Question

If the rate constant for a first order reaction is k, the time required for the completion of 99% of the reaction is given by :
(1) t = 0.693/k
(2) t = 6.909/k
(3) t = 4.606/k
(4) t = 2.303/k

Answer 1

Answer:

Answer is 3 ,see workings below

Explanation:

Using this formula

$k = \frac{1}{t} ln( \frac{a}{a - x} )$

a = 100%

x = 99%

a-x = (100-99)% = 1%

$k = \frac{1}{t} ln( \frac{100}{1} )$

$k \: = \frac{1}{t} 4.606$

Make t the subject of the formula

$t \: = \frac{4.606}{k}$

Answer 2

Time required for completion of 99% of reaction t = 4.606/k

Explanation:

For first order reaction,

the rate constant  $k=\frac{2.303}{t} log\frac{a}{a-x}$

• From the given data,
• the initial concentration of compound a = 100
• the final concentration a-x =100-99=1 (Because 99% reaction completed)
• By substituting these values in above equation, we get

           $k=\frac{2.303}{t} log \frac{100}{100-99}$

              $\Rightarrow k=\frac{2.303}{t} log 100$

              $\Rightarrow k=\frac{2.303}{t} log 10^2$

              $\Rightarrow k=\frac{2.303 \times 2}{t}$                   ($\because log10 = 1$)

               $\therefore t=\frac{4.606}{k}$