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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, \(T_{old} = 295.15\,\mathrm{K}\) \(T_{new} = \frac{(54.0 \times 10^{3})(8.314)(295.15)}{(54.0 \times 10^{3}) - (8.314)(295.15) \ln{7}} 鈮� 352.19\,\mathrm{K}\) **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

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

Consider the following statements: "In general, the rate of a chemical reaction increases a bit at first because it takes a while for the reaction to get 'warmed up.' After that, however, the rate of the reaction decreases because its rate is dependent on the concentrations of the reactants, and these are decreasing." Indicate everything that is correct in these statements, and indicate everything that is incorrect. Correct the incorrect statements and explain.

Define stability from both a kinetic and thermodynamic perspective. Give examples to show the differences in these concepts.

The rate law for the reaction $$ \mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \operatorname{HCl}(g)+\mathrm{CCl}_{4}(g) $$ is $$ \text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right] $$ What are the units for \(k\), assuming time in seconds and concentration in \(\mathrm{mol} / \mathrm{L}\) ?

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.
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