The activation energy of an uncatalyzed reaction is 95 kJ / mol. The addition of a catalyst lowers the activation energy to 55 kJ / mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at125∘C?

Respuesta :

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.

How to calculate the rate of reaction?

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.

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