Question

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 ( \((\mathrm{C})\) 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

To theoretically achieve the same rate for the uncatalyzed hydrogen peroxide decomposition reaction as the catalyzed reaction rate at 20°C, the temperature would need to be raised to approximately \(92.6^{\circ} \mathrm{C}\).

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation relates the rate constant (k) of a chemical reaction to its activation energy (Ea), temperature (T), and the pre-exponential factor (A) as follows: \[k = Ae^{\frac{-Ea}{RT}}\] where R is the gas constant (8.314 J/mol·K)

Set up the equation to relate the rates of uncatalyzed and catalyzed reactions

We're given the activation energies of the uncatalyzed (Ea1) and catalyzed reactions (Ea2), and we want the rates to be equal. Therefore, we'll equate the Arrhenius equations for each reaction, assuming the frequency factors and initial concentrations are the same: \[k_1 = k_2\] \[Ae^{\frac{-Ea1}{R(T+273)}} = Ae^{\frac{-Ea2}{R(20+273)}}\]

Solve for temperature

By equating the expressions in step 2, we can cancel out the pre-exponential factor (A) and solve for temperature (T) using the known values of the activation energies (Ea1 = 70.0 kJ/mol and Ea2 = 42.0 kJ/mol) and the gas constant R: \[e^{\frac{-Ea1}{R(T+273)}} = e^{\frac{-Ea2}{R(20+273)}}\] To solve for T, we'll first take the natural logarithm of both sides: \[\frac{-Ea1}{R(T+273)} = \frac{-Ea2}{R(20+273)}\] Next, we'll multiply both sides by -R and simplify: \[(T + 273)(Ea1) = (20 + 273)(Ea2)\] Now, solve for T: \[T = \frac{(20+273)(Ea2)}{Ea1} - 273\] Plug in the given values of Ea1 and Ea2: \[T = \frac{(20+273)(42.0 \mathrm{kJ/mol})}{70.0 \mathrm{kJ/mol}} - 273\]

Calculate the temperature

Calculate T using the values obtained in step 3: \[T = \frac{(293)(42.0 \mathrm{kJ/mol})}{70.0 \mathrm{kJ/mol}} - 273\] \[T \approx 92.6^{\circ} \mathrm{C}\] So, the temperature to which one would 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°C is approximately 92.6°C.

Key concepts

Activation Energy

In the realm of chemical reactions, activation energy is a pivotal concept. It represents the minimum energy required for reactants to transform into products. Think of it as the hill that reactants need to climb to start a reaction. Only molecules with enough energy can surpass this energy barrier to form products. The Arrhenius equation

• \[ k = Ae^{\frac{-Ea}{RT}} \]is utilized to associate the rate constant \( k \) of a reaction with its activation energy \( Ea \), temperature \( T \), and the pre-exponential factor \( A \).

This energy is typically measured in kilojoules per mole (kJ/mol).

In the original exercise, the uncatalyzed decomposition of hydrogen peroxide has a relatively high activation energy of 70.0 kJ/mol. This means that many reactant molecules don't have enough energy to react unless they are heated up significantly. Adding a catalyst reduces the activation energy to 42.0 kJ/mol, making it easier for the reaction to proceed at lower temperatures.
Understanding how activation energy influences reactions helps chemists manipulate and control reaction rates efficiently.

Catalysis

Catalysis is truly remarkable as it allows chemical reactions to occur more efficiently. A catalyst is a substance that speeds up a reaction without being consumed in the process. It achieves this by providing an alternative reaction pathway with a lower activation energy.

In essence, it reduces the energy required to reach the transition state, making it easier for reactant molecules to convert into products. Using the "magic genie" demonstration from the exercise as an example, when a catalyst is introduced, the activation energy of the hydrogen peroxide decomposition drops from 70.0 kJ/mol to 42.0 kJ/mol.
This significant drop in energy explains why the reaction becomes noticeably faster.

• Without a catalyst, reactions require more energy input to proceed.
• With a catalyst, less energy is needed, translating to faster rates.

What's important to note is that the catalyst does not alter the overall energy change of the reaction but strictly affects the rate. This is why catalysts are vital in industrial processes, helping to produce results quicker and more cost-effectively.

Chemical Kinetics

Chemical kinetics is the study of rates of chemical processes. It delves into how different variables, like concentration, temperature, and catalysts, influence the speed of reactions. Understanding chemical kinetics allows scientists to predict how a reaction will behave under certain conditions and to design experiments accordingly. At the heart of this field lies the Arrhenius equation, which models how temperature and activation energy impact reaction rates.

• Higher temperatures typically increase reaction rates because molecules have more energy and collide more frequently.
• The presence of a catalyst lowers activation energy, accelerating the reaction.

In practical applications, like the uncatalyzed versus catalyzed hydrogen peroxide decomposition, chemical kinetics informs us why adding a catalyst can make reactions viable at significantly lower temperatures.
This understanding is crucial in reducing energy consumption and optimizing reactions in both laboratory and industrial settings.
By studying chemical kinetics, chemists gain the ability to control and manipulate reaction conditions to achieve desired outcomes effectively.