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Chapter 14: Problem 87

The activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

Step by step solution

01

Convert temperatures to Kelvin

To calculate the reaction rates, we must first convert the given temperatures to Kelvin. The conversion formula is: \(T_{K} = T_{C} + 273.15\) For the two given temperatures, the Kelvin values are: \(T_{25^{\circ}\mathrm{C}} = 25 + 273.15 = 298.15\,K\) \(T_{125^{\circ}\mathrm{C}} = 125 + 273.15 = 398.15\,K\)
02

Calculate the reaction rate constants

Using the Arrhenius equation, we can calculate the reaction rate constants for both the catalyzed and uncatalyzed reactions. We have: \(k_{uncatalyzed} = Ae^{-E_a(uncatalyzed) / RT}\) \(k_{catalyzed} = Ae^{-E_a(catalyzed) / RT}\) Since the collision factor A remains the same in both equations, we can find the ratio of catalyzed reaction rate to uncatalyzed reaction rate as: \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-E_a(catalyzed) / RT}}{e^{-E_a(uncatalyzed) / RT}}\)
03

Calculate the rate increase factors

Now, we can plug in the given activation energies and temperature values to calculate the rate increase factors for both temperatures. (a) For \(T = 298.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 298.15)}}{e^{-(95 \times 10^3) / (8.314 \times 298.15)}} ≈ 211.89\) (b) For \(T = 398.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 398.15)}}{e^{-(95 \times 10^3) / (8.314 \times 398.15)}} ≈ 69.40\) Therefore, the catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (c) Which factor is most sensitive to changes in temperature-the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?

Many primary amines, RNH \(_{2},\) where \(R\) is a carbon-containing fragment such as \(C H_{3}, C H_{3} C H_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(C\) atom for \(^{4} \mathrm{R}^{\prime \prime}\) . (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2} .\) The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{s}^{-1} .\) Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from 1.00 \(\mathrm{L}\) of 0.600 \(\mathrm{MN}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of 20.0 \(\mathrm{h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) , what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)
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