The intial concentration of N2O5 in the following first order reaction N2O5g - 2NO2g -1-2O2g was 1-24 - 10-2 mol L-1 at 318 K- The concentration of N2O2 after 60 minutes was 0-20 - 10-2 mol L-1- Calculate the rate constant of the reaction at 318 K-

Given, Initial concentatraion of N_2O_5[R]_1= 1.24×10^ 2 mol L^ 1 Final concentration of nbsp; N_2O_5[R]_2= 0.20×10^ 2 mol L^ 1 Order of reaction = First ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Chemistry

Ionizable and Non-Ionizable Compounds

The intial co...

Question

The intial concentration of $N_{2} O_{5}$ in the following first order reaction $N_{2} O_{5} (g) \to 2 N O_{2} (g) + 1 / 2 O_{2} (g)$ was $1.24 \times 10^{- 2} m o l L^{- 1}$ at $318 K$ . The concentration of $N_{2} O_{2}$ after $60$ minutes was $0.20 \times 10^{- 2} m o l L^{- 1}$ . Calculate the rate constant of the reaction at $318 K$ .

Open in App

Solution

Given,

Initial concentatraion of
$N_{2} O_{5} [R]_{1} = 1.24 \times 10^{- 2} m o l L^{- 1}$

Final concentration of
$N_{2} O_{5} [R]_{2} = 0.20 \times 10^{- 2} m o l L^{- 1}$

Order of reaction = First order
Temperature= $318 K$

Value of rate constant

For a first order reaction,

$l o g \frac{[R]_{1}}{[R]_{2}} = \frac{k (t_{2} - t_{1})}{2.303}$

Putting values in the above formula, we
get,
$k = \frac{2.303}{(t_{2} - t_{1})} log \frac{[R]_{1}}{[R]_{2}}$

$= \frac{2.303}{60 m i n - 0 m i n} log \frac{1.24 \times 10^{- 2} m o l L^{- 1}}{0.20 \times 10^{- 2} m o l L - 1}$

$= \frac{2.303}{60} log 6.2 m i n^{- 1}$

$\Rightarrow= 0.0304 m i n^{- 1}$

Final answer: The rate constant of the
reation at $318 k i s 0.0304 m i n^{- 1}$

Suggest Corrections

Similar questions

Q. The reaction;

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

is of first order for

N_{2} O_{5}

with rate constant

6.2 \times 10^{- 4} s^{- 1}

. What is the value of rate of reaction when

[N_{2} O_{5}] = 1.25 m o l L^{- 1}

Q. The rate of the reaction between hemoglobin (

H b

) and carbon monoxide (

C O

) was studied at

20^{o} C

. The following data were collected with all concentration units

μ m o l / L

(A haemoglobin concentration of

2.21 μ m o l / L

is equal to

2.21 \times 10^{- 6} m o l / L

)

	$[H b]_{0}$ $(μ m o l / L)$	$[C O]_{0}$ $(μ m o l / L)$	Initial rate $(μ m o l L^{- 1} S^{- 1})$
(1)	2.21	1.00	0.619
(2)	4.42	1.00	1.24
(3)	4.42	3.00	3.71

Determine the orders of this reaction with respect to

H b

and

C O

and rate constant.

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 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.

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

In the above first order reaction the initial concentration of

N_{2} O_{5}

is 2.40

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

at 318 K. The concentration of

N_{2} O_{5}

after 1 hour was

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

. The rate constant of the reaction at 318 k is

x \times 10^{- 3} m i n^{- 1}

The value of x (Nearest integer) is

[G i v e n : l o g 3 = 0.447, l o g 5 = 0.699]

Join BYJU'S Learning Program

Explore more

Ionizable and Non-Ionizable Compounds

Standard VII Chemistry

Join BYJU'S Learning Program

The intial concentration of N2O5 in the following first order reaction N2O5(g)→2NO2(g)+1/2O2(g) was 1.24×10−2 mol L−1 at 318 K. The concentration of N2O2 after 60 minutes was 0.20×10−2 mol L−1. Calculate the rate constant of the reaction at 318 K.