Problem 98 - Chemistry The Central Science In Si Units

Question

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O} .\) At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1}\). Calculate the half- life at this temperature. (b) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

Step-by-Step Solution

Answer

(a) The half-life is approximately 21.0 minutes. (b) The new temperature for doubling the rate is approximately 310 K.

Step 1: Understand the Given Data

Identify relevant information from the problem: the reaction is first order with respect to \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) with a rate constant \( k = 3.30 \times 10^{-2} \text{ min}^{-1} \) at \( 300 \text{ K} \). The activation energy \( E_a \) for the reaction is \( 80.0 \text{ kJ/mol} \).

Step 2: First-order Reaction Half-life Calculation

The half-life \( t_{1/2} \) for a first-order reaction is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \] Substituting the given rate constant: \[ t_{1/2} = \frac{0.693}{3.30 \times 10^{-2}} \approx 21.0 \text{ min} \]

Step 3: Use Arrhenius Equation for Temperature Change

To find the temperature at which the rate doubles, use the Arrhenius equation: \( k = Ae^{-E_a/(RT)} \). Set up the equation for the rate constant at two different temperatures \( T_1 \) and \( T_2 \): \( \frac{k_2}{k_1} = e^{-E_a/RT_2 + E_a/RT_1} \). To double the rate, \( \frac{k_2}{k_1} = 2 \).

Step 4: Solve for New Temperature

Rearrange the equation as follows and input \( R = 8.314 \text{ J/mol·K} \) and \( E_a = 80000 \text{ J/mol} \):\[ \ln{2} = \frac{80000}{8.314} \left( \frac{1}{300} - \frac{1}{T_2} \right) \]Calculating yields:\[ \ln{2} \approx 6.9478 \cdot \left( \frac{1}{300} - \frac{1}{T_2} \right) \]Solving gives \( T_2 \approx 310 \text{ K} \).

Step 5: Verify Calculations

Check calculations for any rounding errors or missed steps to ensure accuracy.

Key Concepts

First-order Reaction

When studying chemical kinetics, reactions are often categorized based on their order, which helps us understand how the concentration of reactants affects the rate of the reaction. A first-order reaction is one where the rate depends linearly on the concentration of a single reactant. In the given exercise, the reaction of sucrose with water is first-order with respect to sucrose, represented as \( C_{12}H_{22}O_{11}(aq) \). This means that the rate of decomposition of sucrose is directly proportional to its concentration.For a first-order reaction, the rate law can be expressed as:

• \( ext{Rate} = k[A] \)

where \( k \) is the rate constant and \([A]\) is the concentration of the reactant. The simplicity of first-order reactions makes them relatively easy to analyze, providing clear insights into how reactions proceed over time.

Activation Energy

One of the central factors that influence the rate of a chemical reaction is the activation energy, \( E_a \). This is the minimum amount of energy required for reactants to successfully collide and form products. A high activation energy indicates that a reaction will be slow because fewer molecules have enough energy to overcome the energy barrier.In the original exercise, the activation energy is given as \( 80.0 \, \text{kJ/mol} \). To put this into perspective, think of activation energy as a hill that reactants must "climb" over to become products. If the hill is too high, typical kinetic energy from molecules at room temperature often won't suffice to reach the peak, slowing down the reaction rate.Therefore, the lower the activation energy, the faster the reaction will be under a given set of conditions. Conversely, increasing the temperature can boost reactant energies, facilitating their transformation into products by crossing the activation energy barrier.

Half-life Calculation

The half-life \( t_{1/2} \) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration. For first-order reactions, the half-life is unique because it is constant and does not depend on the initial concentration of the reactant.Given the rate constant \( k = 3.30 \times 10^{-2} \, \text{min}^{-1} \) in the exercise, the formula for first-order half-life is:

• \( t_{1/2} = \frac{0.693}{k} \)

Using this, we find the half-life to be approximately 21.0 minutes at 300 K. This means that every 21 minutes, the concentration of sucrose will be halved, showcasing the steady nature of first-order reactions.

Arrhenius Equation

The Arrhenius equation is a fundamental formula in chemical kinetics that relates the rate constant \( k \) to the temperature \( T \) and activation energy \( E_a \). This equation shows how changes in temperature affect the rate of a chemical reaction.The Arrhenius equation is:

• \( k = Ae^{-E_a/(RT)} \)

where \( A \) is the pre-exponential factor, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.In the context of the given problem, to find the new temperature at which the reaction rate doubles, the equation is rearranged to:

• \( \frac{k_2}{k_1} = e^{-E_a/RT_2 + E_a/RT_1} \)

With the known values of \( E_a \), \( R \), and initial temperature \( T_1 \), solving this yields a new temperature \( T_2 \) of approximately 310 K. This calculation highlights the sensitivity of reaction rates to temperature changes and the utility of the Arrhenius equation in predicting such changes.