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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
At 25°C (298.15 K), the catalyst increases the rate of the reaction by a factor of approximately 9.82. At 125°C (398.15 K), the catalyst increases the rate of the reaction by a factor of approximately 20.64.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures 25°C and 125°C to Kelvin. To convert Celsius to Kelvin, use this equation: Temperature(K) = Temperature(°C) + 273.15 T1 = 25 + 273.15 = 298.15 K T2 = 125 + 273.15 = 398.15 K
02

Set up the Arrhenius equation ratio for each temperature

Let's set up the ratio of the rate constants with the catalyst (kc) to without the catalyst (ku) for each temperature: T1: \(\frac{k_{c1}}{k_{u1}} = \frac{Ae^{\frac{-55}{8.314\times 298.15}}}{Ae^{\frac{-95}{8.314\times 298.15}}}\) T2: \(\frac{k_{c2}}{k_{u2}} = \frac{Ae^{\frac{-55}{8.314\times 398.15}}}{Ae^{\frac{-95}{8.314\times 398.15}}}\)
03

Solve for the ratio of rate constants

We can cancel the collision factor A since it remains the same for both reactions. This leaves us with equations involving only the activation energies and temperatures: T1: \(\frac{k_{c1}}{k_{u1}} = \frac{e^{\frac{-55}{8.314\times 298.15}}}{e^{\frac{-95}{8.314\times 298.15}}}\) T2: \(\frac{k_{c2}}{k_{u2}} = \frac{e^{\frac{-55}{8.314\times 398.15}}}{e^{\frac{-95}{8.314\times 398.15}}}\) Now we can solve for the fractions: T1: \(\frac{k_{c1}}{k_{u1}}\approx 9.82\) T2: \(\frac{k_{c2}}{k_{u2}}\approx 20.64\) These values tell us how much faster the reaction is with the catalyst compared to without. At 25°C (298.15 K), the catalyst increases the rate of the reaction by a factor of approximately 9.82. At 125°C (398.15 K), the catalyst increases the rate of the reaction by a factor of approximately 20.64.

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Most popular questions from this chapter

(a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism?

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q) .\) A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval, in units of \(\mathrm{M} / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min}\), what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M}\) \(\mathrm{HCl}\) occurs according to the reaction \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+\underset{2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)}\) The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the value for the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?(\mathrm{c})\) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

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