Problem 53 - Chemistry: Principles and Reactions

Question

The following gas-phase reaction is second-order. $$2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)$$ Its half-life is \(1.51\) min when \(\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\) is \(0.250 \mathrm{M}\) (a) What is \(k\) for the reaction? (b) How long will it take to go from \(0.187 M\) to \(0.0915 \mathrm{M}\) ? (c) What is the rate of the reaction when \(\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\) is \(0.335 \mathrm{M} ?\)

Step-by-Step Solution

Answer

Answer: The rate of the reaction when the concentration of C2H4 is 0.335 M is approximately 0.297 M·min⁻¹.

Step 1: (a) Find the rate constant (k)

To find the rate constant (k) for the reaction, we first need to use the half-life equation for a second-order reaction: $$t_{1/2} = \frac{1}{k[A]_0}$$ Given \(t_{1/2} = 1.51\ \text{min}\) and \([A]_0 = 0.250\ \text{M}\), we can solve the equation for k: $$k = \frac{1}{t_{1/2}[A]_0} = \frac{1}{1.51\times0.250}$$ $$k \approx 2.64\ \text{M}^{-1}\text{min}^{-1}$$

Step 2: (b) Time for concentration change

To find the time it takes to go from \(0.187\ \text{M}\) to \(0.0915\ \text{M}\), we will use the integrated rate law for second-order reactions: $$\frac{1}{[A]} - \frac{1}{[A]_0} = kt$$ Given the initial concentration \([A]_0 = 0.187\ \text{M}\) and the final concentration \([A] = 0.0915\ \text{M}\), we can solve for the time t: $$\frac{1}{0.0915} - \frac{1}{0.187} = 2.64t$$ $$t \approx 2.15\ \text{min}$$

Step 3: (c) Rate of the reaction

To find the rate of the reaction when \([\mathrm{C}_{2} \mathrm{H}_{4}] = 0.335\ \text{M}\), we will use the rate law for second-order reactions: $$\text{rate} = k[\mathrm{C}_{2} \mathrm{H}_{4}]^2$$ Plugging in the rate constant \(k=2.64\ \text{M}^{-1}\text{min}^{-1}\) and concentration \([\mathrm{C}_{2} \mathrm{H}_{4}] = 0.335\ \text{M}\): $$\text{rate} = 2.64\times(0.335)^2 \approx 0.297\ \text{M}\text{min}^{-1}$$ In summary: a) The rate constant (k) for the reaction is approximately \(2.64\ \text{M}^{-1}\text{min}^{-1}\). b) It takes approximately \(2.15\ \text{min}\) to go from \(0.187\ \text{M}\) to \(0.0915\ \text{M}\). c) The rate of the reaction when \([\mathrm{C}_{2} \mathrm{H}_{4}] = 0.335\ \text{M}\) is approximately \(0.297\ \text{M}\text{min}^{-1}\).

Key Concepts

Understanding the Rate Constant (k) in Second-Order Reactions

The rate constant, symbolized as k, is a crucial factor in the realm of chemical kinetics, acting as the bridge between reaction rate and reactant concentrations. In a second-order reaction, the relationship between these elements is quadratic, implying that a change in the concentration of the reactant affects the rate of reaction exponentially.

To calculate k, we use the half-life formula specific to second-order processes:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
where \( t_{1/2} \) represents the half-life, k is the rate constant, and \( [A]_0 \) is the initial concentration of the reactant. This equation reiterates that half-life in second-order reactions depends on the initial concentration, distinguishing it from first-order reactions where half-life is constant. Understanding and determining k allows students to predict how long a reaction will take to reach certain points of completion, a critical aspect when applying chemistry in real-world scenarios, such as pharmaceutical drug design, where reaction time predictability can be vital.

Integrated Rate Law for Second-Order Reactions

The integrated rate law for second-order reactions is another vital tool for chemists, providing a means to relate reactant concentrations over time. The law is expressed as:
\[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]
Here, \( [A] \) denotes the reactant concentration at time t, and \( [A]_0 \) is the initial concentration. This equation allows us to calculate the time it will take for a reactant's concentration to change from one value to another. It's important to note that unlike zero and first-order reactions where a plot of reactant concentration versus time yields a straight line, for second-order reactions, a plot of the inverse of concentration versus time will be linear. This property is particularly useful in experimental chemistry, where graphing the data can help determine the order of the reaction and the rate constant.

Reaction Half-Life in Second-Order Reactions

Half-life, symbolized as \( t_{1/2} \), is the period required for the concentration of a reactant to lessen to half of its initial value. In the context of second-order reactions, the half-life concept helps students and researchers understand the rate at which reactants are consumed over time.

One interesting aspect of second-order half-life is its direct dependence on the initial concentration of reactants, articulated through:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
As seen from the equation, the half-life increases as the initial concentration decreases. This inverse relationship is a distinctive feature that can be confirmed through experimental data, potentially allowing chemists to validate the reaction order. Being able to calculate and comprehend half-life is especially useful in industries such as pharmaceuticals or environmental chemistry, where the decay of substances over time is key to safety and effectiveness.