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A.$2 \times {10^{ - 4}}{s^{ - 1}}$

B.${10^{ - 4}}{s^{ - 1}}$

C.${10^{ - 6}}{s^{ - 1}}$

D.$4 \times {10^{ - 4}}{s^{ - 1}}$

Answer

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We have to know that the Arrhenius equation is useful in determining the rate of reaction and plays a significant portion in chemical kinetics. We could give Arrhenius equation as,

$K = A{e^{ - {E_a}/RT}}$

Here, K is the rate constant

A is the pre-exponential factor

${E_a}$ is the activation energy

R is the gas constant

T is the temperature (in Kelvin)

The integrated form of Arrhenius equation is,

$\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$

The value of temperature ${T_1}$ is $400K$ .

The value of temperature ${T_2}$ is $500K$ .

The rate constant at $400K$ is ${10^{ - 5}}{s^{ - 1}}$ .

We can write integrated equation as,

$\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$

Let us substitute the known values in the above expression to calculate the rate constant.

We can calculate the rate constant as,

$\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$

$ \Rightarrow 2.303\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$

Now we can substitute the known values we get,

$ \Rightarrow 2.303\log \left( {\dfrac{{{K_2}}}{{{{10}^{ - 5}}}}} \right) = 4606\left[ {\dfrac{1}{{400}} - \dfrac{1}{{500}}} \right]$

On simplifying we get,

${K_2} = {10^{ - 4}}{s^{ - 1}}$

The value of rate constant at $500K$ is ${10^{ - 4}}{s^{ - 1}}$.

We have to know that the fundamentals of the Arrhenius equation is collision theory. According to this theory, a reaction happens because of a collision between two molecules to form an intermediate. The formed intermediate is unstable and stays for a shorter period of time. The unstable intermediate gets broken into two product molecules and the energy used in formation of the intermediate is known as energy of activation. With the use of Arrhenius equation, we could determine the temperature, frequency, presence of catalyst, effect of energy wall and collision orientation.

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