Question

A first order reaction takes 40 min for 30 decomposition. Calculate t1/2

Answer 1

Answer:

k=2.303/t log [Ao]/[A]

k=2.303/40min log 100/70

k=2.303/40*0.155=0.00892 min-1

t1/2=0.693/k

t1/2=0.693/0.00892min

t1/2=77.7min(answer)

thank you it may help you

Answer 2

Answer:

$\Large \bf{Given} -$

t = 40min

30% of the reactants undergoes decomposition that means out of 100 particles in reactant 30 particles has been used

Now,70 particles will undergo out of 100 particles

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

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

$\Large \bf{Answer}$

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

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

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

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

$\text{Now you can use,}$

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

$\Large \sf{k} = \frac{2.303}{40} \log\frac{100}{70}$

$\Large \sf{k} = \frac{2.303}{40} \log\frac{10}{7}$

$\Large \sf{k} = \frac{2.303}{40} \log1.428$

$\Large \sf{k} = \frac{2.303}{40}× 0.1548$

$\Large \sf{k} = \frac{0.3565}{40}$

$\large \sf{k} = 0.008913$

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

$\Large \sf{t_\frac{1}{2}} = \frac{0.693}{0.008913}$

$\Large \sf{t_\frac{1}{2}} = 77.75 min$