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Chapter 12: Problem 27

At \(40^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}_{2}(a q)\) will decompose according to the following reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g}) $$ The following data were collected for the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at various times. $$ \begin{array}{|cc|} \hline \begin{array}{c} \text { Time } \\ (\mathbf{s}) \end{array} & \begin{array}{c} {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 0 & 1.000 \\ \hline 2.16 \times 10^{4} & 0.500 \\ \hline 4.32 \times 10^{4} & 0.250 \\ \hline \end{array} $$ a. Calculate the average rate of decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) between 0 and \(2.16 \times 10^{4} \mathrm{~s}\). Use this rate to calculate the average rate of production of \(\mathrm{O}_{2}(g)\) over the same time period. b. What are these rates for the time period \(2.16 \times 10^{4} \mathrm{~s}\) to \(4.32 \times 10^{4} \mathrm{~s} ?\)

Short Answer

Expert verified
a) The average rate of decomposition of \(H_2O_2\) between 0 and \(2.16 \times 10^4\) seconds is \(-2.31 \times 10^{-5}\, mol/L/s\). The average rate of production of \(O_2\) during this time period is \(1.16 \times 10^{-5}\, mol/L/s\). b) The average rate of decomposition of \(H_2O_2\) between \(2.16 \times 10^4\) and \(4.32 \times 10^4\) seconds is \(-1.16 \times 10^{-5}\, mol/L/s\). The average rate of production of \(O_2\) during this time period is \(5.79 \times 10^{-6}\, mol/L/s\).

Step by step solution

01

Calculate the rate of decomposition of H2O2 in the first time interval

According to the data given: Initial concentration of H2O2: \([H_2O_2]_i = 1.000 \, mol/L\) Initial time: \(t_i = 0\,s\) Final concentration of H2O2: \([H2O2]_f = 0.500 \, mol/L\) Final time: \(t_f = 2.16 \times 10^4 \, s\) Now we can apply the formula for the average rate of reaction: Average rate = \(\frac{[H_2O_2]_f - [H_2O_2]_i}{t_f - t_i}\)
02

Calculate the average rate of production of O2 in the first time interval

To find the average rate of production of O2, analyze the stoichiometry of the reaction: \(2H_2O_2(aq) \longrightarrow 2H_2O(l) + O_2(g)\) The ratio of H2O2 to O2 according to the reaction coefficients is 2:1. Thus, the rate of production of O2 gas will be half the rate of H2O2 decomposition.
03

Calculate the rates for the second time interval

Use the same approach to find the rates for the second time interval (2.16 x 10^4 s to 4.32 x 10^4 s). Initial concentration of H2O2: \([H_2O_2]_i = 0.500\, mol/L\) Initial time: \(t_i = 2.16 \times 10^4\,s\) Final concentration of H2O2: \([H2O2]_f = 0.250 \, mol/L\) Final time: \(t_f = 4.32 \times 10^4 \, s\) Apply the formula for the average rate of decomposition and use stoichiometry to calculate the rate of production of O2 in the second time interval, following the same steps as in the first time interval.

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Key Concepts

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

Rate of Reaction
The rate of reaction refers to how quickly or slowly the reactants in a chemical reaction form into products. It's a measure of the speed at which the concentration of a reactant or product changes over time. For our exercise on the decomposition of hydrogen peroxide (\(\text{H}_2\text{O}_2\)), the rate of reaction tells us how quickly \(\text{H}_2\text{O}_2\) is breaking down into water and oxygen gas.
  • The average rate of reaction can be calculated using the formula:\[ \text{Average rate} = \frac{[\text{H}_2\text{O}_2]_f - [\text{H}_2\text{O}_2]_i}{t_f - t_i}\]
In a practical sense, this means taking the difference in concentration of \(\text{H}_2\text{O}_2\) at two different times, then dividing by the time difference to give a rate in mol/L/s. This tells us how fast the concentration of reactants is changing, and thus, how fast the reaction is occurring. The rates will vary depending on different time intervals as the reaction proceeds.
Stoichiometry
Stoichiometry is a concept that deals with the quantitative relationships of the reactants and products in a chemical reaction. It's all about understanding how much of something is needed or produced based on balanced chemical equations.
  • In this decomposition reaction:\(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\), the stoichiometry is 2:2:1.
  • This means that two moles of \(\text{H}_2\text{O}_2\) decompose to form two moles of water (\(\text{H}_2\text{O}\)) and one mole of oxygen (\(\text{O}_2\)).
Understanding stoichiometry allows us to relate the rates of disappearance of reactants to the rates of appearance of products. In this case, because the ratio of \(\text{H}_2\text{O}_2\) to \(\text{O}_2\) is 2:1, the rate of oxygen gas production is half the rate of hydrogen peroxide decomposition. Stoichiometry ensures we can precisely calculate these relationships.
Decomposition Reaction
A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. It is essentially the opposite of composition or synthesis reactions.
  • The reaction of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) decomposing into water and oxygen can be seen as:\(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\).
  • This is a classic example of a decomposition reaction. The reactant \(\text{H}_2\text{O}_2\) breaks down to form two different products: water (\(\text{H}_2\text{O}\)) and oxygen gas (\(\text{O}_2\)).
Decomposition reactions generally require an input of energy in the form of heat, light, or electricity. In this exercise, the reaction occurs naturally at a moderate temperature of \(40^\circ\text{C}\). Understanding decomposition reactions is essential because they help explain processes such as the decay of organic matter and various industrial applications.

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Most popular questions from this chapter

A reaction of the form $$ \mathrm{aA} \longrightarrow $$ gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach 22.9\(\%\) completion.

Consider the hypothetical reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g),\) where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{L} / \mathrm{mol}\) \(\mathrm{s},\) and at \(508^{\circ} \mathrm{C}\) the rate constant is 0.891 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) . What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C} ?\)

The activation energy for a reaction is changed from 184 \(\mathrm{kJ} /\) \(\mathrm{mol}\) to 59.0 \(\mathrm{kJ} / \mathrm{mol}\) at \(600 . \mathrm{K}\) by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 \(\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at \(37^{\circ} \mathrm{C}\) the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.
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