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Chapter 12: Problem 94

For a first order gas phase reaction \(\mathrm{A} \longrightarrow\) products, \(k=\) \(7.2 \times 10^{-4} \mathrm{s}^{-1}\) at \(660 . \mathrm{K}\) and $k=1.7 \times 10^{-2} \mathrm{s}^{-1}\( at \)720 . \mathrm{K} .$ If the initial pressure of \(\mathrm{A}\) is 536 torr at \(295^{\circ} \mathrm{C},\) how long will it take for the pressure of \(\mathrm{A}\) to decrease to 268 torr?

Short Answer

Expert verified
The pressure of A will decrease to 268 torr in approximately \(1.6 \times 10^{4}\) seconds.

Step by step solution

01

Calculate the Activation Energy (Ea) using the Arrhenius equation

We are given two rate constants k₁ and k₂ at temperatures T₁ and T₂ respectively. We can use these values to find the activation energy (Ea) with the Arrhenius equation: \(k = A\exp{\frac{-E_a}{R \cdot T}}\) Let's consider two scenarios for k and T: \(k_1 = A\exp{\frac{-E_a}{R \cdot T_1}}\) \(k_2 = A\exp{\frac{-E_a}{R \cdot T_2}}\) Where, k₁ = \(7.2 \times 10^{-4} \mathrm{s}^{-1}\) T₁ = 660 K k₂= \(1.7 \times 10^{-2} \mathrm{s}^{-1}\) T₂ = 720 K R = 8.314 J/(mol·K) - Universal Gas constant Divide the first equation by the second equation: \(\frac{k_1}{k_2} = \exp{\frac{E_a}{R} \cdot (\frac{1}{T_1} - \frac{1}{T_2})}\) Now we can find the activation energy (Ea).
02

Determine Ea using the given values

Insert the given values into the equation and solve for Ea: \(\frac{7.2 \times 10^{-4}}{1.7 \times 10^{-2}} = \exp{\frac{E_a}{8.314}(\frac{1}{660} - \frac{1}{720})}\) Simplify the equation and solve for Ea: \(E_a = 8.314 \cdot \ln{(\frac{1.7 \times 10^{-2}}{7.2 \times 10^{-4}})} \cdot (\frac{1}{\frac{1}{660} - \frac{1}{720}})\) \(E_a = 79179.49 \ \mathrm{J \cdot mol}^{-1}\)
03

Calculate the rate constant at the given initial temperature

We are given the initial temperature T = 295°C = 568 K. Now we can find the rate constant k at this temperature using the Arrhenius equation and the obtained Ea: \(k = A\exp{\frac{-E_a}{R \cdot T}}\) \(k = 7.2 \times 10^{-4} \mathrm{s}^{-1} \cdot exp{\frac{-79179.49}{8.314 \cdot 568}}\) Upon calculation, we get: \(k = 4.277 \times 10^{-5} \mathrm{s}^{-1}\)
04

Calculate the time required for the pressure of A to decrease to 268 torr

For a first-order reaction, we can use the following equation: \(\ln{(\frac{P_f}{P_i})} = -kt\) Where, Pᵢ = initial pressure (536 torr) Pf = final pressure (268 torr) t = time in seconds Insert the values into the equation and solve for t: \(\ln{(\frac{268}{536})} = -(4.277 \times 10^{-5})(t)\) \(t = \frac{\ln{(\frac{268}{536})}}{-(4.277 \times 10^{-5})}\) Upon calculation, we get: \(t = 1.6 \times 10^{4} \mathrm{s}\) So, it will take approximately 1.6 x 10⁴ seconds for the pressure of A to decrease to 268 torr.

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Most popular questions from this chapter

The rate law of a reaction can only be determined from experiment. Two experimental procedures for determining rate laws were outlined in Chapter 12. What are the two procedures and how are they used to determine the rate laws?

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0,\) and 40.0 \(\mathrm{min}\) for an experiment in which \([\mathrm{A}]_{0}=0.10 M .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. 80.0 \(\mathrm{min}\) b. 30.0 \(\mathrm{min}\)

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{array}{c}{\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}} \\ {\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}\end{array} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

A first-order reaction has rate constants of $4.6 \times 10^{-2} \mathrm{s}^{-1}\( and \)8.1 \times 10^{-2} \mathrm{s}^{-1}\( at \)0^{\circ} \mathrm{C}\( and \)20 .^{\circ} \mathrm{C},$ respectively. What is the value of the activation energy?
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