Respuesta :
Factor would you expect the reaction rate to increase at 25 C is [tex]1.8*10^{12}[/tex]
Explanation:
Suppose that a catalyst lowers the activation barrier of a reaction from 125 kj/mol to 57 kj/mol . Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical: by what factor would you expect the reaction rate to increase at 25 C?
Catalysis is process of increasing the chemical reaction rate by adding a substance known as catalyst. Reaction rate is the speed at which a chemical reaction proceeds. Activation energy is the energy which must be provided to a chemical or nuclear system with potential reactants to result in chemical reaction, nuclear reaction, etc
The rates of reaction question in Kelvin [tex]25 C = 273 + 25 = 298 K[/tex]
The rate constants under catalysed and non‐catalysed conditions:
Catalysed: [tex]k_{cat} = A e^{\frac{-Ea(cat)}{RT} } = A e^{\frac{‐55000}{(8.314 x 298)}} = A e^{-22.2}[/tex]
Uncatalysed: [tex]k_{uncat} = A e^{\frac{-Ea(uncat)}{RT} } = A e^{\frac{‐125000}{8.314* 298} } = A e^{-50.4}[/tex]
Ratio of catalysed to uncatalysed reaction rates is
[tex]\frac{k_{cat}}{k_{uncat}} = \frac{Ae^{-22.2}}{Ae^{-50.4}}[/tex]
The A values are the same for both processes, therefore [tex]\frac{A}{A} = 1[/tex]
[tex]\frac{k_{cat}}{k_{uncat}} =\frac{e^_{-22.2}}{e^{-50.4}} = 1.8 * 10^{12}[/tex]
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The factor by which reaction rate is more in the presence of catalyst as compared to uncatalysed reaction is 1.8 × 10¹².
What is Arrhenius equation?
From the Arrhenius equation, we can calculate the effect of chnage in the activation energy as well as of temperature on the rate of the reaction.
Arrhenius equation will be represented as:
[tex]k = Ae^{-Ea/RT}[/tex], where
k = rate constant
A = frequency factor = 1 (given)
Ea = activation energy
R = universal gas constant = 8.314 J/ mol.K
T = temperature = 25 degree celsius = 298 K
- Rate of reaction in the presence of catalyst will be calculated as:
Activation energy = 57 kJ/mol = 57000 J/mol
[tex]k1 = e^{-57000/8.314\times 298}[/tex] = [tex]e^{-22} \\[/tex]
- Rate of reaction in the absence of catalyst will be calculated as:
Activation eneregy = 125 kJ/mol = 125000 J/mol
[tex]k2 = e^{-125000/8.314\times 298}[/tex] = [tex]e^{-50.4}[/tex]
Ratio of both the rates will be :
k1/k2 = [tex]e^{-22} \\[/tex] / [tex]e^{-50.4}[/tex] = 1.8 × 10¹²
Hence, the rate is 1.8 × 10¹² times more.
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