What percentage of the initial concentration of reactant is left in 2.0 h for a reaction whose rate constant is `4.25 xx 10^(-5)" s"^(-1)` ?

Explanation

Let the initial concentration of the reactant, `a= 100" mol/L"`
Let, the concentration of the product after 2 h be .x. mol/L.
`therefore` The concentration of the reactant left after
`2h = (100-x)"mol/L"`
we know, `k = 4.25 xx 10^(-5)" s"^(-1)`
and `t=2 h=60 xx 60 xx 2 = 7200"s"`
According to first order rate expression
`k = (2.303)/( t) log"" (a)/( a-x)`
or `4.25 xx 10^(-5) = (2.303)/( 7200) log"" (100)/(100-x)" "...(i)`
or, `log "" (100)/(100-x) =0.1328" "...(ii)`
or `(100)/(100-x) =" antilog" 0.1328 = 1.358`

or `(100)/(100-x) = 1358" "...(iii)`
or `100-x= (100)/(1358) = 73.64`
or `x = 100 - 73.64 = 26.36%`