Graph between log k and 1//T [k rate con...

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

`2.303 xx 2 cal`

`2//2.303 cal`

`2 cal`

None of these

Text Solution

Verified by Experts

The correct Answer is:

`log_(10)K vs.(1)/(T)` is linear,
slope `= (-E_(A))/(2.303 R)` intercept `= log_(10)A`
Since, `|tan theta| = (1)/(2.303)`
`:. (E_(A))/(2.303R ) = (1)/(2.303) rArr E_(a) = R = 2 cal`

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Graph between log k and 1 // T [k is rate constant ( s^(-1) ) and T the temperature (K)] is a straight line with OX = 5, theta = tan^(-1)(1 // 2.303) . Hence- E_(a) will be

`2.303 xx 2` cal

`2 // 2.303` cal

2 cal

none

Graph between log k and 1//T [where K is rate constant in s^(-1) and T is the temperature (in K) is a straight line with Hence, E_(a) will be

`2.303 xx 2` cal

`2/(2.303) cal`

2 cal

None of these

Graph between log k and (1)/(T) (k is rate constant in s^(-1) and T is the temperature in K) is a straight line. As shown in figure if OX = 5 and slope of the line =- (1)/(2.303) then E_(a) is :

`2.303 xx 2 cal`

`(2)/(2.303) cal`

2 cal

none of these

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

Text Solution

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Knowledge Check

Graph between log k and 1 // T [k is rate constant ( s^(-1) ) and T the temperature (K)] is a straight line with OX = 5, theta = tan^(-1)(1 // 2.303) . Hence- E_(a) will be

Graph between log k and 1//T [where K is rate constant in s^(-1) and T is the temperature (in K) is a straight line with Hence, E_(a) will be

Graph between log k and (1)/(T) (k is rate constant in s^(-1) and T is the temperature in K) is a straight line. As shown in figure if OX = 5 and slope of the line =- (1)/(2.303) then E_(a) is :

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A2Z-CHEMICAL KINETICS-Arrhenius Equation, Effect Of Temprature And Effect Of Catalysts

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

Text Solution

Topper's Solved these Questions

Similar Questions

Knowledge Check

Graph between log k and 1 // T [k is rate constant ( s^(-1) ) and﻿ T the temperature (K)] is a straight line with OX = 5,﻿ theta = tan^(-1)(1 // 2.303) . Hence- E_(a) will be

Graph between log k and 1//T [where K is rate constant in s^(-1) and T is the temperature (in K) is a straight line with Hence, E_(a) will be

Graph between log k and (1)/(T) (k is rate constant in s^(-1) and T is the temperature in K) is a straight line. As shown in figure if OX = 5 and slope of the line =- (1)/(2.303) then E_(a) is :

Similar Questions

A2Z-CHEMICAL KINETICS-Arrhenius Equation, Effect Of Temprature And Effect Of Catalysts

Graph between log k and 1 // T [k is rate constant ( s^(-1) ) and T the temperature (K)] is a straight line with OX = 5, theta = tan^(-1)(1 // 2.303) . Hence- E_(a) will be