Problem 18 - Chemistry and Chemical Reactivity

Question

The conversion of cyclopropane to propene (Example \(14.5)\) occurs with a first-order rate constant of \(2.42 \times 10^{-2} \mathrm{h}^{-1} .\) How long will it take for the concentration of cyclopropane to decrease from an initial concentration of \(0.080 \mathrm{mol} / \mathrm{L}\) to \(0.020 \mathrm{mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

It takes approximately 57.3 hours for the concentration to decrease from 0.080 mol/L to 0.020 mol/L.

Step 1: Determine the Rate Law for a First-Order Reaction

The formula for a first-order reaction is given by:\[[A] = [A]_0 \times e^{-kt}\]where \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time. Here, \([A]_0 = 0.080 \, \text{mol/L}\), and \([A] = 0.020 \, \text{mol/L}\).

Step 2: Rearrange the First-Order Equation

We need to solve for \(t\), the time it takes for the concentration to decrease. Rearranging the equation, we get:\[t = \frac{-1}{k} \ln\left(\frac{[A]}{[A]_0}\right)\]

Step 3: Substitute Known Values into the Equation

Substitute \(k = 2.42 \times 10^{-2} \, \text{h}^{-1}\), \([A] = 0.020 \, \text{mol/L}\), and \([A]_0 = 0.080 \, \text{mol/L}\) into the equation to solve for \(t\):\[t = \frac{-1}{2.42 \times 10^{-2}} \ln\left(\frac{0.020}{0.080}\right)\]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm inside the formula:\[\ln\left(\frac{0.020}{0.080}\right) = \ln(0.25) \approx -1.3863\]

Step 5: Calculate the Time

Substitute the value of the natural logarithm back into the equation:\[t = \frac{-1}{2.42 \times 10^{-2}} \times (-1.3863) \approx 57.3 \, \text{hours}\]

Step 6: Interpret the Result

The calculated time, approximately \(57.3\) hours, indicates how long it takes for the concentration of cyclopropane to decrease from \(0.080 \, \text{mol/L}\) to \(0.020 \, \text{mol/L}\) at the given rate constant.

Key Concepts

Rate Constant

In the world of chemical reactions, the rate constant is like a special signal that tells us how fast a reaction happens. This constant is symbolized as \(k\) in reaction equations. For first-order reactions, the rate constant has units of inverse time, such as hours inverse (\(\text{h}^{-1}\)). In our example, the rate constant \(2.42 \times 10^{-2} \, \text{h}^{-1}\) indicates how quickly cyclopropane converts to propene.
A reaction's speed depends on both this rate constant and the concentration of reactants. The bigger the rate constant, the faster the reaction goes. Think of it like the accelerator in your car — the more you press, the faster you drive. Understanding this constant is crucial since it affects the time needed for any observable change in concentration.
In practical terms, knowing the rate constant allows chemists to tailor conditions to speed up or slow down reactions as needed, which is especially important in industrial processes or when synthesizing chemicals.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that looks into the speed or rate of chemical reactions. It's like the science of understanding how reactions "race" to completion. It also studies the way molecules bump together and transform, like cyclopropane changing into propene.
For a first-order reaction like ours, the rate depends on the concentration of a single reactant. In other words, the speed at which cyclopropane turns into propene is directly tied to how much cyclopropane we start with. This is captured in the rate law formula: \[[A] = [A]_0 \times e^{-kt}\]Here, * \([A]_0\) is your starting concentration.* \([A]\) is the concentration at time \(t\).* \(k\) is the rate constant. This formula helps chemists predict how a reaction will play out over time. So if you know the initial concentration and the rate constant, you can find out how much of a reactant will be left after a certain period.

Concentration Change

Concentration change is a key aspect of chemical reactions, showing how much of a substance is used up or formed over time. Let's take our conversion of cyclopropane to propene as an example. We started with \(0.080 \, \text{mol/L}\) of cyclopropane and watched it drop to \(0.020 \, \text{mol/L}\).
The change in concentration is a clue to how the reaction progresses. By rearranging the first-order kinetic equation, we calculated the time it takes for this change.The formula:\[t = \frac{-1}{k} \ln\left(\frac{[A]}{[A]_0}\right) \]allows us to solve for \(t\), giving us insights into the reaction dynamics. In our example, it took around 57.3 hours for the cyclopropane to reach the desired lower concentration.
Concentration changes can help optimize conditions in both laboratory and real-world chemical processes. Recognizing how and why concentrations alter ensures one can control the reaction's direction and final composition effectively. For students, mastering concentration changes involves understanding logarithmic relationships, since many kinetic equations require using logarithms to find time or concentration. So practice is key!