A first order reaction is 25% complete in one hour . At the end of two hours the extent of reaction is :

Options

A. 0.5
B. 0.3125
C. 0.4375
D. 0.75

Answer

C

Explanation

(C ) `k = (2.303)/(t) "log"([A]_(0))/([A])`
`k = (2.303)/(1) "log" ([A]_(0))/(0.75[A]_(0))`
`= (2.303)/(1) "log" 1.33`
`=(2.303)/(1) xx0.12876`
Now `t = (2.303)/(k) "log" ([A]_(0))/([A])`
`2 = (2.303)/(0.2876)"log" ([A]_(0))/([A])`
log `([A]_(0))/([A])= (2xx0.2876)/(2.303)=0.2497 `
or `([A])/([A]_(0))=1.77 `
`([A]_(0))/([A])= (1)/(1.77) == 56 .35 % `
`:. ` Extent of reaction
`= 100 -56.3 = 43. 75% `