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Chapter 14: Problem 54

(a) The activation energy for the reaction $\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\( is \)100 \mathrm{~kJ} / \mathrm{mol}$. Calculate the fraction of the molecule A that has an energy equal to or greater than the activation energy at \(400 \mathrm{~K} .(\mathbf{b})\) Calculate this fraction for a temperature of \(500 \mathrm{~K}\). What is the ratio of the fraction at $500 \mathrm{~K}\( to that at \)400 \mathrm{~K}$ ?

Short Answer

Expert verified
The ratio of the fractions of molecule A with energy equal to or greater than the activation energy at 500 K and 400 K is given by the following formula: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times500)}}{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times400)}}\] Plug in the activation energy and temperatures into the formula and calculate the ratio.

Step by step solution

01

Understand the Boltzmann Distribution formula

The Boltzmann Distribution formula is given by: \[f(E) = A \mathrm{e}^{-E/kT}\] where - \(f(E)\) is the fraction of molecules with energy E, - E is the energy (in this case, the activation energy), - A is a constant, - k is the Boltzmann constant, \(1.38 \times 10^{-23}\mathrm{J\/K}\), - T is the temperature in Kelvin.
02

Plug in values and calculate the fraction at 400 K

At 400 K, the fraction of molecules with energy equal to or greater than the activation energy (100 kJ/mol) is: \[f(E) = A \mathrm{e}^{-100\times10^3\mathrm{J/mol} / (1.38\times10^{-23}\mathrm{J/K} \times 400\mathrm{K})}\] Calculate the value of \(f(E)\) at 400 K: \(f(E_{400}) = A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 400}\)
03

Calculate the fraction at 500 K

At 500 K, the fraction of molecules with energy equal to or greater than the activation energy (100 kJ/mol) is: \[f(E) = A \mathrm{e}^{-100\times10^3\mathrm{J/mol} / (1.38\times10^{-23}\mathrm{J/K} \times 500\mathrm{K})}\] Calculate the value of \(f(E)\) at 500 K: \(f(E_{500}) = A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 500}\)
04

Calculate the ratio of the fractions at 500 K and 400 K

To find the ratio of the fractions at 500 K to 400 K, divide the fraction at 500 K by the fraction at 400 K: \[\frac{f(E_{500})}{f(E_{400})} = \frac{A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 500}}{A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 400}}\] The constant A cancels out in the ratio, so the final expression for the ratio is: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-E/1.38\times10^{-23}\times500}}{\mathrm{e}^{-E/1.38\times10^{-23}\times400}}\] Calculate the ratio using the given activation energy (100 kJ/mol) and temperatures (400 K and 500 K) to get: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times500)}}{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times400)}}\]

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.

In solution, chemical species as simple as \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right]\) of a solution containing an acidcatalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\) concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: $\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .$ This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when $\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\( and \)\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}$

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) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3} .\) (b) Do we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? (d) Is this an example of homogeneous catalysis or heterogeneous catalysis?
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