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The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ ⋅ mol − 1 51.00 kJ⋅mol−1 . Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ ⋅ mol − 1 Ea=136.00 kJ⋅mol−1 Catalyzed: A ⟶ B A⟶B E a = 85.00 k J ⋅ mol − 1 Ea=85.00 kJ⋅mol−1

Respuesta :

The given question is incomplete.The complete question is:

The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ /mol Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ/mol Catalyzed: A ⟶ B A⟶B E a = 85.00 k J/mol. determine the factor by which tha catalysed reaction is faster than the uncatalysed reaction at 289.0 K if all other factors are equal.

Answer: The factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is [tex]1.64\times 10^9[/tex]

Explanation:

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

The expression used with catalyst and without catalyst is,

[tex]\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}[/tex]

[tex]\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}[/tex]

where,

[tex]K_2[/tex] = rate of reaction with catalyst

[tex]K_1[/tex] = rate of reaction without catalyst

[tex]Ea_2[/tex] = activation energy with catalyst

[tex]Ea_1[/tex] = activation energy without catalyst

R = gas constant = [tex]8.314\times 10^{-3}kJ/Kmol[/tex]

T = temperature = [tex]289.0K[/tex]

Now put all the given values in this formula, we get

[tex]\frac{K_2}{K_1}=e^{\frac{51.00}{8.314\times 10^{-3}\times 289.0}}[/tex]

[tex]\frac{K_2}{K_1}=e^{21.22}[/tex]

[tex]\frac{K_2}{K_1}=1.64\times 10^9[/tex]

Therefore, the factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is [tex]1.64\times 10^9[/tex]

The factor by which the catalyzed reaction is faster than the uncatalyzed reaction at 289.0 K is; 1.64 * 10⁹

How to find the factor of a Catalyzed Reaction?

Arrhenius equation for rate constant is expressed as;

K = Ae^(-E_{a}/RT)

where;

K is rate constant

A is pre-exponential factor

E_{a} is activation energy (in the same units as R*T)

R is universal gas constant

T is absolute temperature (in Kelvin)

However, with Catalyst and without catalyst, the formula is;

K₂/K₁ = e^((-E_{a}₁ - E_{a}₂)/RT)

Where;

K₂ = rate of reaction with catalyst

K₁ = rate of reaction without catalyst

E_{a}₂ = activation energy with catalyst

E_{a}₁ = activation energy without catalyst

R = gas constant = 8.314 * 10⁻³ KJ/K.Mol

T = temperature = 289.0 K

Thus;

K₂/K₁ = e^(51/(8.314 * 10⁻³ * 289))

K₂/K₁ = 1.64 * 10⁹

Read more about Catalyzed reaction factor at; https://brainly.com/question/25752442

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