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A popular chemical demonstration is the "magic genie" procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncatalyzed) reaction is \(70.0 \space\mathrm{kJ} / \mathrm{mol}\). When the catalyst is added, the activation energy (at \(20 .^{\circ} \mathrm{C}\) ) is \(42.0 \space\mathrm{kJ} / \mathrm{mol} .\) Theoretically, to what temperature \(\left(^{\circ} \mathrm{C}\right)\) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at \(20 .^{\circ} \mathrm{C} ?\) Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.

Short Answer

Expert verified
To find the temperature at which the rate of the uncatalyzed hydrogen peroxide decomposition reaction is equal to the rate of the catalyzed reaction at 20°C, we used the Arrhenius equation and the given activation energies. After converting the temperatures to Kelvin and plugging in the known values, we calculated that the hydrogen peroxide solution would have to be heated to approximately 212.53°C to achieve the desired reaction rates.

Step by step solution

01

Rearrange the equation for the unknown temperature, Tuncatalyzed

Since we know that the frequency factor A will remain constant for both the reactions, we can eliminate it by dividing both sides of the equation: \(e^{\frac{-E_\text{uncatalyzed}}{RT_\text{uncatalyzed}}} = e^{\frac{-E_\text{catalyzed}}{RT_\text{catalyzed}}}\) Now, we can rearrange for Tuncatalyzed: \(\frac{-E_\text{uncatalyzed}}{RT_\text{uncatalyzed}} = \frac{-E_\text{catalyzed}}{RT_\text{catalyzed}}\) \(\frac{E_\text{catalyzed}}{R} \cdot T_\text{uncatalyzed} = \frac{E_\text{uncatalyzed}}{R} \cdot T_\text{catalyzed}\) \(T_\text{uncatalyzed} = \frac{E_\text{uncatalyzed} \cdot T_\text{catalyzed}}{E_\text{catalyzed}}\)
02

Plug in the known values and solve for Tuncatalyzed

Now, we will substitute the known values into the equation: The given activation energies are Euncatalyzed = 70.0 kJ/mol and Ecatalyzed = 42.0 kJ/mol. To be consistent with the gas constant R we should convert these values to J/mol. Euncatalyzed = 70.0 kJ/mol × 1000 J/kJ = 70,000 J/mol Ecatalyzed = 42.0 kJ/mol × 1000 J/kJ = 42,000 J/mol \(T_\text{uncatalyzed} = \frac{(70,000\space\mathrm{J/mol})(293.15\space\mathrm{K})}{(42,000\space\mathrm{J/mol})}\) Calculating the temperature: \(T_\text{uncatalyzed} = 485.68\space\mathrm{K}\)
03

Convert the temperature back to Celsius

Finally, we convert the temperature from Kelvin back to Celsius: \(T_\text{uncatalyzed} = 485.68\space\mathrm{K} - 273.15 = 212.53 ^{\circ} \mathrm{C}\) Thus, the hydrogen peroxide solution would have to be heated to approximately 212.53°C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at 20°C.

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Key Concepts

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

Catalysis
Catalysis plays a crucial role in increasing the rate of chemical reactions. A catalyst is a substance that accelerates a reaction without being consumed in the process. It achieves this by providing an alternative pathway with a lower activation energy. Activation energy is the minimum energy required for a reaction to occur.

In the context of the hydrogen peroxide decomposition, adding a catalyst significantly reduces the activation energy from 70.0 kJ/mol to 42.0 kJ/mol. This decrease means that at a given temperature, the presence of a catalyst allows a greater number of reactant molecules to have enough energy to overcome the activation energy barrier.
  • Catalysts do not alter the equilibrium position, only the speed at which it is reached.
  • The catalyst is not used up; it participates temporarily and reappears unchanged.
  • Enzymes are biological examples of catalysts that speed up reactions in living organisms.

With the catalyst in place, the hydrogen peroxide decomposes much more swiftly, demonstrating the power of catalysis in making chemical processes more efficient.
Reaction Rate
The speed of a chemical reaction is described by its reaction rate. The reaction rate depends on several factors, including the nature of the reactants, concentration, temperature, and the presence of catalysts. In reactions like the decomposition of hydrogen peroxide, these factors can drastically change how quickly the reaction proceeds.

Temperature has a profound effect on reaction rates. As it increases, reactant molecules move faster, collide more frequently, and with greater energy. More frequent and energetic collisions make it more likely that the molecules can overcome the activation energy barrier, speeding up the reaction. Without a catalyst, temperature must be increased significantly to match the speed of a catalyzed reaction.
  • For the decomposition of hydrogen peroxide, the reaction rate increases with temperature.
  • The catalyst reduces the required energy to achieve a similar reaction rate at a lower temperature.
  • Understanding reaction rates is crucial in various applications, from industrial manufacturing to biological processes.

Effective manipulation of reaction rates is essential in chemical engineering, environmental technology, and pharmacology, among many other fields.
Hydrogen Peroxide Decomposition
Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) naturally decomposes into water (\(\text{H}_2\text{O}\)) and oxygen gas (\(\text{O}_2\)). This decomposition is an exothermic reaction, meaning it releases energy. However, the reaction is relatively slow without a catalyst due to its high activation energy.

The decomposition of hydrogen peroxide can be represented by the equation:\[2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\]Liquid hydrogen peroxide is unstable and can decompose over time if left in storage, releasing gas and energy. To speed up the process, a catalyst such as manganese dioxide (MnO2) or even an enzyme like catalase found in biological systems is often used.
  • Hydrogen peroxide is a common component in household disinfectants and bleaching agents.
  • The rate of decomposition increases at higher temperatures or with the presence of a catalyst.
  • Decomposition is a key consideration when storing hydrogen peroxide for extended periods.

Whether in laboratory settings or industrial applications, understanding how to control hydrogen peroxide decomposition is pivotal for safety and efficiency.

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Most popular questions from this chapter

Two isomers \((A \text { and } B)\) of a given compound dimerize as follows: $$\begin{aligned} &2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{A}_{2}\\\ &2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2} \end{aligned}$$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C} .\) In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} \mathrm{M}\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} \mathrm{M} .\) It was found that after each reaction had progressed for 3.00 min, \([\mathrm{A}]=3.00[\mathrm{B}] .\) In this case the rate laws are defined as $$\begin{array}{l} \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2} \\ \text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2} \end{array}$$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 min. b. Calculate the value of \(k_{2}\) c. Calculate the half-life for the experiment involving A.

Sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) decomposes to sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) by reaction in the gas phase. The following pressure data were obtained when a sample containing \(5.00 \times 10^{-2}\) mol sulfury 1 chloride was heated to \(600 . \mathrm{K}\) in a \(5.00 \times 10^{-1}-\mathrm{L}\) container. Defining the rate as $$-\frac{\Delta\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]}{\Delta t}$$ a. determine the value of the rate constant for the decomposition of sulfuryl chloride at \(600 .\) K. b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{h} ?\)

A first-order reaction is \(75.0 \%\) complete in \(320 .\) s. a. What are the first and second half-lives for this reaction? b. How long does it take for \(90.0 \%\) completion?

The rate law for the reaction $$2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ at some temperature is $$\text { Rate }=-\frac{\Delta[\mathrm{NOBr}]}{\Delta t}=k[\mathrm{NOBr}]^{2}$$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 \space M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to \(0.100 \space\mathrm{M} ?\)

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g)$$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{s}^{-1} ;\) at \(720 . \mathrm{K}, k=1.7 \times 10^{-2} \mathrm{s}^{-1}\) What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C},\) what is the pressure of iodoethane after three half-lives?

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