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Slope of the straight line obtained by p...

Slope of the straight line obtained by plotting `log_(10)` k against 1/T represents what term ?

A

`-E_a`

B

`-2.303E_a//R`

C

`-E_a//(2.303R)`

D

`-E_a//R`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • Slope of the straight line obtained by plotting log_(10)K aainst 1/T represents what term ?

    A
    `-E_a`
    B
    `-2.303 E_(a)//R`
    C
    `-E_(a)//2.303 R`
    D
    `E_a //R`

  • The slope of the straight line obtained by plotting rate versus concentration of reactant in a first order reaction is

    A
    `-k`
    B
    `(-k)/(2.303)`
    C
    `(k)/(2.303)`
    D
    `k`

  • 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`

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