Question

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 $$ \begin{aligned} &k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{~s}^{-1} \\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{~s}^{-1} \end{aligned} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{~B}] ?\)

Short answer

The time required to reach the condition such that the concentration of A is four times the concentration of B is approximately 256.89 seconds.

Step-by-step solution

Write down the given information in terms of concentration and decomposition rates

We are given the rate constants for both reactions: - For substance A: \(k_A = 4.50 \times 10^{-4} \space s^{-1}\) - For substance B: \(k_B = 3.70 \times 10^{-3} \space s^{-1}\) We are also given that the initial concentrations of both substances are equal: \([A]_0 = [B]_0\). Finally, we are given the condition to be reached: \([A] = 4[B]\).

Use the first-order reaction formula for both substances A and B

Using the first-order reaction formula for both substances A and B, we have: - For substance A: \([A] = [A]_0 e^{-k_At}\) - For substance B: \([B] = [B]_0 e^{-k_Bt}\)

Set up an equation for the concentration condition

We need to find the time when the given condition is satisfied: \([A] = 4[B]\). We can use the formulas from step 2 to create an equation involving the concentration of A and B: \([A]_0 e^{-k_At} = 4[B]_0 e^{-k_Bt}\) Since the initial concentrations are equal (\([A]_0 = [B]_0\)), we can simplify the equation: \(e^{-k_At} = 4 e^{-k_Bt}\)

Solve for time t

Now, we want to find the time t when the condition is satisfied. We can use the given rate constants to solve for t: Divide both sides by \(e^{-k_Bt}\): \(e^{(k_B - k_A)t} = 4\) Take the natural logarithm (ln) of both sides: \((k_B - k_A)t = \ln{4}\) Solve for t: \(t = \frac{\ln{4}}{k_B - k_A} = \frac{\ln{4}}{3.70 \times 10^{-3} - 4.50 \times 10^{-4}}\) Now, use a calculator to find t: \(t \approx 256.89 \space s\) So the time required to reach the condition such that the concentration of A is four times the concentration of B is approximately 256.89 seconds.

Key concepts

Rate Constants

Rate constants are fundamental components in the study of chemical kinetics, which is the branch of chemistry that deals with how fast reactions occur. In a first-order reaction, the rate constant determines the speed at which a reactant is consumed or a product is formed. Specifically, for first-order reactions, the rate of the reaction is directly proportional to the concentration of a single reactant. Mathematically, this relationship can be expressed as rate = \( k[Reactant] \), where \( k \) is the rate constant, and [Reactant] denotes the concentration of the reactant.

Understanding the magnitude of the rate constant is crucial. A larger rate constant indicates a faster reaction, with units that ensure the reaction rate is expressed in proper concentration per time, typically M/s in molar concentrations. In the context of the exercise, \( k_A \) and \( k_B \) represent the rate constants for substances A and B, respectively. When comparing these values, we see that substance B has a larger rate constant and, hence, a faster reaction rate under identical conditions.

Chemical Kinetics

Chemical kinetics is the study of the rates of chemical processes. It involves understanding factors that affect these rates and the mechanisms by which reactions occur. Factors include the nature of the reactants, concentrations, temperature, and presence of a catalyst.

Within kinetic analysis, various reaction orders define the relationship between the rate of a reaction and the concentration of reactants. In a first-order reaction, the rate is directly proportional to the concentration of a single reactant. The time-dependent concentration change of a reactant is typically described by the equation \( [Reactant] = [Reactant]_0 e^{-kt} \), where \( [Reactant]_0 \) is the initial concentration, \( k \) is the rate constant, and \( t \) is time. This exponential decay pattern is indicative of first-order kinetics, as seen in the exercise's solutions for substances A and B.

Reaction Rates

The rate of a chemical reaction quantifies how the concentration of a reactant or product changes over time. In essence, it’s the speed at which a reaction progresses. For first-order reactions, the rate is given by the simple formula: rate = \( -d[Reactant]/dt = k[Reactant] \), where \( d[Reactant]/dt \) is the change in concentration of the reactant over time, and \( k \) is the rate constant.

In the textbook solution, the calculation of time \( t \) when the concentration of reactant A becomes four times that of reactant B relies on the understanding of reaction rates. By equating the rates of decay for both reactants and applying the condition \( [A] = 4[B] \), we are able to algebraically solve for \( t \). The equation demonstrates that the reaction rate of B is faster than that of A due to its larger rate constant (\( k_B \)), leading to the need for a concrete time value that satisfies the concentration condition, which is derived using logarithms.