Nitrogen dioxide (NO2) dissociates into nitric oxide (NO) and oxygen (O2) as follows: 2NO2→ 2NO + O2 If the rate of decrease of concentration of NO2 is 6.0 × 10-12 mol L-1 s-1 What will be the rate of increase of concentration of O2 ?

Question

Nitrogen dioxide (NO2) dissociates into nitric oxide (NO) and oxygen (O2) as follows: 2NO2→ 2NO + O2 If the rate of decrease of concentration of NO2 is 6.0 × 10-12 mol L-1 s-1 What will be the rate of increase of concentration of O2 ? (A) 3 × 10-12 mol L-1 s-1 (B) 6 × 10-12 mol L-1 s-1 (C) 1 × 10-12 mol L-1 s-1 (D) 1.5 × 10-12 mol L-1 s-1

Tardigrade Admin · Accepted Answer

For the reaction, 2NO2 arrow 2NO +O2 -(1/2)(d[NO2]/dt)=(1/2)(d[NO]/dt)=(d[O2]/dt) -(d [NO2]/dt)=6×10-12 mol L-1 s-1 (d[O2]/dt)=3×10-12 mol L-1 s-1

Q. Nitrogen dioxide $(N O_{2})$ dissociates into nitric oxide $(NO)$ and oxygen $(O_{2})$ as follows:
$2 N O_{2} \to 2 NO + O_{2}$
If the rate of decrease of concentration of $N O_{2}$ is $6.0 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$ What will be the rate of increase of concentration of $O_{2}$ ?

$3 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$

$6 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$

$1 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$

$1.5 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$

For the reaction, $2 N O_{2} \to 2 NO + O_{2}$
$- \frac{1}{2} \frac{d [ N O _{2} ]}{d t} = \frac{1}{2} \frac{d [ NO ]}{d t} = \frac{d [ O _{2} ]}{d t}$
$- \frac{d [ N O _{2} ]}{d t} = 6 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$
$\frac{d [ O _{2} ]}{d t} = 3 \times 1 0^{- 12} m o l L^{- 1} s^{- 1}$

Q. Nitrogen dioxide (NO2​) dissociates into nitric oxide (NO) and oxygen (O2​) as follows: 2NO2​→2NO+O2​ If the rate of decrease of concentration of NO2​ is 6.0×10−12molL−1s−1 What will be the rate of increase of concentration of O2​ ?

3×10−12molL−1s−1

6×10−12molL−1s−1

1×10−12molL−1s−1

1.5×10−12molL−1s−1