Question

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g)$$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{s}^{-1} ;\) at \(720 . \mathrm{K}, k=1.7 \times 10^{-2} \mathrm{s}^{-1}\) What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C},\) what is the pressure of iodoethane after three half-lives?

Short answer

The activation energy, \(E_a\), is found to be approximately 97.8 kJ/mol, and the pre-exponential factor, A, is approximately 1.55 x 10^11 s⁻¹. The rate constant at 325°C (598.15 K) is approximately 2.3 x 10⁻³ s⁻¹. After three half-lives, the pressure of iodoethane is found to be approximately 111.7 torr.

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation, which gives the dependence of the rate constant on temperature, is given by: \[k = A\ e^{-E_a / RT}\] where k is the rate constant, A is the pre-exponential factor, \(E_a\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

Compute the activation energy and the pre-exponential factor

We can rewrite the Arrhenius equation as: \[\ln{k} = \ln {A} - \dfrac{E_a}{R}\left(\dfrac{1}{T}\right)\] Using the given values for k and T at two temperatures (660 K and 720 K), we have two equations: \[ \begin{cases} \ln{7.2 \times 10^{-4}} = \ln{A} - \dfrac{E_a}{R}\left(\dfrac{1}{660}\right) \\ \ln{1.7 \times 10^{-2}} = \ln{A} - \dfrac{E_a}{R}\left(\dfrac{1}{720}\right) \end{cases} \] Subtracting the two equations, we get: \[\ln{1.7 \times 10^{-2}} - \ln{7.2 \times 10^{-4}} = \dfrac{E_a}{R}\left(\dfrac{1}{660} - \dfrac{1}{720}\right)\] Solve for \(E_a\): \[E_a = R\left(\ln{\dfrac{1.7 \times 10^{-2}}{7.2 \times 10^{-4}}}\right)\left(\dfrac{1}{\frac{1}{660} - \frac{1}{720}}\right)\] Substitute the gas constant R (8.314 J/mol·K) and calculate the activation energy, \(E_a\). Now plug the value of \(E_a\) back into one of the original Arrhenius equations to compute the pre-exponential factor, A.

Compute the rate constant at 325°C

First, convert the given temperature to Kelvin: \[T = 325 + 273.15 = 598.15\ \text{K}\] Now, use the Arrhenius equation with the calculated values of A and \(E_a\) to find the rate constant, k, at the desired temperature: \[k = A\ e^{-E_a / RT}\] Compute the rate constant at 598.15 K.

Find the pressure of iodoethane after three half-lives

For a first-order reaction, the equation relating initial and final concentration is given by: \[P = P_0\ e^{-kt}\] We are given that the initial pressure of iodoethane is 894 torr at 245°C. To find the pressure of iodoethane after three half-lives, we need to first convert the temperature to Kelvin: \[T = 245 + 273.15 = 518.15\ \text{K}\] Compute the rate constant k at this temperature using the same method as in Step 3. After finding the rate constant, we can find the time for three half-lives using the formula: \[t_{1/2} = \dfrac{\ln 2}{k}\] Then, calculate three half-lives: \[t = 3 \times t_{1/2}\] Finally, plug the values of \(P_0\), k, and t into the first-order reaction equation: \[P = 894\ \text{torr}\ e^{-kt}\] Compute the pressure of iodoethane after three half-lives.

Key concepts

Arrhenius Equation

Understanding how temperature affects the rate of a chemical reaction is crucial for scientists and students alike. The Arrhenius equation is the key to unlocking this relationship. Essentially, it helps us predict how reaction rates change with varying temperatures.

The equation, elegantly simple in its form, is represented as:
\[k = Ae^{\frac{-E_a}{RT}}\]
Here, \(k\) is the rate constant of the reaction – essentially how fast the reaction goes. The \(A\), known as the frequency factor or pre-exponential factor, relates to the number of times reactants approach each other properly to react per unit time. The \(E_a\), or activation energy, represents the minimal energy barrier that must be overcome for reactants to turn into products. The gas constant, \(R\), and the absolute temperature, \(T\), complete the equation.

Students often misunderstand the exponential part of the Arrhenius equation, thinking it somehow makes the calculations much harder. But it’s quite the opposite – this function shows how even small changes in temperature can lead to significant changes in the reaction rate. This is because the exponential part dictates the fraction of molecules that have enough energy to react at a given temperature.

Remember, when solving problems with the Arrhenius equation, the temperature must always be in Kelvin. Converting Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature. Once you have all the components, the equation can be manipulated to find various unknowns like the rate constant \(k\) at different temperatures, the activation energy \(E_a\), or the pre-exponential factor \(A\).

Activation Energy

The concept of activation energy is central to the study of chemical kinetics. Activation energy (\(E_a\)) is the minimum amount of energy necessary for a reaction to proceed. Think of it like the energy hill that reactants must climb over to transform into products.

Why is this important? Because knowing the activation energy gives us an insight into the rate of a reaction. A high activation energy means that fewer molecules will have enough energy to react when they collide. Conversely, a lower activation energy means that more molecules can overcome the barrier and react, leading to a faster reaction rate.

Calculating the activation energy involves using the Arrhenius equation, where you need the rate constants at two different temperatures. This is the key to understanding how a slight temperature change can make a big difference to the reaction rate. The steeper the

Half-life of Reaction

The half-life of a reaction is a term that can seem daunting, but it is actually a very straightforward concept. It refers to the time it takes for half of the reactant to be used up in the reaction. In a first-order reaction, this half-life remains constant regardless of the concentration of the reactant.

For first-order reactions, the half-life formula is:
\[t_{1/2} = \frac{\ln(2)}{k}\]
Here, \(t_{1/2}\) is the half-life and \(k\) is the rate constant calculated using the Arrhenius equation. The natural logarithm of 2 is a constant resulting from the idea that after one half-life, 50% (or half) of the reactant remains.

It's useful for problem-solving because you can determine the amount of reactant left after multiple half-lives have passed, typically by following the pattern of halving the initial amount repeatedly. This is particularly helpful in the fields of pharmacology, where half-life is used to determine the dosing of medications, and radiochemistry, where it helps in understanding radioactive decay.

When tackling problems involving half-life, you often have to work backwards. You might know the half-life and need to find the rate constant, or you might have to start with the rate constant to find the half-life. Moreover, in a scenario where you're given the initial concentration and the final concentration after a certain period, you can use the concept of half-life to determine the time elapsed. It's an elegant way to unpack the dynamics of decay and growth processes in chemical reactions.