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Following is the graph between log T(50...

Following is the graph between log `T_(50)` and log a (a = initial concentration) for a given reaction at `27^(@)C`.

Hence, order is

A

1

B

2

C

3

D

0

Text Solution

Verified by Experts

The correct Answer is:
D
`t_(1//2)prop((1)/(a))^(n-1)" or "t_(1//2)=k(a)^(1-n)`
`logt_(1//2)=logk+(1-n)loga`
(It represents straight line equation, y = c + mx)
Slope `=(1-n)=tan45^(@)=1`
`(1-n)=1implies n=0`
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Knowledge Check

  • A graph between log t_(1//2) and log a (abscissa), a being the initial concentration of A in the reaction A rarr product, is depicted in the figure. The rate law is

    A
    `-(d[A])/(dt)=k`
    B
    `-(d[A])/(dt)=k[A]`
    C
    `-(d[A])/(dt)=k[A]^(2)`
    D
    `-(d[A])/(dt)=k[A]^(3)`

  • A graph between log t_((1)/(2)) and log a (abscissa), a being the initial concentration of A in the reaction For reaction Ato Product, the rate law is :

    A
    `-(d[A])/(dt)=K`
    B
    `-(d[A])/(dt)=K[A]`
    C
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    D
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  • A graph between log t_(1 // 2) and log a is shown below, where a is the initial concentration of A. For reaction A to Product, the rate law is

    A
    `(-d[A])/(dt)=k`
    B
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    C
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    D
    `(-d[A])/(dt)=k[A]^(3)`

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