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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°C to 35°C.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures to Kelvin: - 25°C + 273.15 = 298.15 K - 35°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·K): \(E_a = \frac{8.314 \cdot 298.15 \cdot 308.15 \cdot ln(2)}{-10} \) By calculating the expression above, we get: \(E_a ≈\) 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°C to 35°C.

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Most popular questions from this chapter

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.

The activation energy for a reaction is changed from \(184 \mathrm{~kJ} / \mathrm{mol}\) to \(59.0 \mathrm{~kJ} / \mathrm{mol}\) at \(600 . \mathrm{K}\) by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant and assume the initial concentrations are the same.

The reaction $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{~K}\) and \(y\) -intercept of \(33.5\). Assume \(k\) has units of \(\mathrm{s}^{-1}\). a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\). c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\).

Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.) \(\begin{aligned} 2 \mathrm{Ce}^{4+}(a q)+\mathrm{Hg}_{2}^{2+}(a q) & \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+2 \mathrm{Hg}^{2+}(a q) \\\ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\)

Cobra venom helps the snake secure food by binding to acetylcholine receptors on the diaphragm of a bite victim, leading to the loss of function of the diaphragm muscle tissue and eventually death. In order to develop more potent antivenoms, scientists have studied what happens to the toxin once it has bound the acetylcholine receptors. They have found that the toxin is released from the receptor in a process that can be described by the rate law Rate \(=k[\) acetylcholine receptor-toxin complex \(]\) If the activation energy of this reaction at \(37.0^{\circ} \mathrm{C}\) is \(26.2 \mathrm{~kJ} / \mathrm{mol}\) and \(A=0.850 \mathrm{~s}^{-1}\), what is the rate of reaction if you have a \(0.200 M\) solution of receptor-toxin complex at \(37.0^{\circ} \mathrm{C}\) ?
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