The initial concentration of N 2 O 5 in the first order reaction N 2 O 5g - 2NO 2g - 1 - 2 O 2g was 1 1-24- 10 -2 mol L -1 at 318 K- The concentration of N 2 O 5 after 60 minutes was 0-20- 10 -2 mol L -1- Calculate the rate constant of the reaction at 318 K-

k = 2.303t log [A][A x] k = 2.30360 log 1.24 ×10^ 20.2 ×10^ 2 k = 3.04 ×10^ 2min^ 1 ...

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Standard XII

Chemistry

Rate Constant

The initial c...

Question

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.

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Solution

$k = \frac{2.303}{t} l o g \frac{[A]}{[A - x]}$

$k = \frac{2.303}{60} l o g \frac{1.24 \times 10^{- 2}}{0.2 \times 10^{- 2}}$

$k = 3.04 \times 10^{- 2} {m i n}^{- 1}$

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Similar questions

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.

Q. The rate constant

k

, for the reaction

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

1.3 \times 10^{- 2} s^{- 1}

. Which equation given below describes the change of

[N_{2} O_{5}]

with time?

{[N_{2} O_{5}]}_{0}

and

{[N_{2} O_{5}]}_{t}

correspond to concentration of

N_{2} O_{5}

initially and at time

t

Q. The rate constant for the reaction,

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

, is

2.3 \times 10^{- 2} {s e c}^{- 1}

. Which equation given below describes the change of

[N_{2} O_{5}]

with time,

{[N_{2} O_{5}]}_{0}

and

{[N_{2} O_{5}]}_{t}

corresponds to concentration of

N_{2} O_{5}

initially and time

t

respectively?

Q. The reaction

N_{2} O_{5}

(in

C {C l}_{4}

)

\to 2 {N O}_{2} + 1 / 2 O_{2} (g)

is first order in

N_{2} O_{5}

with rate constant

6.2 \times 10^{- 4}

. What is the value of rate of reaction when

[N_{2} O_{5}] = 1.25 m o l e

L^{- 1}

Q. For a first order reaction involving decomposition of

N_{2} O_{5}

, the following information is available :

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

\to

{4 N O}_{2} (g)

O_{2} (g)

Rate = k

[N_{2} O_{5}]

N_{2} O_{5}

(g)

\to

{2 N O}_{2}

(g) +

\frac{1}{2} O_{2}

(g) Rate =

k^{'}

[N_{2} O_{5}]

,
Which of the following expressions is true ?

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The initial concentration of N2O5 in the first order reaction N2O5(g)⟶2NO2(g)+12O2(g) was 11.24×10−2molL−1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20×10−2molL−1. Calculate the rate constant of the reaction at 318 K.

k=2.303tlog[A][A−x]k=2.30360log1.24×10−20.2×10−2k=3.04×10−2 min−1

$k = \frac{2.303}{t} l o g \frac{[A]}{[A - x]}$

$k = \frac{2.303}{60} l o g \frac{1.24 \times 10^{- 2}}{0.2 \times 10^{- 2}}$

$k = 3.04 \times 10^{- 2} {m i n}^{- 1}$