(a) Explain the following terms :
(i) Rate of a reaction
(ii) Activation energy of a reaction
(b) The decompositon of phosphine , PH3, proceeds according to the following equation :
`4 PH_(3) (g) to P_(4)(g) + 6 H_(2) (g)`
It is found that the reaction follows the following rate equation :
Rate = k [`PH_(3)]`.
The half-life of ` PH_(3)" is "37.9" s at "120^(@)C`.
(i) How much time is required for `3//4^("th") "of " PH_(3)` to decompose ?
(ii) What fraction of the original sample of `PH_(3)` remains behind after 1 minute ?

Explanation

(a) (i) Rate of a reaction : The rate of a reaction can be defined as the change in concentration of a reactant or proudct in unit time.
(ii) Activation energy of a reaction : The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. It is denoted by Ea.
(b) (i) `t_(1//2) = 37.9 ` s
Let initial concentration = a
` :. x = 3/4 a`
Using the formula
`t=(2.303)/klog.a/(a-x)`
` t=((2.303)/(0.693))/(t_(1//2))log. a/(a-3/4a)" "[:'k=(0.693)/t_(1//2)]`
`t = 2.303 xxt_(1//2)/(0.693)log.q/(1/4a)`
` t = (2.303xxt_(1//2))/(0.693) xx log 4 =(2.303 xxt_(1//2)xx2log 2)/(0.693)`
`t = (2.303 xx37.9 xx0.3010)/(0.693)" "[:. t_(1//2) = 37.9s]`
` = (52.544)/(0.693) = 75. 82 sec`.
(ii) Here t = 1 min = 60s
` and t_(1//2) = 37.9s`
Using the formula
` t = (2.303)/ k log. ([A]_(0))/([A]_(t))`
` and k = ( 0.693)/(t_(1//2))`
` :. " " t = (2.303)/((0.693)/t_(1//2))log.[A]_(0)/[A]_(t)`
`t=2.303 xx t_(1//2)/(0.693) xx log.[A]_(0)/[A]_(t)`
` 60 = (2.303 xx37.9)/(0.693) log . [A]_(0)/[A]_(t) `
`:. " "log.[A]_(0)/[A]_(t)=(60 xx 0.693)/(2.303xx37.9) = 0.4768`
`[A]_(0)/[A]_(t) = anti-log(0.4768) = 2.997`
` [A]_(t)/[A]_(0) = 1/(2.997) = 0.3337 = 33.37 %`