Chemical Kinetics - Result Question 80

####79. A first order reaction is $20 %$ complete in $10 min$. Calculate (i) the specific rate constant of the reaction, and

(ii) the time taken for the reaction to go to $75 %$ completion.

(1983, 2M)

Show Answer

Solution:

  1. For a first order reaction,

$$ k t=\ln \frac{[A] _0}{[A]} $$

where $[A] _0=$ Initial concentration of reactant

$$ [A]=\text { Concentration of reactant remaining } $$

unreacted at time $t$.

(i) $\Rightarrow k=\frac{1}{t} \ln \frac{[A] _0}{[A]}=\frac{1}{10} \ln \frac{100}{100-20}=\frac{1}{10} \ln \frac{5}{4}$

$$ =\frac{2.303(\log 5-2 \log 2)}{10} \min ^{-1}=0.023 min^{-1} $$

$$ t=\frac{1}{k} \ln \frac{100}{25}=\frac{2 \ln 2}{k}=\frac{2 \times 0.693}{0.023}=60 min $$



NCERT Chapter Video Solution

Dual Pane