Chemical Kinetics Ques 7
7. For the reaction $2 N_2 O_5 \to 4 NO_2+O_2$, rate and rate constant are $1.02 \times 10^{-4} mol lit^{-1} s^{-1}$ and $3.4 \times 10^{-5} s^{-1}$ respectively then concentration of $N_2 O_5$ at that time will be
[2001]
(a) $1.732 M$
(b) $3 M$
(c) $3.4 \times 10^{5} M$
(d) $1.02 \times 10^{-4} M$
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Answer:
Correct Answer: 7.(b)
Solution:
(b) $2 N_2 O_5 \to 4 NO_2+O_2$
From the unit of rate constant, it is clear that the reaction follow first order kinetics. Hence, by rate law equation,
$\quad r=k[N_2 O_5]$ where $r=1.02 \times 10^{-4}, k=3.4 \times 10^{-5}$
$1.02 \times 10^{-4}=3.4 \times 10^{-5}[N_2 O_5]$
$[N_2 O_5]=3 M$
BETA
