Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\), with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l} k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{~s}^{-1} \\ k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{~s}^{-1} \end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{~B}] ?\)

Short Answer

Expert verified
The time required for the concentration of substance A to become equal to 4 times the concentration of substance B is approximately \(t\approx 406.52\,\text{s}\).

Step by step solution

01

Write down the first-order kinetics formula for the concentrations of A and B

The general formula for the concentration of a species after t seconds in a first-order reaction is given by: \[ C(t) = C_0 e^{-kt} \] We are given that the initial concentrations of A and B are equal, and we can denote this as \(C_0\). So, we will apply this formula for A and B: \[ [\mathrm{A}](t) = C_0 e^{-k_\mathrm{A}t} \] \[ [\mathrm{B}](t) = C_0 e^{-k_\mathrm{B}t} \]
02

Write down the condition for the required concentrations of A and B

Given that we need to find a time at which the concentration of substance A becomes equal to 4 times the concentration of substance B, we can write our equation as follows: \[ [\mathrm{A}](t) = 4 [\mathrm{B}](t) \]
03

Substitute the formulas from Step 1 into the condition from Step 2

Substitute the expressions for the concentrations of A and B from Step 1 into the condition equation from Step 2: \[ C_0 e^{-k_\mathrm{A}t} = 4 (C_0 e^{-k_\mathrm{B}t}) \]
04

Solve for t

We now have a single equation with only one variable, t, and we can solve for it. First, divide both sides by \(C_0\): \[ e^{-k_\mathrm{A}t} = 4 e^{-k_\mathrm{B}t} \] Now take the natural logarithm of both sides: \[ \ln (e^{-k_\mathrm{A}t}) = \ln (4 e^{-k_\mathrm{B}t}) \] Using the properties of natural logarithms, we can simplify the equation: \[ -k_\mathrm{A}t = \ln(4) - k_\mathrm{B}t \] Now, isolate t on one side of the equation: \[ t(k_\mathrm{B} + k_\mathrm{A}) = \ln(4) \] Finally, divide by the sum of the rate constants to find the value of t: \[ t = \frac{\ln(4)}{k_\mathrm{B} + k_\mathrm{A}} \] Now, plug in the given values for \(k_\mathrm{A}\) and \(k_\mathrm{B}\): \[ t = \frac{\ln(4)}{(4.50 \times 10^{-4}\,\text{s}^{-1}) + (3.70 \times 10^{-3}\,\text{s}^{-1})} \]
05

Calculate t

Perform the calculation to find the value of t: \[ t \approx \frac{\ln(4)}{4.150\times10^{-3}\,\text{s}^{-1}} \approx 406.52\,\text{s} \] Therefore, approximately 406.52 seconds must pass for the concentration of substance A to become equal to 4 times the concentration of substance B.

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Key Concepts

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

Rate Constant
When studying chemical kinetics, the rate constant is a pivotal component, symbolized by the letter 'k' in mathematical expressions. It's a quantifier of how quickly a reaction proceeds under specific conditions, including temperature and, if present, catalysts. First-order kinetics, in particular, are characterized by a reaction rate that is directly proportional to the concentration of one reactant.
For a reaction like the decomposition of substances A and B in our example, the rate constants, denoted by kA and kB respectively, play a fundamental role in determining the time it takes for the concentrations to change. These constants, intrinsic to each substance, allow us to calculate not just the reaction rate at any given moment, but also predict future concentrations using the formula C(t) = C0 e-kt, where C0 is the initial concentration.
Having different rate constants, such as those given for A and B, indicates that they decompose at different rates, with B decomposing faster due to a higher rate constant. This difference allows us to compare and predict how their concentrations relate over time.
Half-Life
The concept of half-life is particularly important when discussing first-order kinetics. It represents the time it takes for the concentration of a reactant to decrease by half. In mathematical terms, the half-life, usually denoted by t1/2, is given for a first-order reaction as t1/2 = ln(2) / k, where 'k' is the rate constant.
This value is constant for first-order reactions and does not depend on the initial concentration. As such, it is a useful measure for comparing how quickly different substances react over time. For instance, if we were given the task of determining the half-life of substances A and B from our exercise based on their rate constants, we'd find substance B has a quicker half-life implying it reacts and thus depletes more rapidly.
Understanding half-life can aid in analyzing various processes from radioactive decay in physics to medication metabolism in pharmacology. It serves also as a tool for making predictions about reaction progress and is integral in fields like geochronology and environmental science for its role in decay and degradation processes.
Reaction Rate
Reaction rate is a descriptor of the speed at which reactants convert into products in a chemical reaction. It is affected by various factors including concentration, temperature, and the presence of catalysts. In the context of first-order kinetics, the reaction rate is directly proportional to the concentration of a single reactant.
To find this rate, we can use the formula Rate = k * [Reactant], where 'k' is the rate constant, and '[Reactant]' denotes the concentration of the reactant at that moment. As seen in the textbook example involving substances A and B, though they both decompose via first-order kinetics, their rates of reaction vary due to different rate constants.
Additionally, the rate at which the concentration of a reactant decreases is critical for calculating how long it takes to reach a certain concentration, such a scenario was illustrated in the original exercise where we determined the time required for the concentration of A to become four times greater than that of B. Understanding reaction rates is vital in areas like the design of chemical reactors and the pharmaceutical industry where reaction speed is crucial for efficiency and safety.

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Most popular questions from this chapter

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

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?

Consider the reaction $$ 3 \mathrm{~A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}]^{2}[\mathrm{~B}][\mathrm{C}] $$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} M\) a. If after \(3.00 \mathrm{~min},[\mathrm{~A}]=3.26 \times 10^{-5} M\), calculate the value of \(k\). b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of \(\mathrm{A}\) after \(10.0 \mathrm{~min}\).

Consider a reaction of the type \(\mathrm{aA} \longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{~A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 . \mathrm{s}\), what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text { Rate }=\frac{-d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]^{3} $$

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

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