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When the temperature is increased, heat ...

When the temperature is increased, heat is supplied which increases the kinetic energy of the reacting molecules. This shall increase the number of collisions and ultimately the rate of reaction shall be enhanced. Arrhenius suggested a equation which describes `K` as a function of temperature, i.e.
`k=Ae^(-E_(a)//RT)`
where
`k=` rate constant
`A=` a constant (frequency factor)
`E_(a)=` energy of activation
`log_(10)k=log_(10)A-(E_(a))/(2.303R)[(1)/(R )]`
`Y=C+MX`
It is the equation of straight line with negative slope. On plotting `log_(10)k` against `[(1)/(T)]` we get a straight line as shown below :

The slope gives activation energy and intercept gives frequency factor.
Also `log.(k_(2))/(k_(1))=(E_(a))/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))]`
The rate of certain reaction increases by `2.5` times when the temperature is raised from `300K` to `310K`. If `k` is the rate constant at `300K` then rate constant at `310K` will be equal to

A

`k`

B

`2k`

C

`2.5k`

D

`13k`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • When the temperature is increased, heat is supplied which increases the kinetic energy of the reacting molecules. This shall increase the number of collisions and ultimately the rate of reaction shall be enhanced. Arrhenius suggested a equation which describes K as a function of temperature, i.e. k=Ae^(-E_(a)//RT) where k= rate constant A= a constant (frequency factor) E_(a)= energy of activation log_(10)k=log_(10)A-(E_(a))/(2.303R)[(1)/(R )] Y=C+MX It is the equation of straight line with negative slope. On plotting log_(10)k against [(1)/(T)] we get a straight line as shown below : The slope gives activation energy and intercept gives frequency factor. Also log.(k_(2))/(k_(1))=(E_(a))/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))] The activation energy of a reaction is 9 kcal/mol. The increase in rate constant when its temperature is raised from 295 to 300K is approximately

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    `10%`
    B
    `50%`
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    `100%`
    D
    `255%`

  • When the temperature is increased, heat is supplied which increases the kinetic energy of the reacting molecules. This shall increase the number of collisions and ultimately the rate of reaction shall be enhanced. Arrhenius suggested a equation which describes K as a function of temperature, i.e. k=Ae^(-E_(a)//RT) where k= rate constant A= a constant (frequency factor) E_(a)= energy of activation log_(10)k=log_(10)A-(E_(a))/(2.303R)[(1)/(R )] Y=C+MX It is the equation of straight line with negative slope. On plotting log_(10)k against [(1)/(T)] we get a straight line as shown below : The slope gives activation energy and intercept gives frequency factor. Also log.(k_(2))/(k_(1))=(E_(a))/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))] The rate constant of a reaction is increased by 5% when its temperature is raised from 27^(@)C to 28^(@)C . The activation energy of the reaction in kJ/mol is

    A
    `36.6`
    B
    `16.6`
    C
    `46.6`
    D
    `26.6`

  • When the temperature is increased, heat is supplied which increases the kinetic energy of the reacting molecules. This shall increase the number of collisions and ultimately the rate of reaction shall be enhanced. Arrhenius suggested a equation which describes K as a function of temperature, i.e. k=Ae^(-E_(a)//RT) where k= rate constant A= a constant (frequency factor) E_(a)= energy of activation log_(10)k=log_(10)A-(E_(a))/(2.303R)[(1)/(R )] Y=C+MX It is the equation of straight line with negative slope. On plotting log_(10)k against [(1)/(T)] we get a straight line as shown below : The slope gives activation energy and intercept gives frequency factor. Also log.(k_(2))/(k_(1))=(E_(a))/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))] The rate of chemical reaction doubles for every 10^(@)C rise in temperature . If the temperature is increased by 60^(@)C , the rate of reaction increases by about

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    `20` times
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    `32` times
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    D
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