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The second-order rate constant for the decomposition of \(\mathrm{NO}_{2}\) (to \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) ) at \(573 \mathrm{~K}\) is \(0.54 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\). Calculate the time for an initial \(\mathrm{NO}_{2}\) concentration of \(0.20 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to decrease to (a) one-half; (b) one-sixteenth; (c) one-ninth of its initial concentration.

Short Answer

Expert verified
The times for the concentration to decrease to (a) one-half is \[\frac{1}{0.54 \cdot 0.20} \(\approx 9.26 \,s\), (b) one-sixteenth is \[\frac{15}{0.54 \cdot 0.20} \(\approx 138.89 \,s\), and (c) one-ninth is \[\frac{8}{0.54 \cdot 0.20} \(\approx 74.07 \,s\).

Step by step solution

01

Identifying the Type of Reaction

The decomposition of \(\text{NO}_2\) is a second-order reaction, which means the rate is proportional to the square of the concentration of \(\text{NO}_2\). Therefore, we use the second-order rate equation: \[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\] where \[A]_t\] is the concentration at time \[t\], \[A]_0\] is the initial concentration, \[k\] is the rate constant, and \[t\] is the time.
02

Solving for Time When Concentration Halves

To find the time for the concentration to decrease to one-half, substitute \[A]_t = \frac{[A]_0}{2}\] into the rate equation and solve for \[t\]. \[\frac{1}{[A]_0/2} - \frac{1}{[A]_0} = kt\] Simplify and solve for \[t\]: \[2\frac{1}{[A]_0} - \frac{1}{[A]_0} = kt\] \[\frac{1}{[A]_0} = kt\] \[t = \frac{1}{k[A]_0}\] Plugging in the values, we get \[t = \frac{1}{0.54 \,L \cdot mol^{-1} \cdot s^{-1} \cdot 0.20 \,mol \cdot L^{-1}}\], and calculate \[t\].
03

Solving for Time When Concentration Becomes One-Sixteenth

To find the time for the concentration to decrease to one-sixteenth, substitute \[A]_t = \frac{[A]_0}{16}\] into the rate equation and solve for \[t\]. \[\frac{1}{[A]_0/16} - \frac{1}{[A]_0} = kt\] Simplify and find \[t\]: \[16\frac{1}{[A]_0} - \frac{1}{[A]_0} = kt\] \[15\frac{1}{[A]_0} = kt\] \[t = \frac{15}{k[A]_0}\] Use the values of \[k\] and \[A]_0\] to find \[t\].
04

Solving for Time When Concentration Becomes One-Ninth

To find the time for the concentration to decrease to one-ninth, substitute \[A]_t = \frac{[A]_0}{9}\] into the rate equation and solve for \[t\]. \[\frac{1}{[A]_0/9} - \frac{1}{[A]_0} = kt\] Simplify to find \[t\]: \[9\frac{1}{[A]_0} - \frac{1}{[A]_0} = kt\] \[8\frac{1}{[A]_0} = kt\] \[t = \frac{8}{k[A]_0}\] Plug in the values for \[k\] and \[A]_0\] to calculate \[t\].

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Key Concepts

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

Understanding Chemical Kinetics
Chemical kinetics is the study of the rates of chemical processes, and how different variables affect these rates. It examines the speeds at which reactants are converted into products and investigates the mechanisms through which these chemical reactions occur. By understanding kinetics, chemists can control and optimize reactions for industrial processes, biochemical applications, and understand natural phenomena. For students tackling kinetics, it's helpful to visualize reactions as tiny, individual molecular events. Consider how likely molecules are to collide with sufficient energy and the correct orientation to react, which is the essence of reaction kinetics.
Decoding Rate Laws
Rate laws are mathematical expressions that describe the relationship between the concentrations of reactants and the rate of the reaction. In a generic reaction where substance A transforms into products, the rate law can be written as rate = k[A]^n, where 'rate' is the speed of the reaction, 'k' is the rate constant specific to the reaction at a given temperature, and 'n' is the order of the reaction with respect to A. The order is an experimental value that tells us how the reaction rate is affected by the concentration of A. For example, in a second-order reaction like the decomposition of \(\text{NO}_2\), the rate is proportional to the square of the concentration of \(\text{NO}_2\). Understanding and determining rate laws are vital for predicting how the concentration of reactants will change over time.
Reaction Rates and Their Measurement
Reaction rates quantify the speed of chemical reactions in terms of concentration changes over time. They are essential for chemists and engineers to design and control processes safely and efficiently. Rates can be expressed in various units, such as moles per liter per second \(mol/L/s\), depending on the concentrations of reactants or products involved. To measure these rates, you might track the disappearance of a reactant or the appearance of a product over given time intervals. The rate at which \(\text{NO}_2\) decomposes to \(\text{NO}\) and \(\text{O}_2\) is just one example that involves observing the decrease in concentration of \(\text{NO}_2\) with time.
Concentration-Time Relationships in Reaction Kinetics
The relationship between concentrations of reactants or products and time provides valuable insights into the kinetics of a reaction. For second-order reactions, the concentration-time relationship is not linear. Instead, plotting the inverse of the concentration of the reactant versus time will typically yield a straight line from which the rate constant and the order of the reaction can be determined. In practical terms, this means that if you start with a particular concentration of a substance, you can predict how long it will take for that concentration to decrease to half, one-sixteenth, or any other fraction, using specific mathematical relationships derived from the rate law.
Differential Rate Equations
Differential rate equations are the mathematical expressions that relate the rate of the reaction to the concentrations of reactants. For example, in a second-order reaction such as the decomposition of \(\text{NO}_2\), the differential rate equation can be written based on the given rate law. Solving these equations requires a proper understanding of calculus, as they involve rates of change that are dependent on concentrations. In the example provided, the step-by-step solutions demonstrate how to manipulate these equations to find the time required for the reactant's concentration to reach certain values. It's similar to assembling a puzzle – finding the right piece (or in this case, the mathematical manipulation) helps reveal the bigger picture (the time required for certain concentration changes).

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Most popular questions from this chapter

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

The decomposition of A has the rate law Rate \(=k[\mathrm{~A}]^{a}\). Show that for this reaction the ratio \(t_{1 / 2} / t_{3 / 4}\), where \(t_{1 / 2}\) is the halflife and \(t_{3 / 4}\) is the time for the concentration of A to decrease to \(\frac{3}{4}\) of its initial concentration, can be written as a function of \(a\) alone and can therefore be used to make a quick assessment of the order of the reaction in A.

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(0.15 \mathrm{~s}^{-1}\) at \(353 \mathrm{~K}\). (a) What is the half-life (in seconds) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(353 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(2.0 \mathrm{~s}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) For a reaction with a very large equilibrium constant, the rate constant of the forward reaction is much larger than the rate constant of the reverse reaction. (b) At equilibrium, the rate constants of the forward and reverse reactions are equal. (c) Increasing the concentration of a reactant increases the rate of a reaction by increasing the rate constant in the forward direction.

Because partial pressures are proportional to concentrations, rate laws for gas-phase reactions can also be expressed in terms of partial pressures, for instance, as Rate \(=\) \(k P_{\mathrm{X}}\) for a first-order reaction of a gas \(\mathrm{X}\). What are the units for the rate constants when partial pressures are expressed in torr and time is expressed in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions?

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