Determine the rate constant for each of the following firstorder reactions, in each case expressed for the rate of loss of \(A\) : (a) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of \(\mathrm{A}\) decreases to one-half its initial value in \(1000 . \mathrm{s} ;\) (b) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of A decreases from \(0.67 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.53 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) in \(25 \mathrm{~s}\); (c) \(2 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.153 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(115 \mathrm{~s}\) the concentration of \(B\) rises to \(0.034 \mathrm{~mol} \cdot \mathrm{L}^{-1}\).

Short Answer

Expert verified
The rate constants are: (a) \( k = \frac{\ln(2)}{1000} \approx 0.000693 \text{s}^{-1} \), (b) \( k = \frac{\ln(\frac{0.67}{0.53})}{25} \approx 0.0234 \text{s}^{-1} \) and (c) \( k = \frac{\ln(\frac{0.153}{0.153 - 2*0.034})}{115} \approx 0.0292 \text{s}^{-1} \).

Step by step solution

01

- Understanding the First-Order Reaction Rate Constant

The rate constant for a first-order reaction can be determined using the formula \( k = \frac{\ln(\frac{[A]_0}{[A]})}{t} \), where \( [A]_0 \) is the initial concentration of reactant A, \( [A] \) is the concentration of A at time \( t \) seconds, and \( \ln \) is the natural logarithm.
02

- Calculate the Rate Constant for Reaction (a)

We are given that the concentration of A decreases to one-half its initial value in 1000 seconds. We use \( [A] = \frac{1}{2}[A]_0 \) and \( t = 1000 \) s in the formula to find the rate constant \( k \). \( k = \frac{\ln(\frac{[A]_0}{1/2[A]_0})}{1000} = \frac{\ln(2)}{1000} \) (since \( \ln(2) \) is the natural logarithm of 2).
03

- Calculate the Rate Constant for Reaction (b)

For reaction (b), the concentration of A decreases from 0.67 mol/L to 0.53 mol/L in 25 seconds. We input these values into the formula: \( k = \frac{\ln(\frac{0.67}{0.53})}{25} \) and compute the rate constant.
04

- Calculate the Rate Constant for Reaction (c)

For reaction (c), given \( [A]_0 = 0.153 \text{mol/L} \) and the concentration of B (which is also the change in concentration of A, since 2 A produces B) rises to 0.034 mol/L in 115 seconds. Since for every mole of B produced, we lose two moles of A, the change in concentration of A is 2*[B], making the final concentration of A \( [A] = [A]_0 - 2*0.034 = 0.153 - 2*0.034 \) mol/L. We then use this value in our formula \( k = \frac{\ln(\frac{0.153}{0.153 - 2*0.034})}{115} \) to find the rate constant.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of the rates at which chemical reactions proceed and the factors that affect these rates. It's essential for understanding how reactions occur and for designing processes that utilize chemical reactions, such as the manufacture of pharmaceuticals or the combustion of fuels. A fundamental aspect of chemical kinetics is the reaction rate, which measures the speed at which reactants are converted into products. The reaction rate can be affected by various factors, including concentration of reactants, temperature, and the presence of catalysts.

In the context of first-order reactions, they are characterized by their rate depending linearly on the concentration of one reactant. This means the rate at which the reactant is consumed is directly proportional to its concentration at any given moment during the reaction. Consequently, understanding the rate constant for a first-order reaction is crucial since it is a measure of how quickly the reaction occurs and is independent of the reactant concentration.
Reaction Rate
The rate at which a chemical reaction proceeds is known as the reaction rate. In the field of chemical kinetics, it's often expressed in terms of the change in concentration of a reactant or product per unit time. For a first-order reaction, the rate is proportional to the concentration of one reactant, which means the reaction rate will decrease over time as the reactant is consumed. Rates can be calculated using rate laws, which are mathematical expressions that relate the rate of a reaction to the concentration of its reactants.

The rate constant for a first-order reaction, often denoted as 'k', is a crucial parameter as it allows us to calculate how fast the reaction is occurring under specified conditions. The value of 'k' provides insights into the kinetics of the reaction and is influenced by external factors like temperature and the presence of a catalyst. By understanding reaction rates and rate constants, chemists can predict reaction behavior and control conditions to optimize reaction efficiency.
Concentration of Reactants
The concentration of reactants plays a pivotal role in chemical kinetics and determining reaction rates. For many reactions, including first-order reactions, the rate at which reactions proceed is directly proportional to the concentration of one or more reactants. This relationship is encapsulated in the rate law for the reaction.

In the case of first-order reactions, the rate law has the form Rate = k[A], where 'k' is the rate constant and [A] is the concentration of the reactant A. As a reaction proceeds and the concentration of A decreases, the rate of the reaction decreases accordingly. This is why monitoring the concentration of reactants over time is a fundamental method for characterizing the kinetics of a reaction. By doing so, chemists can calculate the rate constant 'k' and further understand the reaction mechanism.
Natural Logarithm
The natural logarithm is a mathematical function that is particularly relevant in the context of first-order reactions in chemical kinetics. Denoted as 'ln', it is the logarithm to the base 'e', where 'e' is an irrational and transcendental number approximately equal to 2.71828. This function is used to transform the exponential decay of reactant concentrations in first-order reactions into a linear form that is easier to analyze.

The rate equation for a first-order reaction integrates to give an expression involving natural logarithms: ln([A]/[A]_0) = -kt. Here, [A] is the concentration of the reactant at time 't', and [A]_0 is the initial concentration of the reactant. By using the principles of the natural logarithm, it simplifies the process of determining the rate constant 'k' from experimental concentration-time data. When plotting ln([A]) against time, a straight line with a slope of -k is obtained, demonstrating the integral role the natural logarithm plays in analyzing and interpreting kinetic data.

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Most popular questions from this chapter

The half-life of a substance taking part in a third-order reaction \(\mathrm{A} \rightarrow\) products is inversely proportional to the square of the initial concentration of A. How can this half-life be used to predict the time needed for the concentration to fall to (a) onehalf; (b) one-fourth; (c) one- sixteenth of its initial value?

(a) From the following mechanism, derive Eq. 19a, which Michaelis and Menten proposed to represent the rate of formation of products in an enzyme-catalyzed reaction. (b) Show that the rate is independent of substrate concentration at high concentrations of substrate. $$ \begin{aligned} &\mathrm{E}+\mathrm{S} \rightleftarrows \mathrm{ES} \quad k_{1}, k_{1}^{\prime} \\ &\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad k_{2} \end{aligned} $$ where \(E\) is the free enzyme, \(S\) is the substrate, ES is the enzyme-substrate complex, and \(P\) is the product. Note that the steady-state concentration of free enzyme will be equal to the initial concentration of the enzyme less the amount of enzyme that is present in the enzyme-substrate complex: \([\mathrm{E}]=[\mathrm{E}]_{0}-[\mathrm{ES}]\)

Each of the following is an elementary reaction. Write its rate law and indicate its molecularity. (a) \(\mathrm{O}+\mathrm{CF}_{2} \mathrm{Cl}_{2} \rightarrow\) \(\mathrm{ClO}+\mathrm{CF}_{2} \mathrm{Cl} ;\) (b) \(\mathrm{OH}+\mathrm{NO}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{HNO}_{3}+\mathrm{N}_{2} ;\) (c) \(\mathrm{ClO}^{-}+\) \(\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}+\mathrm{OH}^{-}\); (d) Which of these reactions might be radical chain propagating?

(a) Using a graphing calculator or graphing software, such as that on the Web site for this book, calculate the activation energy for the acid hydrolysis of sucrose to give glucose and fructose from an Arrhenius plot of the data shown here. (b) Calculate the rate constant at \(37^{\circ} \mathrm{C}\) (body temperature). (c) From data in Appendix 2A, calculate the enthalpy change for this reaction, assuming that the solvation enthalpies of the sugars are negligible. Draw an energy profile for the overall process. $$ \begin{array}{cc} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 24 & 4.8 \times 10^{-3} \\ 28 & 7.8 \times 10^{-3} \\ 32 & 13 \times 10^{-3} \\ 36 & 20 . \times 10^{-3} \\ 40 . & 32 \times 10^{-3} \\ \hline \end{array} $$

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.

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