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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

A

`2.303 xx 2 cal`

B

`2//2.303 cal`

C

`2 cal`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
`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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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

    A
    `2.303 xx 2` cal
    B
    `2 // 2.303` cal
    C
    2 cal
    D
    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

    A
    `2.303 xx 2` cal
    B
    `2/(2.303) cal`
    C
    2 cal
    D
    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 :

    A
    `2.303 xx 2 cal`
    B
    `(2)/(2.303) cal`
    C
    2 cal
    D
    none of these
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