Question

Ethyl chloride vapor decomposes by the first-order reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{HCl} $$ The activation energy is \(249 \mathrm{~kJ} / \mathrm{mol}\), and the frequency factor is \(1.6 \times 10^{14} \mathrm{~s}^{-1} .\) Find the value of the rate constant at \(710 \mathrm{~K}\) What fraction of the ethyl chloride decomposes in 15 minutes at this temperature? Find the temperature at which the rate of the reaction would be twice as fast.

Short answer

Rate constant k at 710 K is calculated using the given activation energy and frequency factor. Then, using first-order kinetics, the fraction that decomposes in 15 minutes is found. Finally, the temperature for a doubled rate is calculated by solving the Arrhenius equation with the new rate constant that is twice the original.

Step-by-step solution

Use the Arrhenius Equation to Find the Rate Constant

The Arrhenius equation relates the rate constant k to the activation energy E_a, the gas constant R, and the temperature T. It is given by: \[ k = Ae^{(-E_a / (RT))} \] where: \( A \) is the frequency factor, \( E_a \) is the activation energy, R is the gas constant (\(8.3145 \mathrm{J/(mol\cdot K)}\) or \(0.0083145 \mathrm{kJ/(mol\cdot K)}\) for energy in kJ/mol), and T is the temperature in Kelvin. Plug in the given values and solve for k.

Calculate the Rate Constant k

Substitute the given values into the Arrhenius equation: \[ k = (1.6 \times 10^{14}) e^{-(249000 \mathrm{~J/mol}) / (8.3145 \mathrm{~J/(mol\cdot K)} \times 710 \mathrm{~K})} \] Solve for k using the exponential function.

Use First-Order Kinetics for Decomposition

For a first-order reaction, the fraction of reactant decomposing over time is given by: \[ \frac{[A]}{[A]_0} = e^{-kt} \] where \([A]\) is the concentration of reactant at time t, \([A]_0\) is the initial concentration of reactant, and k is the rate constant. Use the rate constant from Step 2 and solve for the fraction decomposed after 15 minutes.

Convert Time from Minutes to Seconds

The rate constant k is given in units of s^{-1}, so the time needs to be in seconds. Convert 15 minutes to seconds: \[ 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \]

Calculate the Decomposition Fraction

Substitute the time in seconds and the rate constant k from Step 2 into the first-order kinetics equation to find the fraction decomposed.

Find the Temperature for Doubled Rate Constant

To find the temperature at which the rate constant is doubled, set up an equation where the new rate constant \( k' \) is twice the original \( k \): \[ k' = 2k \] Use the Arrhenius equation to relate the new temperature \( T' \) with the doubled rate constant. Solve for \( T' \).

Set the New Arrhenius Equation with Doubled Rate

The Arrhenius equation for the new rate constant is: \[ 2k = Ae^{(-E_a / (RT'))} \] Isolate \( T' \) and solve the equation. You will compare the two Arrhenius equations (original and doubled rate) and solve for the new temperature \( T' \).

Solve for the New Temperature T'

Since we already have the value of k from the previous steps, we can use the two Arrhenius equations and solve for \( T' \). You will get a mathematical expression involving both temperatures. Solve the equation to find the new temperature \( T' \) that makes the reaction rate twice as fast.

Key concepts

Understanding First-Order Reaction Kinetics

In the realm of chemical reactions, first-order reaction kinetics is a fundamental concept that describes how the rate of reaction is directly proportional to the concentration of the reactant. This implies that if you were to double the concentration of the reactant, the reaction rate would also double.

For a first-order reaction, the rate law can be expressed as \( rate = k[A] \) where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. One of the hallmark features of first-order kinetics is its unique half-life, which is constant regardless of the initial concentration of the reactant.

In our case, the decomposition of ethyl chloride adheres to first-order kinetics. This means that the natural logarithm of the concentration of ethyl chloride decreases linearly over time. To quantify the progress of the reaction, you can calculate the fraction of reactant decomposed at any given time with the equation \( \frac{[A]}{[A]_0} = e^{-kt} \) where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( e \) is the base of the natural logarithm.

Rate Constant Calculation

The rate constant, \( k \) is a crucial part of the kinetic equation, influencing how rapidly a reaction progresses. In first-order reactions, the rate constant has units of \( s^{-1} \) and can be calculated using the Arrhenius equation: \[ k = Ae^{(-E_a / (RT))} \] Here, \( A \) is the frequency factor, or pre-exponential factor, which represents the frequency of collisions leading to a reaction. \( E_a \) is the activation energy necessary for the reaction to occur, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. If we know \( A \) and \( E_a \) as well as the temperature, we can determine \( k \) and then predict how fast the reaction will occur.

At \( 710 \ K \) for ethyl chloride, for instance, inserting the given activation energy and frequency factor into the Arrhenius equation allows us to solve for the rate constant. This is the key to understanding at what speed the decomposition of ethyl chloride will proceed.

Activation Energy

Activation energy \( (E_a) \) is a term that pops up routinely when discussing chemical reactions. It refers to the minimum amount of energy required for reactants to transform into products during a chemical reaction. You can think of it as the 'entry fee' that must be paid for a reaction to occur.

The higher the activation energy, the less likely a reaction will occur at a given temperature because fewer reactant molecules will have the necessary energy to get over the energy barrier. That's why temperature plays such a vital role: increasing the temperature often means a higher reaction rate, as molecules move faster and more of them muster enough energy to meet \( E_a \).

In our exercise involving ethyl chloride, the activation energy is relatively high at \( 249 \, kJ/mol \) suggesting that not all collisions between reactant molecules will result in a reaction, except at sufficiently high temperatures. By using the Arrhenius equation, we can see how this activation energy influences the rate constant and, accordingly, the pace at which ethyl chloride decomposes.