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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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Rate laws generally cannot be written from balanced equations because the stoichiometric coefficients from a balanced equation show the proportional relationship between reactants and products, but they don't provide information about the rates at which these substances participate in the reaction. Rate laws, which describe the effect of reactant concentrations on the reaction rate, depend on reaction orders that are usually determined experimentally. However, in special cases of elementary reactions (single-step reactions without intermediates), the rate law can be written directly from the balanced chemical equation because the reaction orders for the reactants are the same as their stoichiometric coefficients.

Step by step solution

01

Understanding Rate Laws

Rate laws are mathematical expressions that describe the relationship between the concentration of reactants and the rate at which a chemical reaction occurs. They often have the general form: \[rate = k \cdot [A]^m \cdot [B]^n \ldots,\] where \(k\) is the rate constant for the reaction, \([A]\) and \([B]\) represent the concentrations of the reactants, and \(m\) and \(n\) are the reaction orders with respect to each reactant.
02

Understanding Balanced Chemical Equations

A balanced chemical equation represents a chemical reaction and expresses the quantities of reactants needed to produce the quantities of products. By equating the stoichiometric coefficients, a balanced equation shows the proportion of how reactants and products relate to each other and ensures that the law of mass conservation is respected. For example, in a reaction between reactant A and B, producing products C and D, the balanced equation will look like: \[aA + bB \rightarrow cC + dD,\] where \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients.
03

Rate Laws and Balanced Equations

The stoichiometric coefficients from a balanced equation show the proportional relationship between reactants and products. However, they don't provide any information about the rates at which these substances participate in the reaction. The reaction orders (m and n) in the rate law, on the other hand, represent the sensitivity of the reaction rate to the changes in the concentrations of the reactants, so they provide information about the kinetic behavior of the reactants.
04

Why Rate Laws Cannot Generally be Written from Balanced Equations

Since rate laws describe the effect of reactant concentrations on the reaction rate, and balanced chemical equations provide information about the proportionality and mass conservation ratios of reactants and products, the correlation between the stoichiometric coefficients and the reaction orders is not guaranteed. Reaction orders are usually determined experimentally, while stoichiometric coefficients are obtained from the balanced equation following the law of conservation of mass. As such, we cannot generally write rate laws based on balanced chemical equations.
05

When Rate Laws are Directly Related to Balanced Chemical Equations

Rate laws would be directly related to balanced chemical equations in special cases when reactions follow elementary reactions. An elementary reaction is one in which the reaction occurs in a single step, without the formation of any intermediates, and the reaction orders for the reactants are the same as their stoichiometric coefficients. In such cases, the rate law can be written directly from the balanced chemical equation. However, most reactions occur through a sequence of elementary steps, called a reaction mechanism, making it difficult to write rate laws directly from balanced equations for general cases.

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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 how fast chemical reactions occur and what factors affect their rates. It is a cornerstone of physical chemistry, vital for understanding both natural processes, like digestion and combustion, and industrial processes, which include everything from making pharmaceuticals to brewing beer.

At the heart of chemical kinetics is the concept of the reaction rate, which is the speed at which reactants transform into products. Variables that can affect reaction rates include the concentration of reactants, temperature, and the presence of catalysts. Kinetics also involves understanding and applying rate laws, which quantitatively describe the relationship between the concentrations of reactants and the rate of the reaction. These laws are typically unique to each reaction and must often be determined through experiments.

Understanding kinetics is crucial for controlling processes, predicting how systems will change over time, and designing reactions to occur at desired speeds—whether that's as fast as possible, or at a slow, controlled pace.
Balanced Chemical Equations
Balanced chemical equations are vital in chemistry as they ensure the law of conservation of mass is respected in a chemical reaction. The equation demonstrates that atoms are neither created nor destroyed in reactions, only rearranged. By balancing an equation, we ensure that the same number of each type of atom appears on both sides of the reaction.

Within a balanced equation, stoichiometric coefficients tell us the relative amounts of reactants that react and the amount of product produced. For example, consider the equation \[2H_2 + O_2 \rightarrow 2H_2O\]. This equation informs us that two molecules of hydrogen gas (\(H_2\)) react with one molecule of oxygen gas (\(O_2\)) to produce two molecules of water (\(H_2O\)). While these coefficients provide a stoichiometric basis for predicting the amounts of reactants needed and products formed, they do not provide information on how quickly the reaction will happen, which is where reaction kinetics and rate laws come into play.
Reaction Order
The reaction order is an integral part of the rate law that indicates how the rate of reaction is affected by the concentration of each reactant. It is expressed as an exponent in the mathematical rate law expression. For instance, in a rate law like \[rate = k \times [A]^m \times [B]^n\], the exponents \(m\) and \(n\) represent the reaction orders with respect to reactants \(A\) and \(B\), respectively.

Total reaction order is found by summing the orders of all reactants, giving insight into the overall rate change with concentration changes. Reaction order can be zero, meaning the concentration change has no effect on the rate, one, where rate is directly proportional to concentration changes, or two, where rate changes with the square of concentration changes, among other possibilities. Importantly, reaction orders are not typically inferred from the balanced equation, but determined experimentally—a core concept students often need to grasp.
Elementary Reactions
Elementary reactions are the simplest types of reactions on a molecular level and occur in a single step. An important aspect of these reactions is that their rate laws can be directly written from the balanced chemical equation because the coefficients directly equate to the molecularity of the reaction—the number of molecules or atoms that participate in the reaction step.

For example, in a reaction such as \[NO_2 + CO \rightarrow NO + CO_2\], if this is an elementary reaction, the rate law can be written as \[rate = k \times [NO_2] \times [CO]\], with each coefficient representing the actual order with respect to each reactant. However, most chemical reactions are complex and comprise a series of elementary reactions, known as a reaction mechanism. This complexity means the stoichiometric coefficients in a balanced overall equation generally cannot indicate the reaction order in the rate law, highlighting the importance of experimental determination of rate laws.

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Most popular questions from this chapter

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

As described in Exercise \(14.43,\) 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 450 torr, what is the pressure of this substance after \(60 \mathrm{~s} ?\) (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

(a) Consider the combustion of \(\mathrm{H}_{2}(g): 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) \(\longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is burning at the rate of 0.48 \(\mathrm{mol} / \mathrm{s}\), what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of \(\mathrm{NO}\) is decreasing at the rate of 56 torr \(/ \mathrm{min}\), what is the rate of change of the total pressure of the vessel?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

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