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Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?

Short Answer

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In summary, rate laws cannot be directly derived from balanced equations, as they depend on reaction mechanisms and rate-determining steps, which may not correlate with stoichiometric coefficients. However, when a reaction occurs in a single elementary step, the rate law is directly related to the balanced equation.

Step by step solution

01

Introduction to Stoichiometry and Kinetics

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction, as represented by a balanced equation. Balanced equations give the proportion of reactants and products that participate in the reaction, but they do not provide information about the reaction speed or how it occurs on the molecular level. Kinetics is the study of reaction rates and the factors that affect these rates. The rate law of a chemical reaction describes the relationship between the rate of the reaction and the concentrations of reactants. It usually takes the form: rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are the concentrations of the reactants, and m and n are the reaction orders.
02

Reaction Orders

The reaction orders (m and n) are determined experimentally, as they depend on the reaction mechanism, which is the detailed sequence of elementary steps that a reaction undergoes. The reaction orders are not always equal to the stoichiometric coefficients in the balanced equation. This is because the stoichiometric coefficients describe the overall reaction, whereas the reaction orders are determined by the elementary steps which may or may not involve all reactants involved in the balanced equation.
03

Rate-determining step

In a multi-step reaction, the rate-determining step is the slowest elementary step in the reaction mechanism, which dictates the overall rate of the reaction. The rate law is determined by the rate-determining step and may involve reactants that aren't necessarily involved in other steps. This is another reason why a rate law generally cannot be derived directly from a balanced equation because the balanced equation represents the overall reaction, not the rate-determining step.
04

Cases where Rate Law is Related to Balanced Equation

There is one circumstance in which the rate law can be directly related to the balanced equation: when the overall reaction occurs in a single step (elementary reaction) with no intermediate steps. In this case, the rate-determining step is the overall reaction, and the reaction orders directly correspond to the stoichiometric coefficients of the reactants in the balanced equation. This is relatively rare for complex reactions but can be observed in simpler reactions. In conclusion, rate laws generally cannot be written directly from balanced equations because they are determined by the reaction mechanism and rate-determining step, which may not directly correspond to the stoichiometric coefficients in the balanced equation. However, if the overall reaction occurs in a single elementary step, the rate law can be directly related to the balanced equation.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{Cl}(g)\) (b) \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\)

As described in Exercise \(14.37\), the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 375 torr, what is the pressure of this substance after 65 s? (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant. The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has a molar absorptivity constant of \(5.60 \times 10^{3} \mathrm{~cm}^{-1} \mathrm{M}^{-1}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) within \(30.0\) min. Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

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