Problem 17
Question
The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is " \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After \(2.57 \mathrm{h}\) at \(27^{\circ} \mathrm{C}\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)
Step-by-Step Solution
Verified Answer
The rate constant \( k \) is approximately \( 0.037 \text{ h}^{-1} \).
1Step 1: Understand the Rate Equation
The given rate equation is \(-\Delta[\text{sucrose}]/\Delta t=k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]\), indicating a first-order reaction. This means the rate of reaction is directly proportional to the concentration of sucrose.
2Step 2: Determine Change in Concentration
The initial concentration of sucrose \( [\text{sucrose}]_0 \) is 0.0146 M, and the final concentration \([\text{sucrose}]\) is 0.0132 M after 2.57 hours. The change in concentration \( \Delta[\text{sucrose}] = 0.0146 \text{ M} - 0.0132 \text{ M} = 0.0014 \text{ M} \).
3Step 3: Calculate Average Reaction Rate
The average reaction rate is given by \(-\Delta[\text{sucrose}] / \Delta t \). Therefore, \(-\Delta[\text{sucrose}] / \Delta t = 0.0014 \text{ M} / 2.57 \text{ h} \approx 0.000544 \text{ M/h} \).
4Step 4: Solve for Rate Constant
Use the rate equation \(-\Delta[\text{sucrose}] / \Delta t = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}] \) to solve for \(k\). Substitute in the average reaction rate and the initial concentration to find \( k = 0.000544 \text{ M/h} / 0.0146 \text{ M} \approx 0.037 \text{ h}^{-1} \).
Key Concepts
Rate EquationFirst-Order ReactionRate Constant
Rate Equation
In chemical kinetics, the rate equation is a fundamental expression that relates the rate of a chemical reaction to the concentration of reactants. The rate equation signifies how the concentration of a substance changes over time, providing insight into the dynamics of a reaction.
- The general form is: \[ -\frac{d[ ext{A}]}{dt} = k [ ext{A}]^n [ ext{B}]^m \ldots \] where \([\text{A}]\) and \([\text{B}]\) are concentrations, \(k\) is the rate constant, and \(n, m\) are the reaction orders.
- In this specific exercise, we are given the rate equation for sucrose hydrolysis as \[ -\frac{\Delta[\text{sucrose}]}{\Delta t} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]. \]
First-Order Reaction
A first-order reaction is characterized by its direct proportionality between the reaction rate and the concentration of a single reactant. This straightforward relationship simplifies many analyses in chemistry.
- The rate law for a first-order reaction is: \[ -\frac{d[ ext{A}]}{dt} = k [\text{A}] \]This tells us that if the concentration of \([\text{A}]\) doubles, the rate of reaction also doubles.
- In the case of sucrose hydrolysis, the rate equation given is of first-order, meaning the concentration of sucrose is the sole factor influencing how fast the hydrolysis occurs.
- This type of reaction has a unique feature: it follows an exponential decay model, often described with a half-life, which is the time taken for half of the reactant to undergo reaction.
Rate Constant
The rate constant \(k\) is a pivotal factor in the rate equation, representing the intrinsic speed of a reaction under specific conditions.
- It has units that depend on the reaction order, with the first-order reaction having units of \(\text{time}^{-1}\), such as \(\text{h}^{-1}\).
- In our exercise, we calculated the rate constant \(k\) for sucrose hydrolysis to be approximately 0.037 \(\text{h}^{-1}\), indicative of the reaction's speed at the given temperature and concentration.
- The rate constant incorporates the conditions at which the reaction is conducted (e.g., temperature, pressure, and presence of a catalyst), making it a crucial parameter for predicting and controlling reaction behavior.
Other exercises in this chapter
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