(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)`)

Explanation

(a) For first order reaction,
`AtoP`
`t=0" "a" "0`
`t=t" "(a-x)" "x`
Case 1 : If (t) is time required for 99% completion thenx=99% of a (a-x)=1%of
`a=1/100xxa=9/100`
`t=2.030/klog. (9)/(a-x)=2.303/klog. (9xx1000)/a=2.303/klog10^2`
`t=2[2.303/k]`
Cse 2 : If 't' is the required for 90% of completion then
`x=90% of a(a-x)`
`=10%" of " a=10/100a=9/100`
`t=2.303/klog. (9)/(a-x)=2.303/klog. (9.10)/a`
`t=2.303/k`

Therefore, the time required for 99% completion of 1st order reaction is twice the time required for 90% completion.
(b) `logk=logA-E_a/(2.303R)(1/T)`
`E_ato" activation energy"`
The above equation is like `y=mc+c` where if we plot `yv//sx` we get a straight line with slope 'm' and intercept 'c'.
So, slope is equal to `-E_a/(2.303R)`
`rArr (-E_a)/(2.303R)=-4250K`
`E_a=4250xx2.303xx8.314`
`E_a=81,375.35" "35 Jmol^(-1)`
`E_a=81.3753KJmol(-1)`