Question

The decomposition of cyclopropane, was observed at \(500^{\circ} \mathrm{C}\) and its concentration was monitored as a function of time. The data set is given below. What is the order of the reaction with respect to cyclopropane? Time (hour) \(\quad\) [Cyclopropane], M 0 \(1.00 \times 10^{-2}\) 2 \(1.38 \times 10^{-3}\) 4 \(1.91 \times 10^{-4}\) 6 \(2.63 \times 10^{-5}\) a. First b. Second c. Third d. Zero

Short answer

The reaction is first order with respect to cyclopropane.

Step-by-step solution

Understand the Problem

We need to determine the order of the reaction with respect to cyclopropane by analyzing the concentration data over time. Reaction order gives us insight into how the concentration affects the rate of the reaction.

Define the Rate Laws

For a zero-order reaction, the concentration changes linearly over time: \[[A] = [A]_0 - kt\]For a first-order reaction, the concentration follows a natural logarithm relationship:\[ ext{ln}([A]) = ext{ln}([A]_0) - kt\]For a second-order reaction, the relationship is quadratic:\[\frac{1}{[A]} = \frac{1}{[A]_0} + kt\]

Analyze Zero-Order Assumption

Plot concentration \([A]\) vs. time. In a zero-order reaction, this would result in a straight line. Observing the data, concentration decreases sharply, suggesting it's not zero-order as reaction rates are not constant.

Analyze First-Order Assumption

Plot \( ext{ln}([A])\) vs. time. For a first-order reaction, this should result in a straight line. Calculate \( ext{ln}([A])\) for each data point and plot it against time. Here are the calculations:\[\begin{align*} ext{ln}(1.00 \times 10^{-2}) &= -4.605, \ ext{ln}(1.38 \times 10^{-3}) &= -6.581, \ ext{ln}(1.91 \times 10^{-4}) &= -8.561, \ ext{ln}(2.63 \times 10^{-5}) &= -10.545\end{align*}\]

Analyze Second-Order Assumption

Plot \(\frac{1}{[A]}\) vs. time. For a second-order reaction, this should result in a straight line. Calculate \(\frac{1}{[A]}\) for each data point:\[\begin{align*}\frac{1}{1.00 \times 10^{-2}} &= 100, \\frac{1}{1.38 \times 10^{-3}} &= 724, \\frac{1}{1.91 \times 10^{-4}} &= 5235, \\frac{1}{2.63 \times 10^{-5}} &= 38015\end{align*}\]The plot does not result in a straight line.

Compare the Results

Compare the linearity of plots from Step 4 and Step 5. The linear relationship for the natural log plot (Step 4) confirms the reaction is first order, as it's the only plot resulting in a straight line.

Key concepts

first-order reaction

A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of one reactant. This means that if the concentration is doubled, the rate of reaction also doubles.

In mathematical terms, the rate law for a first-order reaction can be expressed as:

• Rate = k[A]

where:

• - Rate is the change in concentration over time.
• - k is the rate constant unique to the reaction.
• - [A] is the concentration of the reactant.

This relationship shows that as the concentration decreases, the rate at which this decrease happens changes proportionately, leading to an exponential decay. To identify a first-order reaction empirically, a concentration vs. time plot will yield a curve, but a plot of the natural logarithm of concentration vs. time will produce a straight line.

rate laws

Rate laws are equations that describe how the rate of a chemical reaction depends on the concentration of its reactants. Rate laws generally take the form:

• Rate = k[A]^m[B]^n

Where:

• - k is the rate constant.
• - [A] and [B] are the concentrations of the reactants.
• - m and n are the orders of reaction with respect to each reactant.

The overall order of a reaction is the sum of these exponents. For example:

• - If m = 1 and n = 0, the reaction is first-order in A and zero-order in B, making it first-order overall.
• - If m = 2 and n = 1, the overall order is third-order.

Rate laws can help us understand how varying concentrations influence the speed of a reaction, and they are vital for predicting reaction behavior under different conditions.

concentration vs. time plot

When studying reactions, plotting concentration against time is a common method to understand the nature of a reaction. This visual representation helps determine the order of the reaction and provides insight into the kinetic behavior of the chemicals involved.

For different reaction orders:

• - Zero-Order: The concentration vs. time plot is a straight line with a negative slope, indicating a constant rate of reaction independent of concentration.
• - First-Order: The natural logarithm of concentration plotted against time results in a straight line with a negative slope, reflective of an exponential decrease in concentration.
• - Second-Order: The reciprocal of the concentration plotted against time shows a straight line, indicating a quadratic relation.

These plots allow chemists to deduce the reaction order and thereby determine the correct rate law that governs their reaction.

cyclopropane decomposition

Cyclopropane is a small cycloalkane with unique properties due to its carbon atoms forming a three-membered ring. At certain conditions, such as elevated temperatures, cyclopropane undergoes decomposition into other hydrocarbons.

This decomposition is important for understanding reaction mechanisms in organic chemistry. In the context provided, cyclopropane decomposes in a manner that demonstrates a first-order kinetic behavior.

The process is typically studied by monitoring the concentration of cyclopropane over time. By analyzing this data and applying knowledge of rate laws, we can confirm that the decomposition of cyclopropane behaves as a first-order reaction. This means the rate of decomposition directly depends on the concentration of cyclopropane present, leading to an exponential decay pattern in concentration with time. This is valuable for both theoretical insights into reaction kinetics and practical applications where cyclopropane modification is involved, such as in fuel chemistry and materials science.