Question

Consider two first-order reactions I and II. The frequency factor of I is 100 times that of II, the activation energy of I is 4.606 kcal higher than that of IL. If the ratio of rate constants for the reactions I and Il is x: 1 at 227°C, then the value of x is: (Given: R = 2 cal/mol K, e46 -0.01).​

Answer 1

Answer:

Explanation:

306 k

As we know,

k=0.693/t  

1/2

​

=0.693/2=0.3465/hr

and E  

a

​

=100kJ/mole

k=Ae  

−E  

a

​

/RT

 so T is 306K.

Answer 2

Answer:

The ratio of rate constants for the reactions I and reaction Il is 0.992.

The value of x is equal to 0.992.

Explanation:

We know that, Arrhenius equation:-

$k= A e^{-\frac{E_a}{RT} }$

Suppose that the frequency factor, activation energy and rate constant of reaction I are A₁, $E_{a1}$ and k₁. Similarly, the frequency factor, activation energy and rate constant of reaction II are A₂, $E_{a2}$ and k₂.

Given, the frequency factor of reaction I,  A₁ = 100 A₂

The activation energy of reaction I, $E_{a1}=E_{a2}+4.606 Kcal$

The gas constant, R = 2cal/mol K, T = 227 + 273 = 500K

Given, x is the ratio of rate constants for the reactions I and reaction Il.

$\frac{K_1}{K_2} =\frac{A_1}{A_2} \;\frac{e^{-\frac{E_{a1}}{RT} }}{e\frac{E_{a2}}{RT}}$

Substitute the value of k, A and activation energy in above equation:

$x= \frac{100A_2}{A_2}\;exp[\frac{1}{RT} (E_{a2}-E_{a1}) ]$

$x= 100 \;exp [\frac{E_{a2}-E_{a2}-4.606\times 10^{3}cal/mol}{2cal\;mol^{-1}K^{-1}\times 500K} ]$

$x= 100e^{-4.606}$

$x= \frac{100}{e^{4.606}}$

$x= \frac{100}{100.08}$

$x=0.992$

Therefore, the value of x is equal to 0.992.

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