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Show that time required for 99% completi...

Show that time required for `99%` completionis twice the time required for the completion of `90%` of reactions for a first order reaction.
The decomposition of hydrocarbon follow the equation
`K=(4.5xx10^(11)5^(-1))e^(-28000K//T)" Calculate Ea"`.

Text Solution

Verified by Experts

`"For first order reaction, t"=(2.303)/(R )log.(a_(0))/(a_(0))`
`rArr" "t_(99%)=2xxt_(90%)`
`k=Ae^(-Ea//RT)`
`(Ea)/(RT)=(28000)/(T )`
`E_(a)=28000xxR`
`=28000xx8.314`
`E_(a)=232.79" KJ mol"^(-1)`
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The decomposition of hydrocarbon follows the equation k=(4.5xx10^(11)s^(-1))e^(-28000K//T) Calculate E_(a) .

For a first order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of reaction.

Show that time required for the completion of 99% of the first order reaction is twice the 90% of completion of the reaction.

(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. (b) Rate constant 'K' of a reaction varies with temperature 'T' according to the equation : logK=logA-E_a/(2.303R)(1/T) Where E_a is the activation energy. When a graph is plotted for log k Vs 1/(T) a straight line with a slope of -4250 K is obtained. Calculate E_a for the reaction. ( R=8.314JK^(-1)mol^(-1) )

Knowledge Check

  • The time of completion of 90% of a first order reaction is approximately

    A
    1.1 times that of half-life
    B
    2.2 times that of half-life
    C
    3.3 times that of half - life
    D
    4.4 times that of half-life