Question: 89. If the rate constant of a reaction is 0.03 \mathrm{~s}^{-1}, how much time does it take for 7.2 \mathrm{~mol} \mathrm{~L}^{-1} concentration of the reactant to get reduced to 0.9 \mathrm{~mol} \mathrm{~L}^{-1} ?
(Given : \log 2=0.301 )
(1) 69.3 s
(2) 23.1 s
(3) 210 s
(4) 21.0 s
Answer: Option (1)
Explanation:
Since the unit of rate constant is \mathrm{s}^{-1},
the reaction follows first order kinetics.
The integrated rate law for a first order reaction is:
t=\frac{2.303}{k}\log\frac{[A]_0}{[A]}Here, [A]_0=7.2 \ \mathrm{mol \ L^{-1}},
[A]=0.9 \ \mathrm{mol \ L^{-1}} and
k=0.03 \ \mathrm{s^{-1}}.
Substituting the values:
t=\frac{2.303}{0.03}\log\frac{7.2}{0.9} t=\frac{2.303}{0.03}\log(8) \log(8)=\log(2^3)=3\log 2=3 \times 0.301=0.903So,
t=\frac{2.303}{0.03}\times 0.903 t=69.3 \ \mathrm{s}Therefore, the time required is 69.3 s.