The conversion of cyclopropane, an anesthetic, to propylene...

The conversion of cyclopropane, an anesthetic, to propylene (see Problem 13.68) has a rate constant $k=1.3 \times$ $10^{-6} \mathrm{~s}^{-1}$ at $673 \mathrm{~K}$ and $k=1.1 \times 10^{-5} \mathrm{~s}^{-1}$ at $703 \mathrm{~K}$. What is the activation energy in $\mathrm{kJ} / \mathrm{mol}$ ? Use the data given at $703 \mathrm{~K}$ to calculate the frequency factor, $A$, for this reaction. What is the rate constant for the reaction at $350^{\circ} \mathrm{C}$ ?

Key Concepts

Temperature Dependence of Reaction Rates

Temperature influences the rate at which chemical reactions occur. As temperature increases, the fraction of molecules with sufficient energy to overcome the activation barrier increases. This concept is central to kinetics and is mathematically described by the Arrhenius equation, highlighting why reaction rates generally increase with temperature.

Frequency Factor

The frequency factor, also known as the pre-exponential factor, represents the number of collisions with proper orientation between reactant molecules per unit time. It is a constant in the Arrhenius expression that provides insight into the likelihood of reactant collisions leading to a reaction, independent of the energy barrier represented by the activation energy.

Rate Constant

The rate constant is a proportionality factor in the rate law that quantifies the speed of a reaction under given conditions. It is specific to a particular reaction and temperature. Through the Arrhenius equation, the rate constant can be adjusted for different temperatures by accounting for the activation energy and the frequency factor, making it a central parameter in predicting reaction behavior.

Arrhenius Equation

The Arrhenius equation describes the temperature dependence of reaction rates. It relates the rate constant (k) to temperature (T), activation energy (E_a), and the frequency factor (A) by the equation k = A exp(–E_a/RT). This equation is fundamental in chemical kinetics as it connects the molecular energy barrier for a reaction with observable reaction rates at different temperatures.

Activation Energy

Activation energy is the minimum energy required for a reaction to occur. It is the energy barrier that reactants must overcome to be transformed into products. In the context of the Arrhenius equation, a higher activation energy corresponds to a slower reaction rate at a given temperature, and determining its magnitude is crucial for understanding the reaction’s sensitivity to temperature changes.

Transcript

00:01 This question, similar to the previous question, contains multiple parts.

00:05 First, given the rate constants at two separate temperatures, you're asked to solve for the activation energy.

00:18 To do this, we'll take the arranius equation provided in the textbook at two different temperatures and two different rate constants.

00:28 Rearrange this equation to solve for ea.

00:32 With a little bit of algebraic rearrangement, we get ea is equal to r, by the natural log of k2 over k1 divided by 1 over t1 minus 1 over t 2 we then plug in all of our values r is a constant at 8 .314 i have my k2 value at the higher temperature divided by k1 at the lower temperature.

00:53 I'll then divide by 1 over the kelvin temperature that is lower minus 1 over the kelvin temperature that is higher and i get 280 ,000 joules per mole or 280 kilojoules per mole.

01:11 Then to solve for the frequency factor, i'll take a different form of the arraneous equation that only has one rate constant and one temperature.

01:23 Using this equation and rearranging it from your textbook, we'll get the frequency factor is equal to the rate constant divided by e to the negative ea over rt.

01:35 I can then solve for the frequency factor at the temperature they suggested the temperature of 703.

01:45 So a will be equal to the rate constant at 703 divided by e so the negative ea which i determined up here has to be in joules don't use kilojoules divided by r and the temperature they suggested the 703 kelvin temperature and i get a frequency factor at 703 kelvin of 7 .0 times 10 to the 15 for the last part of this question, they want you to solve for the rate constant at a different temperature at 350 degrees celsius...

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