Question

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2h2o2(aq)⟶2h2o(l)+o2(g) the activation energy for this reaction is 75 kj·mol−1. the enzyme catalase, found in blood, lowers the activation energy to 8.0 kj·mol−1. at what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the enzyme-catalyzed reaction at 25 °c?

Answer 1

the temperature of the non- catalyzed reaction is = 2793.75 K

Step-by-step explanation:

The reaction of the spontaneous decomposition of hydrogen peroxide to give water and oxygen is given as:

`2H_2O_(2(aq)) ----> 2H_(2(l)) + O_2_((g))`

The activation energy of non-catalyzed reaction
`E{a_1} = 75 kJ/mol`

The activation energy of metal catalyzed reaction
`E{a_2} = 8 kJ/mol`

The temperature of metal catalyzed reaction
`T_2 = 25^0C` = (25+273)K = 298 K

The rate constant of the non-catalyzed reaction can be expressed as:

`k_1 = Ae^{{-Ea_1}/RT_1}` ----- equation (1)

The rate constant of the metal catalyzed reaction can be expressed as:

`k_2 = Ae^{{-Ea_2}/RT_2}`

Then
`k_1 = k_2`

`Ae^{{-Ea_1}/RT_1}=Ae^{{-Ea_2}/RT_2}`

`e^{{-Ea_1}/RT_1}=e^{{-Ea_2}/RT_2}`

`(Ea_1)/(RT_1)}=(Ea_2)/(RT_2)}`

`T_1 = (Ea_1*T_2)/(Ea_2)`

`T_1 = (75*298)/(8)`

`\\T_1 = 2793.75 \ K\\`

Thus; the temperature of the non- catalyzed reaction is = 2793.75 K