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Chapter 12: Problem 75

A certain reaction has an activation energy of 54.0 $\mathrm{kJ} / \mathrm{mol}\( . As the temperature is increased from \)22^{\circ} \mathrm{C}$ to a higher temperature, the rate constant increases by a factor of 7.00 . Calculate the higher temperature.

Short Answer

Expert verified
The higher temperature of the reaction is approximately 62.25 °C. This is calculated using the Arrhenius equation and the given activation energy, initial temperature, and rate constant increase factor.

Step by step solution

01

Identify the Given Variables

The given variables are: - Activation energy (Ea) = 54 kJ/mol - Initial temperature (T1) = 22°C = 295.15 K (convert to Kelvin by adding 273.15) - Rate constant increases by a factor = 7
02

Write Down the Arrhenius Equation

The Arrhenius equation is expressed as: \[ k = Ae^{-\frac{Ea}{RT}}\] Where: - k is the rate constant, - A is the pre-exponential or frequency factor, - Ea is the activation energy, - R is the gas constant, and - T is the temperature in Kelvin.
03

Write Down the Arrhenius Equation for Both Temperatures

Let \(k_1\) be the initial rate constant at temperature \(T_1\) and \(k_2\) be the final rate constant at the higher temperature \(T_2\). Then we can write the Arrhenius equation for both temperatures: \[ k_1 = Ae^{-\frac{Ea}{RT_1}}\] and \[ k_2 = Ae^{-\frac{Ea}{RT_2}}\]
04

Relate the Rate Constants and Solve for T2

The rate constant increases by a factor of 7 i.e., \(k_2 = 7k_1\). Divide the second Arrhenius equation by the first to eliminate A: \[\frac{k_2}{k_1} = \frac{Ae^{-\frac{Ea}{RT_2}}}{Ae^{-\frac{Ea}{RT_1}}}\] Replace \(k_2\) with \(7k_1\) and simplify the equation: \[7 = e^{\frac{Ea}{R}(\frac{1}{T_1} - \frac{1}{T_2})}\] Now, we need to solve for \(T_2\).
05

Convert the Activation Energy to J/mol

In order to use the gas constant R in J/(mol·K), we need to convert the activation energy from kJ/mol to J/mol: \(Ea = 54\,kJ/mol \times 1000\,J/kJ = 54000\,J/mol\)
06

Insert the Given Variables and Solve for T2

Insert the values of Ea, T1, R, and rate constant increase factor into the equation to solve for T2: \[7 = e^{\frac{54000\,J/mol}{8.314\,J/(mol\cdot K)}(\frac{1}{295.15\,K} - \frac{1}{T_2})}\] To find \(T_2\), you can follow these steps: 1. Take the natural logarithm (ln) of both sides. 2. Solve for \(\frac{1}{T_2}\). 3. Take the reciprocal of the result to find \(T_2\). After solving, you should find the value of the higher temperature, \(T_2\approx 335.4 K\).
07

Convert the Higher Temperature to Celsius

Finally, convert the higher temperature from Kelvin to Celsius: Higher temperature = 335.4 K - 273.15 = 62.25 °C

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