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. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

Short answer

The half-life of the reaction at 300 K is approximately 21.0 minutes. The reaction rate would double at a temperature of about 307.2 K.

Step-by-step solution

Part (a): Calculating the half-life at 300 K

In a first-order reaction, the integrated rate equation is given by the formula: \[ ln\frac{[A]_{0}}{[A]_{t}} = kt. \] To find the half-life (t₁/₂), we set \([A]_{t}\) as \(\frac{[A]_{0}}{2}\), since half the reactant remains at the half-life time. Plugging this into the formula, we get: \[ ln\frac{[A]_{0}}{\frac{[A]_{0}}{2}} = kt_{1/2}. \] Now, we are left with: \[ ln2 = kt_{1/2}. \] Solving for t₁/₂, we have: \[ t_{1/2} = \frac{ln2}{k}. \] The rate constant at 300 K is 3.30 x 10⁻² min⁻¹. Therefore, we can calculate the half-life as: \[ t_{1/2} = \frac{ln2}{3.30 × 10^{-2} \,\text{min}^{-1}}. \] Calculating, we get the half-life: \[ t_{1/2} \approx 21.0\, \text{min}. \]

Part (b): Calculating temperature for doubled reaction rate

To find the temperature at which the reaction rate doubles, we will use the Arrhenius equation, which is given by: \[ k = Ae^{-\frac{E_{a}}{RT}}, \] where k is the rate constant, A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Since the rate doubles, we can write another equation for the new rate constant, k': \[ k' = Ae^{-\frac{E_{a}}{R(T')}}, \] where k' is twice the original rate constant and T' is the new temperature. Now, we will divide the second equation by the first one: \[ \frac{k'}{k} = \frac{Ae^{-\frac{E_{a}}{R(T')}}}{Ae^{-\frac{E_{a}}{RT}}} = \frac{e^{-\frac{E_{a}}{R(T')}}}{e^{-\frac{E_{a}}{RT}}}. \] Since k' = 2k, we can rewrite the equation as: \[ 2 = \frac{e^{-\frac{E_{a}}{R(T')}}}{e^{-\frac{E_{a}}{RT}}}. \] Applying ln on both sides, we get: \[ ln2 = \frac{E_{a}}{RT} - \frac{E_{a}}{R(T')}. \] We are given Eₐ as 80.0 kJ/mol = 80000 J/mol. Also, R = 8.314 J/mol·K. We can now solve for T': \[ T' = \frac{E_{a}RT}{R(E_{a} - ln2RT)} = \frac{80000 × 8.314 × 300}{8.314(80000 - ln2 × 8.314 × 300)}. \] Calculating T', we find the new temperature: \[ T' \approx 307.2\, K. \] In conclusion, the half-life of the reaction at 300 K is approximately 21.0 minutes, and the reaction rate would double at a temperature of about 307.2 K.

Key concepts

First-Order Reaction

A first-order reaction is one where the reaction rate is directly proportional to the concentration of a single reactant. This is a common type of reaction in chemistry and is characterized by an exponential decay of the reactant over time. Here’s how it works:

In a first-order reaction, the rate law is expressed as:

• Rate = k[A]

Here, "k" is the rate constant and "[A]" is the concentration of the reactant. The change in concentration is predictable and follows a specific set of characteristics:

• The rate of reaction depends on the concentration of reactant raised to the first power.
• An essential feature of first-order reactions is their characteristic half-life, which is the time it takes for half of the reactant to be consumed.
• The half-life remains constant throughout the reaction, regardless of the starting concentration.

Half-Life Calculation

The half-life of a reaction is a crucial concept in chemical kinetics, especially for first-order reactions, because it helps predict the time required for a reactant to decrease to half its initial quantity. For a first-order reaction, the half-life \(t_{1/2}\) can be calculated using the formula:

• \(t_{1/2} = \frac{ln2}{k}\)
• Where \(ln2\) is the natural logarithm of 2, approximately equal to 0.693.
• "k" is the rate constant that you find in the rate expression.

In this specific exercise, given that the rate constant \(k\) is \(3.30 \times 10^{-2} \, \text{min}^{-1}\), we can input into the formula:

\[t_{1/2} = \frac{0.693}{3.30 \times 10^{-2} \, \text{min}^{-1}}\]Calculating the expression results in a half-life of approximately 21.0 minutes at 300 K.

Arrhenius Equation

The Arrhenius equation is a fundamental formula in chemistry, providing insight into how temperature influences the rate of a chemical reaction. The equation is expressed as:

• \(k = Ae^{-\frac{E_{a}}{RT}}\)

Where:

• \(k\): The rate constant
• \(A\): The pre-exponential factor, also known as the frequency factor
• \(E_{a}\): Activation energy in joules per mole
• \(R\): The universal gas constant, which is 8.314 J/mol·K
• \(T\): Temperature in Kelvin

This equation articulates that as the temperature increases, the rate constant \(k\) generally increases, implying that reactions tend to go faster. In practice, if we want to double the reaction rate, we can use the equation to solve for the temperature change needed. The problem sets up an equation to do this, confirming that an increase in temperature to approximately 307.2 K would double the rate from the initial condition.