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The exponents in a rate law have no relationship to the coefficients in the overall balanced equation for the reaction. Give an example of a balanced equation and the rate law for a reaction that clearly demonstrates this.

Short Answer

Expert verified
Coefficients do not equal exponents; see \([\text{H}_2\text{O}_2]^1\) in the rate law for its decomposition.

Step by step solution

01

Identify a Balanced Reaction

Let's consider a simple chemical reaction: the decomposition of hydrogen peroxide into water and oxygen. The balanced equation for this reaction is: \[ 2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g) \] Here, the coefficients are '2' for \( \text{H}_2\text{O}_2 \) and '2' for \( \text{H}_2\text{O} \), and '1' for \( \text{O}_2 \).
02

Determine the Rate Law

The rate law for a reaction is determined experimentally and expresses the rate of the reaction as a function of the concentration of the reactants. For the decomposition of hydrogen peroxide, one possible rate law is:\[ \text{Rate} = k [\text{H}_2\text{O}_2] \]In this rate law, the reaction is first-order with respect to hydrogen peroxide, as indicated by the exponent '1' on the concentration term [\( \text{H}_2\text{O}_2 \)].
03

Compare Exponents to Coefficients

Notice in the balanced equation, the coefficient for \( \text{H}_2\text{O}_2 \) is '2'. However, in the rate law, the exponent for [\( \text{H}_2\text{O}_2 \)] is '1'. This shows that the coefficient in the balanced equation does not directly relate to the exponent in the rate law. The exponents must be determined through experimental data, not based on the balanced equation.

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Key Concepts

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

Reaction Order
The concept of reaction order is a fundamental aspect of chemical kinetics. Reaction order refers to the power to which the concentration of a reactant is raised in the rate law. This provides insight into how the concentration of reactants affects the rate of a chemical reaction.
For example, in the decomposition of hydrogen peroxide (\( 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \)), the reaction rate can be expressed as \( \text{Rate} = k [\text{H}_2\text{O}_2]^1 \), making this a first-order reaction with respect to hydrogen peroxide. Here, the term "first-order" implies that a change in the concentration of hydrogen peroxide leads to a proportional change in the reaction rate.
  • If the concentration of hydrogen peroxide is doubled, the rate of reaction also doubles.
  • This linear relationship indicates a direct dependence, characteristic of first-order reactions.
It's crucial to note that the overall reaction order is the sum of all individual orders of reactants involved in the rate equation, which can be different from the stoichiometric coefficients in the balanced chemical equation.
Balanced Chemical Equation
A balanced chemical equation offers a visual and quantitative description of a chemical reaction. It ensures the conservation of mass by having the same number of atoms of each element on both sides of the equation. For instance, in the balanced equation \( 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \), coefficients '2' and '1' indicate the ratio in which substances react and are formed.
However, these coefficients should not be mistaken for the exponents found in a rate law. The coefficients in a balanced equation serve solely to maintain mass balance and do not influence the kinetics of the reaction directly.
  • Coefficients indicate the stoichiometry of the reaction.
  • They help in calculating the amounts of reactants needed and products formed.
It's important to differentiate between these coefficients and the exponents in the rate law, as the actual rate of reaction is dependent on experimental conditions, not just the balanced chemical equation.
Experimental Determination of Rate Laws
To determine a rate law, experimental data is crucial. Unlike balanced equations, which are straightforward to write using chemical stoichiometry, the rate law must be derived from empirical data. This involves monitoring the change in concentration of reactants or products over time and determining how these changes affect the reaction rate.
Steps involved in experimental determination usually include:
  • Conducting experiments to measure reaction rates under various concentrations of reactants.
  • Analyzing data to establish a relationship between concentration and rate.
  • Deriving the exponents in the rate law that indicate the reaction order.
For example, in the hydrogen peroxide decomposition reaction, using experimental data allows us to establish that it is first-order with respect to hydrogen peroxide. This understanding is vital, as it allows chemists to predict and control reaction conditions and predict reaction behavior in different scenarios.

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Most popular questions from this chapter

In the presence of excess thiocyanate ion, \(\mathrm{SCN}^{-}\), the following reaction is first order in iron(III) ion, \(\mathrm{Fe}^{3+}\); the rate constant is \(1.27 / \mathrm{s}\). $$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Fe}(\mathrm{SCN})^{2+}(a q)$$ What is the half-life in seconds? How many seconds would be required for the initial concentration of \(\mathrm{Fe}^{3+}\) to decrease to each of the following values: \(25.0 \%\) left, \(12.5 \%\) left, \(6.25 \%\) left, \(3.125 \%\) left? What is the relationship between these times and the half-life?

Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q)$$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$\text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]{m}$$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n}\). Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming firstand then second-order kinetics. [Data from H. Fukutomi and G. Gordon, J. Am. Chem. Soc., 89,1362 (1967).] \(\begin{array}{cl}\text { Time (s) } & {\left[\text { ClO }_{2}\right] \text { (moll } \text { L) }} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6}\end{array}\)

You perform some experiments for the reaction \(\mathrm{A} \longrightarrow\) \(\mathrm{B}+\mathrm{C}\) and determine the rate law has the form $$\text { Rate }=k[\mathrm{~A}]^{x}$$ Calculate the value of exponent \(x\) for each of the following. a. \([\mathrm{A}]\) is tripled and you observe no rate change. b. [A] is doubled and the rate doubles. c. \([\mathrm{A}]\) is tripled and the rate goes up by a factor of \(27 .\)

Rate constants for reactions often follow the Arrhenius equation. Write this equation and then identify each term in it with the corresponding factor or factors from collision theory. Give a physical interpretation of each of those factors.

Benzene diazonium chloride, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl}\), decomposes by a first-order rate law. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{N}_{2}(g)$$ If the rate constant at \(20^{\circ} \mathrm{C}\) is \(4.3 \times 10^{-5} / \mathrm{s}\), how long will it take for \(75 \%\) of the compound to decompose?

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