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Chapter 12: Problem 65

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

Step by step solution

01

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**
02

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**
03

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**
04

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**
05

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 \(79.04^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g) $$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{~s}^{-1}\); at 720. \(\mathrm{K}, k=1.7 \times 10^{-2} \mathrm{~s}^{-1}\). What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C}\), what is the pressure of iodoethane after three half-lives?

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