A first order reaction is 20% complete in the 10 minutes. Calculate the time period for 75% completion of the reaction.

Explanation

We know that for first order reaction ,
`t = ( 2. 303 )/( k ) log "" (a)/( a -x )`
From (i) we have
`k = ( 2. 303 )/( 10 ) log "" ( 100)/( 80) = 0. 2003 log "" (10)/( 8)`
`= 0. 2303 [ log 10 - log 8]= 0. 2303 [1- log 2 ^(3) ]`
`= 0. 2303 [1-3 log 2] = 0.2303 [-3 xx 0.30 10 ] = 0.2303 [1- 0.9030]`
`or k = 0. 2303 xx 0.09 70 = 0.0223391 = 22. 33 91 xx 10 ^(-3) min ^(-1).`
Calculate of `t _( 75%)`
Here `a = 100, x = 75 a - x = 25`
`t _( 75%) = ( 2. 303 )/( 0. 022 33 91) log "" (100)/( 25) = ( 2. 303 xx 10 ^(3))/( 22. 33 91 ) log 2 ^(2) = 103. 09 xx 2 xx 0. 3010 = 62.06min.`