Question

Part 1. When we are given, concentration decreased to 27 % from the initial concentration then , t = ? We know, k = 0.17 min-1 , [HOF]o = 0.54 M

So at 27 % , [HOF]t = 0.146 M

4.60*102 s this is the answer but I do not know how .146M was calculated. What is the formula (it is first order)

Answer 1

Given

k = 0.17 min-1 , [HOF]o = 0.54 M

[HOF]t = 0.146 M

For first order reaction, use the equation

ln Ct/Co=-kt

where Ct=concentration at time=t

Co=initial concentration ,at time=0

K=rate constant

Plug in the given values,

ln (0.146M/0.54M)=-0.17 min-1 *t

or,ln 0.27=-0.17 min-1 *t

or,-1.31=0.17 min-1 *t

or,t=1.3/0.17=7.65 min=7.65*60s=458.82S=4.6*10^2S

alternate way of solving,

Ct=0.27Co

ln (0.27Co/Co)=-0.17 min-1 *t

or, ln 0.27=-0.17 min-1 *t

or,-1.3=-0.17 min-1 *t

or,t=1.3/0.17=7.65 min=7.65*60s=458.82S=4.6*10^2S