The rate constant for the reaction, 2N2O5 → 4NO2 + O2 is 2 × 10-5 s-1. If rate of reaction is 1.4 × 10-5 mol L-1 s-1, what will be the concentration of N2O5 in mol L-1 ?

Question

The rate constant for the reaction, 2N2O5 → 4NO2 + O2 is 2 × 10-5 s-1. If rate of reaction is 1.4 × 10-5 mol L-1 s-1, what will be the concentration of N2O5 in mol L-1 ? (A) 0.8 (B) 0.7 (C) 1.2 (D) 1

Tardigrade Admin · Accepted Answer

Rate =k[N2O5] (first order as unit of rateconstant is s-1) [N2O5]=(rate/k)=(1.4×10-5mol L-1 s-1/2×10-5 s-1) =0.7 mol L-1

Q. The rate constant for the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ is $2 \times 1 0^{- 5} s^{- 1}$ . If rate of reaction is $1.4 \times 1 0^{- 5} m o l L^{- 1} s^{- 1}$ , what will be the concentration of $N_{2} O_{5}$ in mol $L^{- 1}$ ?

$0.8$

$0.7$

$1.2$

$1$

Rate $= k [N_{2} O_{5}]$ (first order as unit of rateconstant is $s^{- 1}$ )
$[N_{2} O_{5}] = \frac{r a t e}{k} = \frac{1.4 \times 1 0 ^{- 5} m o l L ^{- 1} s ^{- 1}}{2 \times 1 0 ^{- 5} s ^{- 1}}$
$= 0.7 m o l L^{- 1}$

Q. The rate constant for the reaction, 2N2​O5​→4NO2​+O2​ is 2×10−5s−1. If rate of reaction is 1.4×10−5molL−1s−1, what will be the concentration of N2​O5​ in mol L−1 ?

0.8

0.7

1.2

1

Rate =k[N2​O5​] (first order as unit of rateconstant is s−1) [N2​O5​]=krate​=2×10−5s−11.4×10−5molL−1s−1​ =0.7molL−1

Q. The rate constant for the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ is $2 \times 1 0^{- 5} s^{- 1}$ . If rate of reaction is $1.4 \times 1 0^{- 5} m o l L^{- 1} s^{- 1}$ , what will be the concentration of $N_{2} O_{5}$ in mol $L^{- 1}$ ?

$0.8$

$0.7$

$1.2$

$1$

Rate $= k [N_{2} O_{5}]$ (first order as unit of rateconstant is $s^{- 1}$ )
$[N_{2} O_{5}] = \frac{r a t e}{k} = \frac{1.4 \times 1 0 ^{- 5} m o l L ^{- 1} s ^{- 1}}{2 \times 1 0 ^{- 5} s ^{- 1}}$
$= 0.7 m o l L^{- 1}$