Question

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

The higher temperature at which the rate constant increases by a factor of 7 is approximately \(79.04^{\circ}\mathrm{C}\).

Step-by-step solution

Convert Celsius temperatures to Kelvin

We need to work with the temperatures in Kelvin. To convert the initial temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature: \(T_{old} = 22^{\circ}\mathrm{C} + 273.15 = 295.15\mathrm{K}\) **Step 2: Write the Arrhenius equation for the old and new rate constants**

Write the Arrhenius equation

The Arrhenius equation is: \(k = A \exp{(-E_a / RT)}\) We can write it for the old and new temperatures: \(k_{old} = A \exp{(-E_a / R T_{old})}\) \(k_{new} = A \exp{(-E_a / R T_{new})}\) **Step 3: Use the fact that the rate constant increases by a factor of 7**

Rate constant ratio equation

We know that \(k_{new} = 7 * k_{old}\), so we can set up the following equation: \(\frac{k_{new}}{k_{old}} = \frac{A \exp{(-E_a / R T_{new})}}{A \exp{(-E_a / R T_{old})}} = 7\) **Step 4: Simplify the equation and solve for the new temperature**

Simplify and solve for \(T_{new}\)

Simplify the equation above: \(\frac{\exp{(-E_a / R T_{new})}}{\exp{(-E_a / R T_{old})}} = 7\) Take the natural logarithm of both sides: \(-\frac{E_a}{R T_{new}} + \frac{E_a}{R T_{old}} = \ln{7}\) Now, solve for the new temperature \(T_{new}\): \(T_{new} = \frac{E_a R T_{old}}{E_a - R T_{old} \ln{7}}\) Insert the known values: - Activation energy, \(E_a = 54.0 \times 10^3\,\mathrm{J/mol}\) - Gas constant, \(R = 8.314\,\mathrm{J/(mol\cdot K)}\) - Old temperature, \(T_{old} = 295.15\,\mathrm{K}\) \(T_{new} = \frac{(54.0 \times 10^{3})(8.314)(295.15)}{(54.0 \times 10^{3}) - (8.314)(295.15) \ln{7}} ≈ 352.19\,\mathrm{K}\) **Step 5: Convert the final answer back to Celsius**

Convert the temperature back to Celsius

Convert the new temperature in Kelvin to Celsius: \(T_{new} = 352.19\,\mathrm{K} - 273.15 = 79.04^{\circ}\mathrm{C}\) The higher temperature at which the rate constant increases by a factor of 7 is approximately \(79.04^{\circ}\mathrm{C}\).