Chemical Kinetics Question 12
Question 12 - 31 January - Shift 2
The rate constant for a first order reaction is $20 min^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is $\times 10^{-2} min$. (Nearest integer)
(Given : $\ln 10=2.303$
$
\log 2=0.3010)
$
$C=\frac{C_o}{2^{n}}=\frac{C_o}{32}$ $n=5$ $t=5 t _{1 / 2}$ $=\frac{5 \times 0.693}{20}=\frac{0.693}{4}$ $=0.17325 min=17.325 \times 10^{-2} min$.
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Answer: (17)
Solution:
