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Chapter 12: Problem 66

Chemists commonly use a rule of thumb that an increase of \(10 \mathrm{~K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The activation energy must be approximately 46.7 kJ/mol for the rate of the reaction to double when the temperature increases from 25掳C to 35掳C.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures to Kelvin: - 25掳C + 273.15 = 298.15 K - 35掳C + 273.15 = 308.15 K
02

Write equations using the Arrhenius equation

We are given that the rate of the reaction doubles when the temperature increases by 10K. Using the Arrhenius equation with the given temperatures, we can write two equations: \( k_1 = A \cdot e^{-E_a / (R \cdot 298.15)} \) \( 2 k_1 = A \cdot e^{-E_a / (R \cdot 308.15)} \)
03

Divide the equations

Now we can divide the second equation by the first equation to cancel out the pre-exponential factor A and the rate constant k: \( 2 = \frac{e^{-E_a / (R \cdot 308.15)}}{e^{-E_a / (R \cdot 298.15)}} \)
04

Simplify the equation

We can simplify this equation further by taking the natural logarithm of both sides: ln(2) = ln(\( \frac{e^{-E_a / (R \cdot 308.15)}}{e^{-E_a / (R \cdot 298.15)}} \)) Then, we can use the property of logarithms ln(a/b) = ln(a) - ln(b): ln(2) = \(\frac{-E_a}{R \cdot 308.15} - \frac{-E_a}{R \cdot 298.15} \)
05

Solve for the activation energy \(E_a\)

Now, we can solve for the activation energy \(E_a\): ln(2) = \( \frac{E_a (298.15 - 308.15)}{R \cdot 298.15 \cdot 308.15} \) \(E_a = \frac{R \cdot 298.15 \cdot 308.15 \cdot ln(2)}{-10} \) Insert the value of R as 8.314 J/(mol路K): \(E_a = \frac{8.314 \cdot 298.15 \cdot 308.15 \cdot ln(2)}{-10} \) By calculating the expression above, we get: \(E_a 鈮圽) 46.7 kJ/mol So, the activation energy must be approximately 46.7 kJ/mol for the rate of the reaction to double when the temperature increases from 25掳C to 35掳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.

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

The following data were obtained for the reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) where \(\quad\) Rate \(=-\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\)

In an effort to become more environmentally friendly, you have decided that your next vehicle will run on biodiesel that you will produce yourself. You have researched how to make biodiesel in your own home and have decided that your best bet is to use the following chemical reaction: $$ \mathrm{Oil}+\mathrm{NaOH} \text { (in methanol) } \longrightarrow \text { biodiesel }+\text { glycerin } $$ You performed a test reaction in your kitchen to study the kinetics of this process. You were able to monitor the concentration of the oil and found that the concentration dropped from \(0.500 M\) to \(0.250 \mathrm{M}\) in \(20.0\) minutes. It took an additional \(40.0\) minutes for the concentration of the oil to further drop to \(0.125 M\). How long will it take for you to convert \(97.0 \%\) of the oil to biodiesel?

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in \(\mathrm{A}\) and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} M\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{~s}\) has elapsed.
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