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In an effort to become more environmentally friendly, you have decided that your next vehicle will run on biodiesel that you will produce yourself. You have researched how to make biodiesel in your own home and have decided that your best bet is to use the following chemical reaction: $$ \mathrm{Oil}+\mathrm{NaOH} \text { (in methanol) } \longrightarrow \text { biodiesel }+\text { glycerin } $$ You performed a test reaction in your kitchen to study the kinetics of this process. You were able to monitor the concentration of the oil and found that the concentration dropped from \(0.500 M\) to \(0.250 \mathrm{M}\) in \(20.0\) minutes. It took an additional \(40.0\) minutes for the concentration of the oil to further drop to \(0.125 M\). How long will it take for you to convert \(97.0 \%\) of the oil to biodiesel?

Short Answer

Expert verified
It takes \(100\) minutes for \(97\%\) of the oil to be converted to biodiesel, using the given chemical reaction.

Step by step solution

01

Write down the first-order reaction equation and data

The first-order reaction equation can be represented as follows: \[ -ln(\frac{[A]}{[A]_0}) = kt \] Where: - \(\frac{[A]}{[A]_0}\) is the fraction of remaining oil at time t (Molarity of remains, divided by initial molarity); - k is the rate constant; - t is time. Given data: Initial concentration: \([A]_0 = 0.500\,\text{M}\) Final concentration after 20 min: \([A]_1 = 0.250\,\text{M}\) Final concentration after 40 min: \([A]_2 = 0.125\,\text{M}\) 97% of the oil should be converted to biodiesel, which means remaining \([A]_x = 0.03 \times [A]_0\). We need to find the time it takes for this conversion to happen.
02

Calculate the rate constant (k)

To calculate the rate constant, let's use the given concentrations and time from the data. We can write the equation for the first concentration drop (from \(0.500\,\mathrm{M}\) to \(0.250\,\mathrm{M}\)) in 20 minutes: \[ -ln(\frac{[A]_1}{[A]_0}) = k(20) \] Plug in the given values and solve for k: \[ -ln(\frac{0.250}{0.500}) = k(20 \,\text{min}) \] Now, calculate k: \[ k = \frac{-ln(0.250/0.500)}{20\,\text{min}} = \frac{ln(2)}{20\,\text{min}} = 0.03466\,\text{min}^{-1} \]
03

Find the time it takes for 97% oil conversion

Now that we have the rate constant k, we can find the time it takes for 97% of the oil to be converted to biodiesel. Use the equation for 97% conversion: \[ -ln(\frac{[A]_x}{[A]_0}) = k t \] We know that from the given data, \([A]_x = 0.03 \times [A]_0\). So plug in the values and solve for t: \[ -ln(\frac{0.03 \times [A]_0}{[A]_0}) = (0.03466\,\text{min}^{-1}) t \] Simplify the equation and solve for t: \[ -ln(0.03) = (0.03466) t \] \[ t = \frac{-ln(0.03)}{0.03466\,\text{min}^{-1}} = 100\,\text{min} \]
04

Conclusion

It takes 100 minutes for 97% of the oil to be converted to biodiesel, using the given chemical reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Reaction Equation
The notion of a first-order reaction is pivotal for understanding numerous chemical processes, including biodiesel production. Essentially, a first-order reaction is one where the rate is directly proportional to the concentration of one reactant. In simple terms, it means that if you were to double the amount of that reactant, the reaction rate would also double.

The mathematical representation of a first-order reaction is expressed as: \[-ln(\frac{[A]}{[A]_0}) = kt\] 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 elapsed time. This equation essentially tells us that the natural logarithm of the ratio of the concentration of reactant at any time to the initial concentration is equal to the rate constant \(k\) times the time \(t\).

By understanding this equation, you can predict how the concentration will change over time and thus control your biodiesel production effectively.
Rate Constant Calculation
The rate constant, often symbolized by \(k\), is a crucial part of the first-order reaction equation. It provides insight into how quickly a reaction proceeds. Calculating \(k\) is more than just a mathematical exercise; it has practical uses in predicting the time required for a reaction to complete to a certain degree.

For a first-order reaction, the rate constant is calculated by rearranging the reaction equation: \[k = \frac{-ln(\frac{[A]}{[A]_0})}{t}\]. Once you know the initial concentration (\([A]_0\)) and the concentration at a specific time (\([A]\)), you can calculate \(k\) using the time elapsed. As seen in the biodiesel example, the rate constant helps us forecast the time it will take for a certain percentage of the reactant to convert into products.
Chemical Reaction Kinetics
Chemical reaction kinetics delves into the speed or rate at which a chemical reaction occurs. This aspect of chemistry is vital for engineers and scientists who design and control processes, such as the production of biodiesel.

Reaction kinetics is not just about the speed, but also the pathway taken by a reaction and the factors affecting both rate and pathway, such as temperature, pressure, and concentration of reactants. Understanding kinetics is fundamental in optimizing a chemical process to be both efficient and economical. In the case of biodiesel production, recognizing how factors affect the rate of reaction can help in scaling the process from your kitchen to a commercial level.
Concentration and Time Relationship
In accordance with chemical kinetics, the relationship between concentration and time lets us predict how reactant concentrations change during a reaction. For first-order reactions, this relationship is particularly straightforward. It is based on the logic that as time progresses, the concentration of reactants decreases exponentially.

This relationship is not just a theoretical construct but is directly observable and measurable in a laboratory or production setting. Through the application of the first-order reaction equation, you can determine how much time is needed for a reactant, in this case, oil, to reach a certain concentration, thus allowing for efficient planning and control of the biodiesel production process.

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Most popular questions from this chapter

For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\), explain at least two ways in which the rate law could be zero order in chemical A.

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at \(273 \mathrm{~K}\) is \(2.3 \times 10^{-12} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\), and the activation energy is \(111 \mathrm{~kJ} / \mathrm{mol}\). How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of \(2.5\) atm to \(1.5\) atm at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

The reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$ \text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t} $$

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

Experiments have shown the average frequency of chirping of individual snowy tree crickets (Oecanthus fultoni) to be \(178 \mathrm{~min}^{-1}\) at \(25.0^{\circ} \mathrm{C}, 126 \mathrm{~min}^{-1}\) at \(20.3^{\circ} \mathrm{C}\), and \(100 . \mathrm{min}^{-1}\) at \(17.3^{\circ} \mathrm{C}\). a. What is the apparent activation energy of the reaction that controls the chirping? b. What chirping rate would be expected at \(15.0^{\circ} \mathrm{C}\) ? c. Compare the observed rates and your calculated rate from part b to the rule of thumb that the Fahrenheit temperature is 42 plus \(0.80\) times the number of chirps in \(15 \mathrm{~s}\).

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