The following reaction was carried out at 44 o C- N 2 O 5 - 2N O 2 - 1 2 O 2The concentration of N O 2 is 6-0- 10 -3 M after 10 minutes of the start of the reaction- Calculate the rate of production of N O 2 over the first ten minutes of the reaction-

N _ 2 O _ 5 ⟶ 2 NO _ 2 + 1 2 O _ 2 d[ N _ 2 O _ 5 ] #160; dt = 1 2 d[ NO _ 2 ] #160; dt = 2d[ O _ 2 ] #160; dt d[ NO _ 2 ] ...

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Standard XII

Chemistry

Rate of Reaction

The following...

Question

The following reaction was carried out at $44^{o} C$ :
$N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$
The concentration of $N O_{2}$ is $6.0 \times 10^{- 3} M$ after $10$ minutes of the start of the reaction. Calculate the rate of production of $N O_{2}$ over the first ten minutes of the reaction.

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Solution

$N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}$

$\frac{d [N_{2} O_{5}]}{d t} = \frac{1}{2} \frac{d [{N O}_{2}]}{d t} = \frac{2 d [O_{2}]}{d t}$

$\frac{d [{N O}_{2}]}{d t} = \frac{6 \times 10^{- 3}}{10} = 6 \times 10^{- 4} m o l L^{- 1} {m i n}^{- 1}$

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Similar questions

Q. The following reaction was carried out at

44^{o} C : N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

The concentration of

N O_{2}

6.0 \times 10^{- 3} M

after

10

minutes of the start of the reaction. Calculate the rate of production of

N O_{2}

over the first ten minutes of the reaction.

Q. The initial concentration of

N_{2} O_{5}

in the first order reaction

N_{2} O_{5 (g)} ⟶ {2 N O}_{2 (g)} + \frac{1}{2} O_{2 (g)}

was 1

1.24 \times 10^{- 2} m o l L^{- 1}

at 318 K. The concentration of

N_{2} O_{5}

after 60 minutes was

{0.20 \times 10}^{- 2} m o l L^{- 1}

. Calculate the rate constant of the reaction at 318 K.

Q. For the reaction,

2 N_{2} O_{5} (g) ⟶ 4 N O_{2} (g) + O_{2} (g)

, if the concentration of

N O_{2}

increases by

5.2 \times 10^{- 3} M

100

sec, then the rate of reaction is:

Q. The following reaction was carried out in water,

{C l}_{2} + 2 I^{-} ⟶ I_{2} + 2 {C l}^{-}

.
The initial concentration of

I^{-}

was

0.25

m o l

L^{- 1}

and the concentration after

10

minutes was

0.23

m o l

L^{- 1}

. Calculate the rate of disappearance of

I^{-}

and appearance of

I_{2}

Q. The decomposition of

N_{2} O_{5}

at 318 K according to the following equation follows first order reaction:

N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)

The initial concentration of

N_{2} O_{5}

was

1.24 \times 10^{- 2}

mol

L^{- 1}

and after 60 minutes was

0.20 \times 10^{- 2}

L^{- 1}

, Calculate the rate constant of the reaction at 318 K.

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The following reaction was carried out at 44oC:N2O5→2NO2+12O2The concentration of NO2 is 6.0×10−3M after 10 minutes of the start of the reaction. Calculate the rate of production of NO2 over the first ten minutes of the reaction.

N2O5⟶2NO2+12O2d[N2O5]dt=12d[NO2]dt=2d[O2]dtd[NO2]dt=6×10−310=6×10−4mol L−1min−1

The following reaction was carried out at $44^{o} C$ :
$N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$
The concentration of $N O_{2}$ is $6.0 \times 10^{- 3} M$ after $10$ minutes of the start of the reaction. Calculate the rate of production of $N O_{2}$ over the first ten minutes of the reaction.

$N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}$

$\frac{d [N_{2} O_{5}]}{d t} = \frac{1}{2} \frac{d [{N O}_{2}]}{d t} = \frac{2 d [O_{2}]}{d t}$

$\frac{d [{N O}_{2}]}{d t} = \frac{6 \times 10^{- 3}}{10} = 6 \times 10^{- 4} m o l L^{- 1} {m i n}^{- 1}$