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Standard XII

Chemistry

Rate Constant

From the foll...

Question

From the following data for the decomposition of $N_{2} O_{5}$ at $30^{0} C$ , find out the rate constant (in min $^{- 1}$ )?
Volume of $O_{2}$ after 10 mins. of the reaction $=$ 90ml.
Volume of $O_{2}$ after completion of the reaction $=$ 100ml

2.303

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0.2303

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0.02303

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23.03

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Solution

The correct option is B 0.2303
$k = \frac{2.303}{t} \times l o g (\frac{a}{a - x}) = \frac{2.303}{10} l o g (\frac{100}{10}) = 0.2303.$

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Similar questions

Q. The half life of decomposition of

N_{2} O_{5}

is a first order reaction represented by

N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2}

After 15 min the volume of

O_{2}

produced is 9mL and at the end of the reaction 35 mL. The rate constant is equal to:

Q. The composition of

N_{2} O_{5}

is first order reaction represented by:

N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2} .

After 20 min the volume of

O_{2}

poduced is 10 mL and at the end of the reaction 40 mL. The rate constant is equal to:

Q. The half life decomposition of

N_{2} O_{5}

in a first order reaction represented by

N_{2} O_{5} \to N_{2} O_{4} + \frac{1}{2} O_{2}

After 20 min the volume of

O_{2}

produced is 10 ml and at the end of the reaction 40 ml. The value of rate constant is:

[G i v e n 2.303 log (\frac{4}{3}) = 0.28]

Q. For a reaction

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

;
Initial concentration of

N_{2} O_{5}

7.2

mol

L^{- 1}

and after

20

minutes

[N O_{2}] = 2.4

mol

L^{- 1}

, then what will be rate of decomposition of

N_{2} O_{5}

in mol

L^{- 1}

m i n^{- 1}

Q. The specific rate constant of the decomposition of

N_{2} O_{5}

0.008 m i n^{- 1}

. The volume of

O_{2}

collected after 20 minute is 16 mL. The volume in ml that would be collected at the end of reaction.

N O_{2}

formed is dissolved in

C C l_{4}

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From the following data for the decomposition of N2O5 at 300C, find out the rate constant (in min−1)?Volume of O2 after 10 mins. of the reaction=90ml.Volume of O2 after completion of the reaction=100ml

The correct option is B 0.2303k=2.303t×log(aa−x)=2.30310log(10010)=0.2303.

From the following data for the decomposition of $N_{2} O_{5}$ at $30^{0} C$ , find out the rate constant (in min $^{- 1}$ )?
Volume of $O_{2}$ after 10 mins. of the reaction $=$ 90ml.
Volume of $O_{2}$ after completion of the reaction $=$ 100ml

The correct option is B 0.2303
$k = \frac{2.303}{t} \times l o g (\frac{a}{a - x}) = \frac{2.303}{10} l o g (\frac{100}{10}) = 0.2303.$