Question

Hydrogen peroxide decomposes to water and oxygengas with the aid of a catalyst (MnO2). The activationenergy of the uncatalyzed reaction is 70.0 kJ/mol. Whenthe catalyst is added, the activation energy at 20.0 ^oC is42.0 kJ/mol. Theoretically, to what temperature (8C)would one have to heat the hydrogen peroxide solutionso that the rate of the uncatalyzed reaction is equal tothe rate of the catalyzed reaction at 20.0 ^oC? Assume thefrequency factor A is constant, and assume the initialconcentrations are the same.

Short answer

The temperature,of the hydrogen peroxide solution,so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reactionis 12^oC.

Step-by-step solution

 Explaining the hydrogen peroxide reaction to form water and oxygen

Two moles of Hydrogen peroxide (H₂O₂)decomposeto form water and oxygen gas while MnO₂ is used as a catalyst.

$2H_2{O}_2\to 2H_{2}O+ O_2$

The reaction activation energy is 42.0 kJ/mol.

Explaining the temperature to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction

Given Data

The reaction isunimolecular i.e. first order.

The activation energy of the catalyzed reactionis 42.0 kJ/mol.

The activation energy of the uncatalyzed reactionis 70.0 kJ/mol.

Final temperature= 20^oC

The pre-exponential factor $\left(A\right)$is constant over the reaction.

At 20⁰ C rate of catalysed reaction is equal to the rate of the catalysed reaction. i.e.

$$\begin{gathered} r_1= r_2 \\ {k}_1{{\left[H_{2}O2\right]}^1}_0=k_2{{\left[H_{2}O2\right]}^2}_0 \end{gathered}$$

We assume that the initial concentrations areconstant.

$k_1=k_2$

The Arrhenius equation is as follows:

$\ln k=\ln A-\frac{E}{RT}$

On the W catalyst surface with a rate constant $\left(k_1\right)$,

$\ln k_1=\ln A-\frac{E_1}{R{T}_1}......\left(1\right)$

On the W catalyst surface with a rate constant $\left(k_2\right)$

$\ln k_2=\ln A-\frac{E_2}{R{T}_2}......\left(2\right)$

Subtracting (2) From (1), we get

$\ln \left(\frac{k_1}{k_2}\right)=\frac{E_2}{R{T}_2}-\frac{E_1}{R{T}_1}$

But$k_1=k_2$,we get

$$\begin{gathered} \ln \left(1\right)=\frac{1}{R}\left(\frac{E_2}{T_2}-\frac{E_1}{T_1}\right) \\ 0=\frac{1}{R}\left(\frac{E_2}{T_2}-\frac{E_1}{T_1}\right) \end{gathered}$$

$$\begin{gathered} 0=\left(\frac{E_2}{T_2}-\frac{E_1}{T_1}\right) \\ \frac{E_2}{T_2}=\frac{E_1}{T_1} \\ \frac{70}{20}=\frac{42}{T_1} \\ {T}_1=12°C \end{gathered}$$

The temperature to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction is 12^oC.