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Chapter 12: Problem 118

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at 273 \(\mathrm{K}\) is $2.3 \times 10^{-12} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\( , and the activation energy is 111 \)\mathrm{kJ} / \mathrm{mol}$ . How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

Short Answer

Expert verified
To find the time it takes for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at 500K, we can follow these steps: 1. Use the Arrhenius equation to find the rate constant, \(k_{500}\), at 500K. 2. Calculate the reaction rate using the rate law and given reaction order, and convert the given partial pressures to concentrations using the ideal gas law. 3. Calculate the average concentration between the initial and final concentrations. 4. Use the expressions for reaction rate to find the time it takes for the reaction. After following these steps and plugging in the given values, we can calculate the time, t, which is how long it will take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at 500K.

Step by step solution

01

1. Using the Arrhenius equation

We can use the Arrhenius equation to find the rate constant, k, at 500K. The equation is: \(k = k_0 e^{(-E_a / R T)}\) where: \(k\): rate constant at temperature T \(k_0\): rate constant at a reference temperature (273K in this case) \(E_a\): activation energy \(R\): gas constant (8.314 J/molK) \(T\): temperature in Kelvin First, convert the activation energy from kJ/mol to J/mol and the rate constant to L/mol·s: \(E_a = 111,000 \, J / mol\) Now, we need to rearrange the Arrhenius equation to solve for \(k_0\): \(k_0 = k / e^{(-E_a / R T)}\) Plug in the values and calculate \(k_0\): \(k_0 = (2.3 \times 10^{-12}) / e^{(-111,000 / (8.314 \times 273))}\) Now we have the value for \(k_0\), so we can plug it back into the Arrhenius equation to find the rate constant, \(k_{500}\), at 500K: \(k_{500} = k_0 e^{(-E_a / R T)}\) \(k_{500} = k_0 e^{(-111,000 / (8.314 \times 500))}\)
02

2. Calculate the reaction rate

We are given that the reaction is second-order with respect to \(\mathrm{NO}_2\). The rate law for this reaction is: \(rate = k_{500}[NO_2]^2\) Since we are given the initial and final partial pressures, we can convert them to concentrations using the ideal gas law: \(PV = nRT \Rightarrow [NO_2] = \frac{n}{V} = \frac{P}{RT}\) Therefore, the initial and final concentrations of \(NO_2\) are: \([NO_2]_i = \frac{2.5 \,atm}{(0.0821 \, L/atm \cdot mol \cdot K)(500 \, K)}\) \([NO_2]_f = \frac{1.5 \,atm}{(0.0821 \, L/atm \cdot mol \cdot K)(500 \, K)}\) Now we can find the average concentration between the initial and final concentrations: \([NO_2]_{avg} = \frac{[NO_2]_i + [NO_2]_f}{2}\)
03

3. Calculate the time taken for the reaction

Now that we have the average concentration and the rate constant at 500K, we can find the reaction rate: \(rate = k_{500}[NO_2]_{avg}^2\) We also know that the rate is equal to the change in concentration divided by the change in time: \(rate = \frac{[NO_2]_i - [NO_2]_f}{t}\) By equating both expressions for rate, we get: \(k_{500}[NO_2]_{avg}^2 = \frac{[NO_2]_i - [NO_2]_f}{t}\) Rearrange the equation to solve for time, t: \(t = \frac{[NO_2]_i - [NO_2]_f}{k_{500}[NO_2]_{avg}^2}\) Plug in the values for \([NO_2]_i, [NO_2]_f, k_{500}\), and \([NO_2]_{avg}\), then calculate the time, t. This is how long it will take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at 500K.

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Most popular questions from this chapter

Consider the reaction $$ 3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ An experiment is carried out where $[\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}$ and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t :\) $$ \begin{aligned}-\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\\\-& \frac{\Delta \mathrm{B}}{\Delta t}=0.0120 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

A certain substance, initially present at \(0.0800 M,\) decomposes by zero-order kinetics with a rate constant of $2.50 \times 10^{-2} \mathrm{mol} / \mathrm{L}$ . s. Calculate the time (in seconds required for the system to reach a concentration of 0.0210\(M .\)

The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate $=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )$
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