# Question f875f

Feb 12, 2017

We know that the equation of rate constant for 1st oder reaction is as follows

$\textcolor{b l u e}{k = \frac{1}{t} \ln \left({C}_{0} / {C}_{t}\right) \ldots \ldots \left[1\right]}$

where

$k \to \text{rate constant}$
$t \to \text{the time for which change being observed}$
${C}_{0} \to \text{concentration of the reactant at t=0}$

${C}_{t} \to \text{concentration of the reactant at t th instant}$

For 50% completion of the reaction time is 30 min

so for $t = 30 \min \text{ } {C}_{0} / {C}_{t} = \frac{100}{50} = 2$

So inserting these in [1] we get

$k = \frac{1}{30} \ln 2. \ldots . . \left[2\right]$

Let the reaction becomes 90% complete in $T \min$.This means 10% reactant remains.
So ${C}_{0} / {C}_{T} = \frac{100}{10} = 10$

The equation [1] then becomes

$k = \frac{1}{T} \ln \left(10\right) \ldots \ldots \ldots \left[3\right]$

Comparing [2] and [3] we get

$\frac{1}{30} \ln 2 = \frac{1}{T} \ln \left(10\right)$

=>T=(30ln10)/ln2~~99.65 min#