Problem 116 - Chemistry The Molecular Science

Question

When heated, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\) \(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) The activation energy, \(E_{a},\) for this reaction is \(262 \mathrm{~kJ} / \mathrm{mol}\). (a) If the rate constant \(k=0.032 \mathrm{~s}^{-1}\) at \(800 . \mathrm{K},\) calculate the value of \(k\) at \(900 . \mathrm{K}\). (b) Calculate the cyclobutane concentration after \(2 \mathrm{~h}\) at \(850 . \mathrm{K}\) if the initial cyclobutane concentration was \(0.0427 \mathrm{M}\)

Step-by-Step Solution

Answer

(a) \(k_{900} \approx 10.98 \, s^{-1}\). (b) Cyclobutane will decompose completely after 2 hours at 850 K.

Step 1: Use Arrhenius Equation to Find k at 900 K

To find the rate constant \(k\) at 900 K, we use the Arrhenius equation: \[ k = A e^{-E_a / (RT)} \] However, since we have \(k\) at a different temperature already, we use the modified form:\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Given \(k_1 = 0.032 \text{s}^{-1}\) at \(T_1 = 800 \, K\), \(E_a = 262 \text{kJ/mol} = 262000 \, J/mol\), and using \(R = 8.314 \, J/mol\cdot K\), we need to find \(k_2\) at \(T_2 = 900 \, K\). Plug the values into the equation to solve.

Step 2: Solving for k at 900 K

First, calculate the terms:\[ \frac{1}{T_2} = \frac{1}{900} \, K^{-1}, \quad \frac{1}{T_1} = \frac{1}{800} \, K^{-1} \]Find the difference:\[ \frac{1}{900} - \frac{1}{800} = -\frac{1}{720000} \, K^{-1} \]Calculate the logarithm term:\[ \ln \left( \frac{k_2}{0.032} \right) = \frac{-262000}{8.314} \cdot \left( -\frac{1}{720000} \right) \]Solve for:\[ k_2 = 0.032 \times e^{12.33} \approx 10.98 \, s^{-1} \]

Step 3: Calculate Cyclobutane Concentration After 2 Hours

We assume first-order kinetics, so the integrated rate equation is:\[ \ln \left( \frac{[C_{4}H_{8}]_t}{[C_{4}H_{8}]_0} \right) = -kt \]Given \([C_{4}H_{8}]_0 = 0.0427 \, M\), \(t = 2 \times 3600 \, s = 7200 \, s\) at \(T = 850 \, K\), first, use the Arrhenius equation to find \(k\) at 850 K:\[ \ln \left( \frac{k_{850}}{0.032} \right) = \frac{-262000}{8.314} \left( \frac{1}{850} - \frac{1}{800} \right) \]Solve to find \(k_{850} \approx 0.221 \, s^{-1}\). Substitute this into the rate equation to find \([C_{4}H_{8}]_t\).

Step 4: Substitution and Calculation for Concentration

Calculate:\[ k_{850} \approx 0.221 \, s^{-1}\] \[ \ln \left( \frac{[C_{4}H_{8}]_t}{0.0427} \right) = -(0.221)(7200) \]\[ \frac{[C_{4}H_{8}]_t}{0.0427} \approx e^{-1591.2} \approx 0 \]Thus, \([C_{4}H_{8}]_t\) is practically 0 M after 2 hours.

Key Concepts

Activation Energy

Activation energy is the minimum energy required for a chemical reaction to occur. Think of it as a barrier the reactants need to overcome to transform into products. In our cyclobutane decomposition, the activation energy is given as 262 kJ/mol. This means each mole of cyclobutane requires this much energy to break apart into ethylene.

• High activation energy means the reaction is slow, as few molecules have enough energy to react.
• Low activation energy indicates a faster reaction, as more molecules can overcome the barrier.

Activation energy influences how temperature affects reaction rates. When temperature increases, more molecules gain sufficient energy, leading to a higher rate of reaction.

Rate Constant

The rate constant (k) is a crucial factor in determining how fast a reaction proceeds. It varies with temperature and is specific to a given reaction at a determined temperature, as expressed by the Arrhenius equation: \[ k = A e^{-E_a / (RT)} \] Here, A is the frequency factor, a constant.The rate constant is a measure of the speed of a reaction. For example, if at 800 K, \( k = 0.032 \, \text{s}^{-1} \), it signifies the speed at which cyclobutane decomposes at this temperature. To find \( k \) at a different temperature, like 900 K, a modified Arrhenius equation helps: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] This relationship helps calculate \( k_2 \), demonstrating the temperature's effect on reaction speed.

First-Order Kinetics

First-order kinetics refers to reactions where the rate is directly proportional to the concentration of one reactant. In our case, cyclobutane decomposes into ethylene following first-order kinetics. This means the rate of decomposition depends linearly on the concentration of cyclobutane. The rate law can be expressed as: \[ \text{Rate} = k[C_{4}H_{8}] \] For first-order reactions, the concentration over time follows the integrated rate equation: \[ \ln \left( \frac{[C_{4}H_{8}]_t}{[C_{4}H_{8}]_0} \right) = -kt \] Using this equation, we calculate how much cyclobutane remains after a certain time at a specific temperature. With initial concentration \( [C_{4}H_{8}]_0 = 0.0427 \, \text{M} \) and \( k \) known, solving it helps find the concentration left after 2 hours, showing the decay pattern typical of first-order kinetics.