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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