Question

The activation energies are 45. 3 kj mol^-1 for k1 and 69. 8 kj mo1-1 for k2. If the rate constants are equal at 320 k, at what temperature will k1/k2 = 2. 00?

Answer 1

At temperature 298K, the value of k1/k2 will equal to 2.00

According to Arrhenius's equation

Sometimes the Arrhenius equation is written as k = Ae-$^{E/RT}$, where k is the rate of the chemical reaction, A is a constant that varies depending on the chemicals involved, E is the activation energy, and R is the universal gas constant, and T is the temperature.

$\frac{k1}{k2} = \frac{A_{1} e^{-E_{a1}/ RT} }{A_{2} e^{-E_{a2}/ RT} }$

    = $\frac{A1}{A2} e^{(E_{a2}-E_{a1)}/RT }$

 

Given

Eₐ₂ = 69.8 × 10³ J/mol

Eₐ₁ = 45.3 × 10³ J/mol

R = 8.314J/kmol³

Now if the rate constant is equal means K1/K2 = 1 at T = 320 K

So if we put the values of K1/K2, Ea2,Ea1, T, and R into the formula

On solving

We get

A1/A2 = 1.001 × 10⁻⁴

Now when K1/K2 = 2

Then if we put the value of A1/A2, Ea1, Ea2, R, K1/K2 in the same equation,

On solving we get

T = 298 K

Hence, At temperature 298K k1/k2 will equal to 2.00

Learn more about Arrhenius's equation here  brainly.com/question/26724488

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