3][\mathrm{OH}^-]\).

• This means that the rate is directly proportional to the concentration of each reactant.
• The constant k is known as the rate constant, and its value tells us how fast the reaction proceeds at a given temperature.
• For this reaction, \(k = 0.14\,\mathrm{L\,mol}^{-1}\,\mathrm{s}^{-1}\)

Understanding the rate law helps chemists predict how changes in concentration will affect the reaction rate. This is especially useful in industrial applications where controlling the speed of a reaction can impact product yield and quality.
To determine the initial reaction rate using the rate law, insert the known concentrations into the expression to calculate the rate effectively.

Reaction Rate

The reaction rate refers to the speed at which reactants are converted into products in a chemical reaction. In the context of ester hydrolysis, we are interested in how fast methyl acetate reacts with hydroxide ions under given conditions.

• The initial rate can be found using the rate law: \(\text{rate} = k[\mathrm{CH}_3\mathrm{COOCH}_3][\mathrm{OH}^-]\).
• For example, with concentrations of \([\mathrm{CH}_3\mathrm{COOCH}_3] = 0.025\,\mathrm{M}\) and \([\mathrm{OH}^-] = 0.025\,\mathrm{M}\), the initial rate is \(8.75 \times 10^{-5}\,\mathrm{M\,s}^{-1}\)

Calculating the reaction rate provides valuable insights into the efficiency of the process. A faster reaction can mean quicker production times in a manufacturing process, but it might also necessitate careful control to avoid unwanted side effects or thermal issues.

Ester Hydrolysis

Ester hydrolysis is a chemical reaction where an ester bond in a molecule is broken down by the action of water and generally involves a base such as hydroxide ions.In the case of methyl acetate (\(\mathrm{CH}_3\mathrm{COOCH}_3\)) hydrolysis, the reaction breaks an ester bond to form methyl alcohol (\(\mathrm{CH}_3\mathrm{OH}\)) and an acetate ion.

• This specific ester hydrolysis is a common reaction to produce alcohols and carboxylates, and it involves nucleophilic substitution.
• Every molecule of methyl acetate reacts with a hydroxide ion to yield one molecule of methyl alcohol.

This reaction is particularly important in organic chemistry and industrial applications. Understanding the specifics of this hydrolysis process can aid in designing better industrial systems for making specific products from ester-based raw materials.