Answer:
The rate of the catalyzed reaction will increase by a 1.8 × 10⁵ factor.
Explanation:
The rate of a reaction (r) is proportional to the rate constant (k). We can find the rate constant using the Arrhenius equation.
[tex]k=A.e^{-Ea/R.T}[/tex]
where,
A: collision factor
Ea: activation energy
R: ideal gas constant
T: absolute temperature (125°C + 273 = 398 K)
For the uncatalized reaction,
[tex]kU=A.e^{-95\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=3.4\times 10^{-13}A[/tex]
For the catalized reaction,
[tex]kC=A.e^{-55\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=6.0\times 10^{-8}A[/tex]
The ratio kC to kU is 6.0 × 10⁻⁸A/3.4 × 10⁻¹³A = 1.8 × 10⁵
The catalyzed reaction is increased by the factor of 1.8 × 10⁵. The rate of reaction is directly proportional to the rate constant.
The rate of reaction can be calculated by the Arrhenius equation.
[tex]k = A.e^{-Ea/R.T}[/tex]
Where,
[tex]A[/tex] - collision factor
[tex]Ea[/tex] - activation energy
[tex]R[/tex] - ideal gas constant
[tex]T[/tex]- absolute temperature 125°C = 398 K
For the uncatalyzed reaction:
[tex]k_u = A.e^{95/8.314\times398}\\\\k_u = 3.4\times 10^{-13}[/tex]
For the catalyzed reaction,
[tex]k_c = A.e^{55/8.314\times398}\\\\k_c = 6\times 10^{-8}[/tex]
After calculating the ratio of catalyzed and uncatalyzed reaction is 1.8 × 10⁵.
Therefore, the catalyzed reaction is increased by the factor of 1.8 × 10⁵.
Learn more about activation energy:
https://brainly.com/question/864515
