Question

a first order reaction takes 30 minutes for 20% calculate t1/2 (log 2= 0.3010)​

Answer 1

Answer:

Explanation:

For the first order reaction

k = (2.303 / t) log (a / a – x)

When x = (20 / 100) a = 0.2 a

t = 30 minutes (given)

Therefore, k = (2.303 / 30) log (a / a – 0.2 a)

k = (2.303 / 30) log (1 / 0.8)

  = 0.007439 min-1

Hence the value of the rate constant is 0.007439 min-1

t1/2 = 0.693/k = 0.693/0.007439 = 93 min

Answer 2

Answer:

$\Large \bf{Given} -$

t = 30min

20% of the reactants undergoes decomposition that means out of 100 particles in reactant 20 particles has been used

Now,80 particles will undergo out of 100 particles

$\Large \bf{To\:find} -$

$\Large \sf{t_\frac{1}{2}} = \Large{?}$

$\Large \bf{Answer}$

$\text{You know for a first order reaction}$ $\text{half life is -}$

$\Large \sf{t_\frac{1}{2}}$ = $\Large \frac{0.693}{k}$

$\text{You can directly put half life formula }$ $\text{but k is missing}$

$\text{So,now you have to bring k in order to }$ $\text{solve}$

$\text{Now you can use,}$

$\Large \sf{k} = \frac{2.303}{t} \log\frac{[R]_o}{[R]}$

$\Large \sf{k} = \frac{2.303}{30} \log\frac{100}{80}$

$\Large \sf{k} = \frac{2.303}{30} \log\frac{5}{4}$

$\Large \sf{k} = \frac{2.303}{30} \log1.25$

$\Large \sf{k} = \frac{2.303}{30}× 0.0969$

$\Large \sf{k} = \frac{0.2231}{30}$

$\large \sf{k} = 0.00743$

$\text{Now put k in half-life formula}$

$\Large \sf{t_\frac{1}{2}} = \frac{0.693}{0.00743}$

$\Large \sf{t_\frac{1}{2}} = 93.27 min$