For the reaction -2 N 2 O 5- 4 NO 2- O 2Rate and rate constant 1-02 - 10-4 and 3-4 - 10-5 s -1 respectively- then concentration of N 2 O 5 at that time will beA- 1-732B- 3- 1-02 - 10-4D- 3-4 - 105

The correct option is B 3Rate = k[N_2O_5] 1.02 × 10^ 4= 3.4 × 10^ 5[N_2O_5] ∴ [N_2O_5]=1.02× 10^ 4/3.4× 10^ 5=3 ...

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Byju's Answer

Standard VIII

Chemistry

First Order Reaction

For the react...

Question

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

1.732

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

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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

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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}

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rate at a particular time and rate constant are

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Q. For a reaction

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For the reaction : 2N2O5→4NO2+O2 Rate and rate constant 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 3Rate=k[N2O5]1.02×10−4=3.4×10−5[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 $1.02 \times 10^{- 4}$ 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 $3$
$R a t e = k [N_{2} O_{5}] 1.02 \times 10^{- 4} = 3.4 \times 10^{- 5} [N_{2} O_{5}] ∴ [N_{2} O_{5}] = \frac{1.02 \times 10^{- 4}}{3.4 \times 10^{- 5}} = 3$