Question

A first order reaction takes 25 min for 25 % completion calculate t 1/2

Answer 1

Given that,

Time = 25 min

Percentage = 25 %

A first order reaction takes 25 minutes for 25% decomposition

We need to calculate the value of k

Using formula of k

$k=\dfrac{2.303}{t}\log(\dfrac{a}{a-x})$

Put the value into the formula

$k=\dfrac{2.303}{25}\log(\dfrac{a}{a-x})$

$k=\dfrac{2.303}{25}\log(\dfrac{a}{a-(0.25a)})$

$k=\dfrac{2.303}{25}\log(\dfrac{1}{0.75})$

$k=0.0115$

When 75% of the reaction will be completed.

We need to calculate the value of $t_{\frac{1}{2}}$

Using formula for $t_{\frac{1}{2}}$

$k=\dfrac{2.303}{t}\log(\dfrac{a}{a-x})$

$t=\dfrac{2.303}{k}\log(\dfrac{a}{a-x})$

Put the value into the formula

$t=\dfrac{2.303}{0.0115}\log(\dfrac{a}{a-(0.75a)})$

$t=\dfrac{2.303}{0.0115}\log(\dfrac{1}{0.25})$

$t =120.5\ min$

Hence, The time is 120.5 min.