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First order reaction, A to B requires ac...

First order reaction, `A to B` requires activation energy of `70" kJ mol"^(-1)`. When a 20% solution of A was kept at 25°C for 20 min. 25% decomposition took place. What will be percentage decomposition in the same time in a 30% solution maintained at 40°C? Assume that activation energy remains constant in this range of temperature.

Text Solution

Verified by Experts

We know that the fraction of reactants reacted in a first order reaction is independent of initial concentration. So, we need not worry about the terms 20% solution and 30% solution.
For first order reaction, `k = (2.303)/( t) log"" (a)/( a-x)`
Given, `t= 20" min", a = 100 , (a-x) = 100 - 25 = 75`
`therefore " "k = (2.303)/( 20) log"" (100)/( 75) = 1.44 xx 10^(-2)" min"^(-1)`
Let, `k_2` be the rate constant of this reaction at 40°C. According to Arrhenius equation,
`log"" (k_2)/( k_1) = (E_a)( 2.303 R) [(1)/(T_1) - (1)/(T_2) ]`
`log k_2 = (70 xx 30^(3) "J mol"^(-1) )/( 2.303 xx 8.314" JK"^(-1)" mol"^(-1) ) [ (1)/(298"K") - (1)/(313"K") ]`
`k_2 = 5.56 xx 10^(-2)" min"^(-1)`
Now, again using first order reaction
`log"" (a)/(a-x) = (k)/( 2.303), t = (5.56 xx 10^(-2)" min"^(-1) )/( 2.303) xx 20" min" = 0.33`
`therefore` Percentage of A remaining `= 33%`
Percentage of A reacted `=100-33= 67%`
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