Question

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}(g) $$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{s}^{-1} ;\) at \(720 . \mathrm{K}, k=1.7 \times 10^{-2} \mathrm{s}^{-1}\) What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C},\) what is the pressure of iodoethane after three half-lives?

Short answer

The rate constant for the first-order decomposition of iodoethane at 325°C is approximately \( 2.13 \times 10^{-3} \, s^{-1} \). After three half-lives, the pressure of iodoethane is approximately \( 111.75 \, torr \).

Step-by-step solution

Convert the desired temperature to Kelvin.

To work with the Arrhenius equation, we need the temperature in Kelvin. Convert the desired temperature of 325°C to Kelvin: \[ T_{3} = 325 + 273.15 = 598.15 \, K \]

Apply the Arrhenius equation to find the activation energy (Ea).

The Arrhenius equation is: \[ k = A \times e^{-Ea/RT} \] where k is the rate constant, A is the pre-exponential factor (frequency factor), Ea is the activation energy, R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin. We'll first solve for Ea using the given rate constants and temperatures. Let's set up two equations: \[ k_{1} = A \times e^{-Ea/R \times T_{1}} \] \[ k_{2} = A \times e^{-Ea/R \times T_{2}} \] Divide the second equation by the first equation to eliminate A: \[ \frac{k_{2}}{k_{1}} = e^{Ea/R \times (1/T_{1} - 1/T_{2})} \] Now, let's plug in the values for \(k_1\), \(k_2\), \(T_1\), and \(T_2\): \[ \frac{1.7 \times 10^{-2} \, s^{-1}}{7.2 \times 10^{-4} \, s^{-1}} = e^{Ea/(8.314 \, J \cdot mol^{-1} \cdot K^{-1}) \times (1/660 \,K - 1/720 \,K)} \]

Solve for the activation energy (Ea)

We can solve for Ea in the above equation: Calculate the fraction within the exponent: \[ (1/660 \, K - 1/720 \, K) = -1.97 \times 10^{-5} \, K^{-1} \] Now, find the natural logarithm of the rate constants fraction: \[ ln(23.61) ≈ 3.1642 \] So, \[ 3.1642 = -Ea/(8.314 \, J \cdot mol^{-1} \cdot K^{-1} \times -1.97 \times 10^{-5} \, K^{-1})\) \] \[ Ea ≈ 51669.77 \, J \cdot mol^{-1} \]

Find the rate constant at 325°C (598.15 K)

Now that we know the activation energy (Ea), we can find the rate constant (k) at 598.15 K: Plug in the values for \(k_1\), \(T_1\), and Ea into the Arrhenius equation: \[ k = 7.2 \times 10^{-4} \, s^{-1} \times e^{(51669.77 \, J \cdot mol^{-1})/(8.314 \, J \cdot mol^{-1} \cdot K^{-1}) \times (1/598.15 \, K - 1/660 \, K)} \] \[ k ≈ 2.13 \times 10^{-3} \, s^{-1} \] So, the rate constant for this first-order decomposition at 325°C is approximately 2.13 x 10^-3 s^-1.

Calculate the final pressure of iodoethane after three half-lives.

Convert initial temperature to Kelvin: \[ T_{1'} = 245 + 273.15 = 518.15 \, K \] We'll use the first-order decay formula for pressure: \[ P_{final} = P_{initial} \times (1/2)^{n} \] where n is the number of half-lives. In our case, n = 3, and the initial pressure is 894 torr. So, \[ P_{final} = 894 \, torr \times (1/2)^3 = 894 \, torr \times 1/8 \] \[ P_{final} ≈ 111.75 \, torr \] The final pressure of iodoethane after three half-lives is approximately 111.75 torr.

Key concepts

First-order Reactions

In chemistry, a first-order reaction is one where the rate of the reaction is directly proportional to the concentration of one of the reactants. This means that the reaction speed depends on how much of that particular substance is present. For our decomposition of iodoethane, it follows the equation \( C_2H_5I(g) \longrightarrow C_2H_4(g) + H(g) \).
In mathematical terms, the rate law for a first-order reaction can generally be expressed as:

• The rate \( = k[A] \),
• Where \( k \) is the rate constant and \([A]\) is the concentration of reactant \( A \).

When looking at the decay of chemicals over time, one common calculation for first-order reactions is the half-life, the time it takes for half of the substance to decompose. First-order reactions have the unique property that their half-lives are constant, no matter how much of the reactant you start with.

Rate Constants

The rate constant \( k \) is a crucial parameter in the study of chemical kinetics. It provides insight into the reaction speed and is affected by different conditions such as temperature and the presence of a catalyst.
To illustrate this, consider the decomposition of iodoethane, where the temperature changes from 660K to 720K, causing \( k \) to increase from \( 7.2 \times 10^{-4} \, s^{-1} \) to \( 1.7 \times 10^{-2} \, s^{-1} \). This demonstrates how temperature can significantly alter the rate at which a reaction proceeds.

• At higher temperatures, molecules move faster and collide more often with more energy, increasing \( k \) and speeding up reactions.
• The value of \( k \) not only tells us about the spontaneity of the reaction but also allows calculations needed to predict reactant and product concentrations over time.

Arrhenius Equation

The Arrhenius equation is a formula that expresses how the rate constant \( k \) is related to the temperature and the activation energy \( Ea \). It's given by the formula:\[k = A \times e^{-Ea/RT}\]

• Where \( A \) is the pre-exponential factor, \( e \) is a constant (Euler's number), \( Ea \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
• The equation shows that as temperature increases, \( k \) increases, indicating faster reactions.

This equation is essential for predicting how a reaction will behave under different conditions. By comparing the rate constants at different temperatures, we estimated the activation energy for the reaction—critical for understanding the energy barrier that the reactants must overcome to transform into products.

Activation Energy

Activation energy \( Ea \) is the minimum amount of energy required for a chemical reaction to occur. Think of it as the barrier that reactants must overcome to transform into products.
In our discussion, we determined the activation energy for the decomposition of iodoethane using the Arrhenius equation:

• This energy is not easily visible in the reaction but plays a crucial role in controlling how slowly or quickly the reaction proceeds.
• A high \( Ea \) suggests that the reaction is slow, as fewer molecules have sufficient energy to overcome this barrier.
• Conversely, a low \( Ea \) indicates a faster reaction.

To find \( Ea \), we used the change in rate constants at two different temperatures, helping us visualize how temperature and energy needs align. Understanding \( Ea \) is vital for manipulating reaction speeds in industrial and laboratory settings.