# For a reaction consider the plot of $\ln K$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400K$ is ${10^{ - 5}}{s^{ - 1}}$ , then the rate constant at $500K$ is:

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

Verified

174.9k+ views

**Hint:**We have to know that the rate constant could be calculated using the Arrhenius equation. We have to derive the rate constant using the rate constant at $400K$, both the temperatures and value of slope. The value of slope is $ - 4606K$.

**Complete step by step answer:**

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

**Therefore, the option (B) is correct.**

**Note:**

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE