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?

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

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 \(T_2\): \[ \frac{1.7 \times 10^{-2} \, s^{-1}}{7.2 \times 10^{-4} \, s^{-1}} = e^{Ea/(8.314 \, J \cdot mol^{-1} \cdot K^{-1}) \times (1/660 \,K - 1/720 \,K)} \]

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})\) \] \[ Ea ≈ 51669.77 \, J \cdot mol^{-1} \]

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.

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.