Question

For the reaction $2N_2O_5 \to 4NO_2 + O_2$, rate and rate constant are $1.02 \times 10^{-4}$ mol $lir^{-1} \sec^{-1}$ and $3.4 \times 10^{-5} \sec^{-1}$ respectively then concentration of $N_2O_5$ at that time will be

• A. $1.732\, M$
• B. $3 \,M$
• C. $3.4\times 10^{-5}M$
• D. $1.02 \times 10^{-4}M$

Answer 1

Answer: (B)

$2N_2O_5 \to 4NO_2 + O_2$
from the imit of rate constant it is clear that the reaction follow first order kinetics.
Hence
by rate law equation, $r = k [N_2O_5]$
wherer $= 1.02 \times 10^{-4}, k= 3.4 \times 10^{-5}$
$1.02 \times 10^{-4} = 3.4 \times \left[N_{2}O_{5}\right]$
$\left[N_{2}O_{5}\right] = 3M$