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 \(70.0 \mathrm{~kJ} / \mathrm{mol}\). When the catalyst is added, the activation energy (at \(20 .{ }^{\circ} \mathrm{C}\) ) is \(42.0 \mathrm{~kJ} / \mathrm{mol}\). 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.

We will use the Arrhenius equation for the catalyzed reaction. \[k_{catalyzed} = Ae^{(-\frac{E_{a_{catalyzed}}}{RT_{catalyzed}})}\] Given the activation energy of the catalyzed reaction \(E_{a_{catalyzed}} = 42.0 \mathrm{~kJ} / \mathrm{mol}\), and the temperature \(T_{catalyzed} = 20°C + 273.15 = 293.15K\). The gas constant can be expressed as \(R = 8.314 \mathrm{~J} / (\mathrm{mol} \cdot \mathrm{K})\). We can substitute the known values into the Arrhenius equation: \[k_{catalyzed} = Ae^{(-\frac{42000}{8.314 \cdot 293.15})}\] Since we will be comparing with the uncatalyzed reaction, we will keep k (the reaction rate) and A (the frequency factor) in the equation as variables.

Similarly, we will use the Arrhenius equation for the uncatalyzed reaction. \[k_{uncatalyzed} = Ae^{(-\frac{E_{a_{uncatalyzed}}}{RT_{uncatalyzed}})}\] Given the activation energy of the uncatalyzed reaction \(E_{a_{uncatalyzed}} = 70.0 \mathrm{~kJ} / \mathrm{mol}\), and the unknown temperature is \(T_{uncatalyzed}\). We can write the equation for the uncatalyzed reaction: \[k_{uncatalyzed} = Ae^{(-\frac{70000}{8.314 \cdot T_{uncatalyzed}})}\]

Now, we equate the two reaction rates: \[k_{catalyzed} = k_{uncatalyzed}\] \[Ae^{(-\frac{42000}{8.314 \cdot 293.15})} = Ae^{(-\frac{70000}{8.314 \cdot T_{uncatalyzed}})}\] Since we are given that the A is constant in the problem, we can cancel it from both sides: \[e^{(-\frac{42000}{8.314 \cdot 293.15})} = e^{(-\frac{70000}{8.314 \cdot T_{uncatalyzed}})}\] Now, we can take the natural logarithm of both sides of the equation to get: \[-\frac{42000}{8.314 \cdot 293.15} = -\frac{70000}{8.314 \cdot T_{uncatalyzed}}\] Solve the equation for \(T_{uncatalyzed}\): \[\frac{70000}{42000} = \frac{293.15}{T_{uncatalyzed}}\] \[T_{uncatalyzed} = \frac{293.15}{\frac{70000}{42000}}\] \[T_{uncatalyzed} = 418.941K\] Now, convert it back to Celsius: \(T_{uncatalyzed} (°C) = T_{uncatalyzed} (K) - 273.15 = 418.941 - 273.15 = 145.79 °C\) Therefore, one would have to heat the hydrogen peroxide solution to approximately\(145.79^{\circ} \mathrm{C}\) so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at \(20^{\circ} \mathrm{C}\).