/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 The conversion of \(\mathrm{NO}_... [FREE SOLUTION] | 91影视

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The conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) can occur through the following reaction: $$\mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ The activation energy for this reaction is \(111 \mathrm{kJ} \mathrm{mol}^{-1}\) and the pre-exponential factor is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Assume that these quantities are temperature independent. a. What is the rate constant for this reaction at \(298 \mathrm{K} ?\) b. What is the rate constant for this reaction at the tropopause where \(\mathrm{T}=225 \mathrm{K}\).

Short Answer

Expert verified
a. At 298 K, the rate constant \(k\) is approximately \(9.91 \times 10^{-12} \mathrm{M}^{-1} \mathrm{s}^{-1}\). b. At 225 K, the rate constant \(k\) is approximately \(2.57 \times 10^{-14} \mathrm{M}^{-1} \mathrm{s}^{-1}\).

Step by step solution

01

Calculate k at 298K

We are given the activation energy (Ea = 111 kJ mol鈦宦) and the pre-exponential factor (A = 2.0 x 10鈦烩伖 M鈦宦 s鈦宦). At 298 K, we can calculate the rate constant (k) using the Arrhenius equation: \(k = Ae^{-\frac{E_{a}}{RT}}\) where R = 8.314 J mol鈦宦 K鈦宦 and T = 298 K. To use compatible units, we need Ea in J mol鈦宦, so Ea = 111,000 J mol鈦宦. $$k = 2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1} e^{-\frac{111,000 \mathrm{J} \mathrm{mol}^{-1}}{(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1})(298 \mathrm{K})}}$$ Calculate the exponent and then solve for k.
02

Calculate k at 225K

Now we follow the same steps as before but with T = 225 K. \(k = Ae^{-\frac{E_{a}}{RT}}\) $$k = 2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1} e^{-\frac{111,000 \mathrm{J} \mathrm{mol}^{-1}}{(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1})(225 \mathrm{K})}}$$ Calculate the exponent and then solve for k.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Activation Energy
Activation energy is a term that defines the minimum amount of energy necessary for a chemical reaction to occur. Think of this as a barrier that reactants need to overcome to transform into products. The concept of activation energy is crucial for understanding why some reactions proceed at a fast rate while others are much slower.

For example, striking a match initiates a reaction. The energy you apply via friction is enough to overcome the activation energy barrier which allows the chemicals in the match head to react and ignite. In the case of our textbook exercise, the activation energy for the conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) is stated to be 111 kJ/mol. This means that for this specific reaction to proceed, a significant amount of energy is needed, which impacts the rate of reaction at different temperatures.
Rate Constant
The rate constant, represented by the letter \(k\), is an essential factor in chemical kinetics, which provides the speed at which a chemical reaction proceeds. It's unique for each reaction and is influenced by conditions such as temperature and the presence of a catalyst. A higher rate constant indicates a faster reaction.

The Arrhenius equation allows us to calculate the rate constant based on the known variables such as the activation energy and the temperature. In our exercise, understanding the rate constant at different temperatures, such as \(298 K\) and \(225 K\), highlights how even a slight change in temperature can greatly affect how fast the reaction takes place. The exercise shows that with the provided activation energy and pre-exponential factor, we can solve for the rate constant at any given temperature.
Pre-exponential Factor
The pre-exponential factor, often denoted as \(A\), appears in the Arrhenius equation and is sometimes referred to as the frequency factor. It relates to the frequency of collisions and the orientation of reactant molecules that result in successful reactions. It's important because it reflects how often particles collide in the right way to react.

In simpler terms, if we think about a room full of people moving randomly鈥攕ome collisions result in a handshake (successful reaction), others do not. The pre-exponential factor could represent how many handshakes happen in a set amount of time if everyone had perfect aim. In the exercise at hand, the value given for \(A\) is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\), which we use along with the activation energy to determine the rate constant for the reaction.
Chemical Kinetics
Chemical kinetics is the study of the speed or rate at which chemical reactions occur. It's a crucial field in chemistry because it not only helps us understand reaction rates but also the mechanisms behind them鈥攈ow reactants become products over time. Factors like temperature, concentration, surface area, and presence of catalysts can influence these rates.

For students, the study of kinetics can often seem daunting, but it is vital for predicting how long reactions will take and under what conditions they might occur more readily. The textbook exercise involving the decomposition of \(\mathrm{NO}_{2}(g)\) serves as a practical example of applying chemical kinetics principles to understand and calculate specific aspects of a reaction such as the rate constant at different temperatures.

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Most popular questions from this chapter

Calculate the ratio of rate constants for two thermal reactions that have the same Arrhenius preexponential term but have activation energies that differ by \(1.0,10 .,\) and \(30 . \mathrm{kJ} / \mathrm{mol}\) for \(T=298 \mathrm{K}\).

In the limit where the diffusion coefficients and radii of two reactants are equivalent, demonstrate that the rate constant for a diffusion controlled reaction can be written as $$k_{d}=\frac{8 R T}{3 \eta}$$

Consider the gas phase thermal decomposition of \(1.0 \operatorname{atm}\) of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COOC}\left(\mathrm{CH}_{3}\right)_{3}(g)\) to acetone \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(g)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)(\mathrm{g}),\) which occurs with a rate constant of \(0.0019 \mathrm{s}^{-1} .\) After initiation of the reaction, at what time would you expect the pressure to be 1.8 atm?

P35.10 (Challenging) The first-order thermal decomposition of chlorocyclohexane is as follows: \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{Cl}(g) \rightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{10}(g)+\mathrm{HCl}(g) .\) For a constant volume system the following total pressures were measured as a function of time: $$\begin{array}{rccr}\text { Time }(s) & P(\text { Torr }) & \text { Time }(s) & P(\text { Torr }) \\\\\hline 3 & 237.2 & 24 & 332.1 \\\6 & 255.3 & 27 & 341.1 \\\9 & 271.3 & 30 & 349.3 \\\12 & 285.8 & 33 & 356.9 \\\15 & 299.0 & 36 & 363.7 \\\18 & 311.2 & 39 & 369.9 \\\21 & 322.2 & 42 & 375.5\end{array}$$ a. Derive the following relationship for a first-order reaction: \\[P\left(t_{2}\right)-P\left(t_{1}\right)=\left(P\left(t_{\infty}\right)-P\left(t_{0}\right)\right) e^{-k t_{1}}\left(1e^{-k\left(t_{2}-t_{1}\right)}\right)\\] In this relation, \(P\left(t_{1}\right)\) and \(P\left(t_{2}\right)\) are the pressures at two specific times; \(\mathrm{P}\left(t_{0}\right)\) is the initial pressure when the reaction is initiated, \(P\left(t_{\infty}\right)\) is the pressure at the completion of the reaction, and \(k\) is the rate constant for the reaction. To derive this relationship do the following: i. Given the first-order dependence of the reaction, write the expression for the pressure of chlorocyclohexane at a specific time \(t_{1}\) ii. Write the expression for the pressure at another time \(t_{2},\) which is equal to \(t_{1}+\Delta\) where delta is a fixed quantity of time. iii. Write expressions for \(P\left(t_{\infty}\right)-P\left(t_{1}\right)\) and \(P\left(t_{\infty}\right) P\left(t_{2}\right)\) iv. Subtract the two expressions from part (iii). b. Using the natural log of the relationship from part (a) and the data provided in the table given earlier in this problem, determine the rate constant for the decomposition of chlorocyclohexane. (Hint: Transform the data in the table by defining \(t_{2}-t_{1}\) to be a constant value, for example, 9 s.

One loss mechanism for ozone in the atmosphere is the reaction with the \(\mathrm{HO}_{2} \cdot\) radical: Using the following information, determine the rate law expression for this reaction: $$\begin{array}{ccc}\text { Rate }\left(\mathrm{cm}^{-3} \mathrm{s}^{-1}\right) & {\left[\mathrm{HO}_{2} \cdot\right]\left(\mathrm{cm}^{-3}\right)} & {\left[\mathrm{O}_{3}\right]\left(\mathrm{cm}^{-3}\right)} \\ \hline 1.9 \times 10^{8} & 1.0 \times 10^{11} & 1.0 \times 10^{12} \\\9.5 \times 10^{8} & 1.0 \times 10^{11} & 5.0 \times 10^{12} \\\5.7 \times 10^{8} & 3.0 \times 10^{11} & 1.0 \times 10^{12}\end{array}$$

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