Problem 10 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Cyclobutane decomposes to form ethane according to: \\[ \mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g}) \\] A quantity of cyclobutane was sealed in a container and exerted a pressure of \(53.30 \mathrm{kPa}\) at \(700 \mathrm{K}\). The pressure changed during the reaction as follows. At the end of the reaction, the pressure was \(106.60 \mathrm{kPa}\) $$\begin{array}{ll} \hline \text { Time/s } & \text { Total pressure/kPa } \\ \hline 0 & 53.30 \\ 2000 & 64.53 \\ 4000 & 73.59 \\ 6000 & 80.53 \\ 8000 & 85.99 \\ 10000 & 90.39 \end{array}$$ Assuming the gases behave ideally, show graphically that the reaction is first order with respect to cyclobutane. Calculate the rate constant and the half life for the reaction. (Section \(9.4)\)

Step-by-Step Solution

Answer

Plot ln(pressure) vs. time to confirm first-order kinetics, calculate k from the slope, then use k to find the half-life.

Step 1: Understand the Reaction and Data

The reaction involves the decomposition of cyclobutane \(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g})\) to form ethylene \(\mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g})\). Initially, the pressure of cyclobutane is 53.30 kPa. Over time, this pressure changes as shown in the table, and at the end, the pressure is 106.60 kPa. We need to determine if this reaction is first order with respect to cyclobutane by analyzing the change in total pressure over time.

Step 2: Calculate Initial and Final Pressures for Species

The initial pressure represents only cyclobutane, so initially \(P_{\text{C}_4\text{H}_8} = 53.30\, \text{kPa}\). At the end, the reaction pressure is entirely from ethylene with \(P_{\text{total}} = 106.60\, \text{kPa}\). Since one mole of cyclobutane produces two moles of ethylene, the pressure of ethylene at the end should be double the decrease in cyclobutane. Therefore, the cyclobutane's final pressure is zero, and all pressure increase is due to ethylene formation.

Step 3: Determine Change in Pressures

Since the pressure increased by 53.30 kPa from the decomposition of cyclobutane, this matches the doubling expected from two moles of ethylene for each cyclobutane decomposed. Thus, for each unit decrease in cyclobutane, a total of 2 times pressure increase occurs due to formation of ethylene.

Step 4: Calculate Partial Pressures of Cyclobutane

To determine if the reaction is first order, calculate the partial pressures of \(\mathrm{C}_4\mathrm{H}_8\) based on the pressure change due to its decomposition. Use the reaction stoichiometry where pressure change indicates ethylene formation: each 26.65 kPa pressure increase represents complete reaction towards final ethylene accumulation.

Step 5: Determine if First Order Kinetics

For a first-order reaction, the natural log of the cyclobutane pressure (\(\ln P_{\text{C}_4\text{H}_8}\)) versus time should be linear. Convert the skill-appropriate pressure values to the natural log format and plot these against time to check for linearity.

Step 6: Calculate Rate Constant (k)

If the plot in Step 5 is linear, its slope is used to calculate the rate constant \(k\). Use the formula: \[ \text{slope} = -k \] which can be derived from the line equation of first-order kinetics: \(\ln P = -kt + C\). Find the slope and hence \(k\).

Step 7: Find Half-Life (t₁/₂)

For a first-order reaction, the half-life (\(t_{1/2}\)) is independent of the initial concentration and given by \[ t_{1/2} = \frac{0.693}{k} \]. Use the \(k\) found in Step 6 to calculate the half-life of cyclobutane decomposition.

Key Concepts

Cyclobutane Decomposition

Cyclobutane is a fascinating chemical compound, known for its ability to break down into smaller molecules. In the decomposition of cyclobutane (\(\mathrm{C}_4\mathrm{H}_8\)), it transforms into ethylene (\(\mathrm{C}_2\mathrm{H}_4\)). This reaction is important because it is a part of understanding organic chemistry and how larger hydrocarbons can break down into simpler substances. During this reaction, one molecule of cyclobutane decomposes to yield two molecules of ethylene. This stoichiometry is crucial when analyzing the pressure changes in a closed reaction system. Understanding these stoichiometric relationships helps in predicting the behavior of gases involved in chemical reactions.

Reaction Kinetics

Reaction kinetics is the study of rates at which chemical processes occur. Knowing the order of a reaction, like first order in the case of cyclobutane decomposition, allows us to predict how quickly a reaction proceeds based on the concentration of its reactants. For a first-order reaction, the rate is directly proportional to the concentration of one reactant. This means if the concentration of cyclobutane decreases over time, the reaction rate is proportional to this change. Analyzing the decomposition of cyclobutane focuses on how the concentration (or pressure, in this case) of the reactants changes over time. By observing these changes, we can categorize the reaction as being of a specific order. This understanding is vital because it lays the groundwork for controlling and predicting chemical processes in various applications, from industrial synthesis to biological systems.

Ideally Behaving Gases

The concept of ideally behaving gases simplifies our calculations and predictions regarding gas reactions. An ideal gas is one that follows the ideal gas laws, which means its molecules do not interact with each other, except for elastic collisions, and occupy a negligible volume. In the cyclobutane decomposition exercise, assuming ideal gas behavior allows us to use pressure changes as a direct representation of concentration changes. This assumption simplifies the math, as pressure is directly proportional to the number of moles of gas present, according to the ideal gas law: \[ PV = nRT \] This equation relates pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas (\(n\)), with the constant (\(R\)) known as the universal gas constant. By assuming the gases involved in the reaction behave ideally, we can apply these relationships directly, allowing for straightforward analysis of changes in the system.

Rate Constant Calculation

The rate constant, denoted by \( k \), is a critical component in the rate law for a reaction. For first-order reactions like the decomposition of cyclobutane, the rate law can be expressed as \[ ext{Rate} = k [ ext{Cyclobutane}] \] where \([\text{Cyclobutane}]\) is the concentration of cyclobutane. Calculating the rate constant involves plotting the natural logarithm of the concentration (or pressure, for gases) of cyclobutane against time. If this plot is a straight line, the slope of that line is \(-k\). Once the rate constant is known, it provides valuable insight into how fast the reaction occurs under set conditions.Additionally, for first-order reactions, the half-life (\( t_{1/2} \)) can be determined using the formula: \[ t_{1/2} = \frac{0.693}{k} \]. This half-life is the time required for half of the cyclobutane to decompose, which remains constant regardless of the initial concentration. Understanding and calculating these parameters is essential for predicting and controlling the outcomes of chemical reactions.