For the reaction- 2 N 2 O 5 - 4N O 2 - O 2 rate and rate constant are 1-02- 10 -4 and 3-4- 10 -5 s -1 respectively- then concentration of N 2 O 5 at that time will be

The correct option is B 3 2 N _ 2 O _ 5 → 4N O _ 2 + O _ 2 d[ N _ 2 O _ 5 ] #160; dt =k[ N _ 2 O _ 5 ] 1.02× 10 ^ 4 =3.4× 10 ^ ...

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Standard XII

Chemistry

Factors Affecting Rate of Reaction

For the react...

Question

For the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ rate and rate constant are $1.02 \times 10^{- 4}$ and $3.4 \times 10^{- 5} s^{- 1}$ respectively, then concentration of $N_{2} O_{5}$ at that time will be

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1.02 \times 10^{- 4}

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3.4 \times 10^{5}

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Solution

The correct option is B $3$
$2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$

$\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$

$1.02 \times 10^{- 4} = 3.4 \times 10^{- 5} s^{- 1} \times [N_{2} O_{5}]$

$∴ [N_{2} O_{5}] = \frac{1.02 \times 10^{- 4}}{3.4 \times 10^{- 5}} = 3$

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Similar questions

Q. For the reaction :

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

Rate and rate constant

1.02 \times 10^{- 4}

and

3.4 \times 10^{- 5} s^{- 1}

respectively, then concentration of

N_{2} O_{5}

at that time will be

Q. For the reaction :

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

Rate and rate constant

1.02 \times 10^{- 4}

and

3.4 \times 10^{- 5} s^{- 1}

respectively, then concentration of

N_{2} O_{5}

at that time will be

Q. For a reaction

2 N_{2} O_{5} (g)

\to 4 N O_{2} (g) + O_{2} (g)

, the rate and rate constant are

1.02 \times 10^{- 4}

m o l L^{- 1} s^{- 1}

and

3.4 \times 10^{- 5} s^{- 1}

respectively. The concentration of

N_{2} O_{5}

at this time will be:

Q. For the reaction

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

rate at a particular time and rate constant are

1.02 \times 10^{- 4} M s^{- 1}

and

3.4 \times 10^{- 5} M s^{- 1}

respectively. Then the concentration of

N_{2} O_{5}

at that time will be:

Q. For the reaction,

2 N_{2} O_{5} ⟶ 4 N O_{2} + O_{2}

the value of rate constant and rate of reaction are respectively

3 \times 10^{- 5} s^{- 1}

and

2.4 \times 10^{- 5} M s^{- 1}

. The concentration of

N_{2} O_{5}

(in mol L

^{- 1}

) will be:

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

The correct option is B 32N2O5→4NO2+O2−d[N2O5]dt=k[N2O5]1.02×10−4=3.4×10−5s−1×[N2O5]∴[N2O5]=1.02×10−43.4×10−5=3

For the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ rate and rate constant are $1.02 \times 10^{- 4}$ and $3.4 \times 10^{- 5} s^{- 1}$ respectively, then concentration of $N_{2} O_{5}$ at that time will be

The correct option is B $3$
$2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$

$\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$

$1.02 \times 10^{- 4} = 3.4 \times 10^{- 5} s^{- 1} \times [N_{2} O_{5}]$

$∴ [N_{2} O_{5}] = \frac{1.02 \times 10^{- 4}}{3.4 \times 10^{- 5}} = 3$