Question

Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), forms \(\cdot \mathrm{CH}_{3}\) radicals at \(700 .^{\circ} \mathrm{C}\) in a firstorder reaction, for which \(k=1.98 \mathrm{~h}^{-1}\). (a) What is the half-life for the reaction? (b) Calculate the time needed for the amount of ethane to fall from \(1.15 \times 10^{-3} \mathrm{~mol}\) to \(2.35 \times 10^{-4} \mathrm{~mol}\) in a 500.-mL reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\). (c) How much of a 6.88-mg sample of ethane in a \(500 .-\mathrm{mL}\) reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\) will remain after \(45 \mathrm{~min}\) ?

Short answer

a) Half-life is 0.350 h. b) Time needed is 1.386 h. c) After 45 min, 1.93 mg of ethane will remain.

Step-by-step solution

Determine the half-life for a first-order reaction

The half-life (\( t_{1/2} \) of a first-order reaction is determined using the equation \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Plug in the given rate constant \( k = 1.98 \, \text{h}^{-1} \) into the equation to calculate the half-life.

Calculate the time for concentration decrease

For a first-order reaction, the time \( t \) it takes for the concentration of a reactant to decrease from \( [A]_0 \) to \( [A] \) is given by the equation \( \ln \left( \frac{[A]}{[A]_0} \right) = -kt \). Solve for \( t \) using the initial concentration \( [A]_0 = 1.15 \times 10^{-3} \, \text{mol} \) and final concentration \( [A] = 2.35 \times 10^{-4} \, \text{mol} \), along with the rate constant \( k = 1.98 \, \text{h}^{-1} \).

Determine the remaining ethane after a given time

We can use the first-order kinetics equation \( [A] = [A]_0 e^{-kt} \) to find the remaining concentration of ethane, \( [A] \), after 45 min. First, convert the mass of ethane to moles, then use the rate constant \( k \) and time \( t = 45 \, \text{min} = 0.75 \, \text{h} \) to find \( [A] \).

Key concepts

Half-Life Calculation

Understanding the half-life of a reaction is crucial for mastering chemical kinetics, especially for reactions that are described as first-order. The half-life, denoted as \( t_{1/2} \), represents the time required for half of the reactant to be converted into products. In the context of a first-order reaction, the half-life is independent of the initial concentration of the reactant, making it a particularly convenient measure.

For a first-order reaction, the half-life is calculated using the formula: \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. This elegant relationship shows that for any given first-order reaction, knowing the rate constant allows you to determine how quickly the concentration will decrease by half. This is especially helpful in predicting the behavior of reactions over time.

Concentration Decrease Over Time

If you're wondering how the concentration of a reactant in a first-order reaction changes as time progresses, the answer lies in a straightforward logarithmic relationship. Specifically, the time \( t \) it takes for the concentration to decrease from an initial value \( [A]_0 \) to a later value \( [A] \) can be described by the equation: \( \ln \left( \frac{[A]}{[A]_0} \right) = -kt \).

This expression is incredibly useful for calculating not just how much time has passed, but also for predicting future concentrations at a given time. By rearranging the equation to solve for either \( t \) or \( [A] \), we can track the decrease of reactant concentration over time and make accurate predictions about the reaction's progress, which is a central aspect of chemical kinetics.

Rate Constant

Key to mastering the concept of chemical kinetics, the rate constant, symbolized as \( k \), is a proportionality factor that connects the reaction rate to the reactant concentrations. It is specific to each reaction and conditions such as temperature. The rate constant determines the speed at which a chemical reaction proceeds.

For first-order reactions, the rate constant has units of \( \text{time}^{-1} \), and its value is crucial for calculating both the half-life and the time-dependent concentration of reactants. The higher the value of \( k \), the faster the reaction occurs, which is why knowing \( k \) provides valuable insights into the reaction's kinetics. Changes in temperature or the use of a catalyst can significantly alter the rate constant and thereby the reaction rate.

Chemical Kinetics Equations

The equations of chemical kinetics serve as mathematical models describing the rate at which reactants are transformed into products. In the realm of first-order reactions, the rate of the reaction is directly proportional to the concentration of the single reactant.

One of the fundamental equations for first-order reactions is: \( [A] = [A]_0 e^{-kt} \), where \( [A] \) is the concentration at time \( t \), \( [A]_0 \) is the initial concentration, and \( e \) is the base of the natural logarithm. This equation is invaluable for determining the concentration of a reactant at any point in time, as it incorporates the exponential decay pattern typical of first-order kinetics.

Understanding these equations not only allows for prediction of reaction progress but also equips students to interpret and manipulate kinetic data accurately in practical scenarios.