For the reaction 2N2O5 → 4NO2 + O2, rate and rate constant are 1.02 × 10-4 mol lir-1 sec-1 and 3.4 × 10-5 sec-1 respectively then concentration of N2O5 at that time will be

Question

For the reaction 2N2O5 → 4NO2 + O2, rate and rate constant are 1.02 × 10-4 mol lir-1 sec-1 and 3.4 × 10-5 sec-1 respectively then concentration of N2O5 at that time will be (A) 1.732 M (B) 3 M (C) 3.4× 10-5M (D) 1.02 × 10-4M

Tardigrade Admin · Accepted Answer

2N2O5 → 4NO2 + O2 from the imit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation, r = k [N2O5] wherer = 1.02 × 10-4, k= 3.4 × 10-5 1.02 × 10-4 = 3.4 × [N2O5] [N2O5] = 3M

Q. For the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , rate and rate constant are $1.02 \times 1 0^{- 4}$ mol $l i r^{- 1} sec^{- 1}$ and $3.4 \times 1 0^{- 5} sec^{- 1}$ respectively then concentration of $N_{2} O_{5}$ at that time will be

$1.732 M$

$3 M$

$3.4 \times 1 0^{- 5} M$

$1.02 \times 1 0^{- 4} M$

Q. For the reaction 2N2​O5​→4NO2​+O2​, rate and rate constant are 1.02×10−4 mol lir−1sec−1 and 3.4×10−5sec−1 respectively then concentration of N2​O5​ at that time will be

1.732M

3M

3.4×10−5M

1.02×10−4M

2N2​O5​→4NO2​+O2​ from the imit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation, r=k[N2​O5​] wherer =1.02×10−4,k=3.4×10−5 1.02×10−4=3.4×[N2​O5​] [N2​O5​]=3M

Q. For the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , rate and rate constant are $1.02 \times 1 0^{- 4}$ mol $l i r^{- 1} sec^{- 1}$ and $3.4 \times 1 0^{- 5} sec^{- 1}$ respectively then concentration of $N_{2} O_{5}$ at that time will be

$1.732 M$

$3 M$

$3.4 \times 1 0^{- 5} M$

$1.02 \times 1 0^{- 4} M$