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Chapter 11: Problem 81

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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Most popular questions from this chapter

One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the following reaction: $$\mathrm{CCl}_{2} \mathrm{F}_{2}(g) \stackrel{h v}{\longrightarrow} \mathrm{CF}_{2} \mathrm{Cl}(g)+\mathrm{Cl}(g)$$ Chlorine atoms can act as a catalyst for the destruction of ozone. The activation energy for the reaction $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)$$is \(2.1 \mathrm{kJ} / \mathrm{mol} .\) Which is the more effective catalyst for the destruction of ozone, Cl or NO? (See Exercise 75.)

The reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of \(\mathrm{O}_{3}\). The results were as follows \(\left(\left[\mathrm{O}_{3}\right]\right.\) remains effectively constant at \(1.0 \times 10^{14}\) molecules/cm \(^{3}\) ): In the second experiment [NO] was held constant at \(2.0 \times 10^{14}\) molecules/cm \(^{3}\). The data for the disappearance of \(\mathbf{O}_{3}\) are as follows: a. What is the order with respect to each reactant? b. What is the overall rate law? c. What is the value of the rate constant from each set of experiments? $$\text { Rate }=k^{\prime}[\mathrm{NO}]^{x} \quad \text { Rate }=k^{\prime \prime}\left[\mathrm{O}_{3}\right]^{y}$$ d. What is the value of the rate constant for the overall rate law? $$\text { Rate }=k[\mathrm{NO}]^{\mathrm{x}}\left[\mathrm{O}_{3}\right]^y$$

Consider two reaction vessels, one containing A and the other containing \(\mathrm{B},\) with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$\begin{aligned} &k_{A}=4.50 \times 10^{-4} \mathrm{s}^{-1}\\\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1} \end{aligned}$$how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{B}] ?\)

The activation energy for some reaction $$\mathrm{X}_{2}(g)+\mathrm{Y}_{2}(g) \longrightarrow 2 \mathrm{XY}(g)$$ is \(167 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(+28 \mathrm{kJ} / \mathrm{mol} .\) What is the activation energy for the decomposition of XY?

Hydrogen reacts explosively with oxygen. However, a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) can exist indefinitely at room temperature. Explain why \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) do not react under these conditions.
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