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The question is incomplete, complete question is :
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.
[tex]\frac{K_1}{K_2}=?[/tex]
Activation energy of the reaction 1 ,[tex]Ea_1[/tex] = 14.0 kJ/mol
Activation energy of the reaction 2,[tex]Ea_1[/tex] = 11.9 kJ/mol
Answer:
0.4284 is the ratio of the rate constants.
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 constant reaction -1
[tex]K_1[/tex] = rate constant reaction -2
Activation energy of the reaction 1 ,[tex]Ea_1[/tex] = 14.0 kJ/mol = 14,000 J
Activation energy of the reaction 2,[tex]Ea_1[/tex] = 11.9 kJ/mol = 11,900 J
R = gas constant = 8.314 J/ mol K
T = temperature = [tex]25^oC=273+25=298 K[/tex]
Now put all the given values in this formula, we get
[tex]\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340[/tex]
0.4284 is the ratio of the rate constants.
The activation energy is the minimum energy required by the substance to adds convert to the product. The ratio of the reaction rate constant for the two reactions is 2.3340.
What is the Arrhenius equation?
The Arrhenius equation contributes to the relationship between the activation energy and the rate constant for the reaction. The equation is given as:
[tex]K=A\;\times\;e^\frac{Ea}{RT}[/tex]
The activation energy for reaction 1 is 14 kJ/mol, and reaction 2 is 11.9 kJ/mol.
The ratio of the rate constant (K) for the reaction is given as:
[tex]\dfrac{K_1}{K_2} =\dfrac{A\;\times\;e^\frac{Ea1}{RT} }{A\;\times\;e^\frac{Ea2}{RT} }[/tex]
Simplifying the equation and substituting the values of activation energy:
[tex]\dfrac{K_1}{K_2} =e^\frac{Ea1-Ea2}{RT}\\ \\\dfrac{K_1}{K_2} =e^\dfrac{14,000-11900\;\text J}{8.314\;\text{J/mol.K}\;\times\;398\;\text K} \\\\\\\\\dfrac{K_1}{K_2} =2.3340[/tex]
The ratio of rate constant for the reaction is 2.3340.
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