Problem 13 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The acid-catalysed hydrolysis of sucrose shows first order kinetics. The half life for the reaction at room temperature was found to be 190 min. Calculate the rate constant for the reaction under these conditions. (Section \(9.4)\)

Step-by-Step Solution

Answer

The rate constant \\(k\\) is approximately \\(0.00365\, \text{min}^{-1}\\).

Step 1: Understanding the relationship

For a first-order reaction, the relationship between the half-life \(t_{1/2}\) and the rate constant \(k\) can be given by the equation \[ t_{1/2} = \frac{0.693}{k} \]. This formula allows us to find the rate constant if the half-life is known.

Step 2: Rearranging the formula

Rearrange the formula to solve for \(k\): \[ k = \frac{0.693}{t_{1/2}} \] where \(t_{1/2}\) is the given half-life of the reaction.

Step 3: Substituting the values

Substitute the given half-life value into the formula: \[ k = \frac{0.693}{190\, \text{min}} \] Now simplify the expression to calculate the rate constant.

Step 4: Calculating the rate constant

Carry out the division to find \(k\): \[ k = \frac{0.693}{190} \] This results in a rate constant, \(k \approx 0.00365\, \text{min}^{-1}\).

Key Concepts

Understanding the Rate Constant in First-Order Kinetics

In chemical reactions, the rate constant, denoted as \(k\), is a crucial parameter that determines the speed of the reaction. For reactions that follow first-order kinetics, the rate of reaction is directly proportional to the concentration of a single reactant. This means that the greater the concentration of the reactant, the faster the reaction will proceed. The concept of the rate constant is integral to understanding how quickly a reaction reaches completion.

• **Rate constant (\(k\))**: A measure of the speed of a reaction under certain conditions.
• **First-order kinetics**: Describes reactions where the rate is proportional to the concentration of one reactant.

For such reactions, the formula linking the rate constant to the half-life is \( t_{1/2} = \frac{0.693}{k} \), where \(t_{1/2}\) is the time taken for half of the reactant to be consumed. By knowing the half-life, we can easily rearrange the formula to solve for the rate constant.

Decoding Half-Life in Chemical Reactions

Half-life is a concept frequently encountered in the study of reactions where the rate follows first-order kinetics. It refers to the time required for half of the reactant to be used up or converted during the reaction. Understanding half-life helps in predicting how the concentration of a reactant decreases over time.

• In first-order reactions, each successive half-life period consumes the same proportion of reactant.
• Its uniformity simplifies calculations and predictions about the behavior of a reaction over time.

A key feature of the half-life of a first-order reaction is that it remains constant regardless of the initial concentration of the reactant. This differs from other reactions where the half-life can vary with concentration changes. For the acid-catalysed hydrolysis of sucrose, the half-life is found to be 190 minutes.

Exploring Acid-Catalysed Hydrolysis

Acid-catalysed hydrolysis is a reaction where an acid is used to accelerate the breakdown of a compound with the addition of water. In the given exercise, sucrose undergoes this form of hydrolysis to break down into simpler sugars due to the presence of an acid catalyst.

• **Acid catalyst**: A substance that increases the rate of reaction by lowering the activation energy.
• **Sucrose breakdown**: Hydrolysis in the presence of acid catalyzes the conversion of sucrose into glucose and fructose.

Such reactions often exhibit first-order kinetics, where the rate is dependent solely on the concentration of the sucrose, not the concentration of the catalyst. This makes the hydrolysis process predictable and quantifiable using the rate constant derived from its half-life.