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Standard XI

Chemistry

First Order Reaction

The decomposi...

Question

The decomposition of $N_{2} O_{5}$ according to the equation $2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
is a first order reaction. After 2 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 500 mm Hg. On complete decomposition, the total pressure is 800 mm Hg. If the rate constant of the reaction is $y m i n^{- 1}$ then, what is the value of y? (given log 2 = 0.3, log 3 = 0.5)

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Solution

The given reaction is
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
$\begin{matrix} p^{0} & 0 & 0 & i n i t i a l l y p^{0} - 2 p & 4 p & p & a t 2 m i n 0 & 2 p^{0} & \frac{p^{0}}{2} & a f t e r l o n g t i m e \end{matrix}$
Given:
$p^{0} + 3 p = 500 m m H g$ and
$2.5 p^{0} = 800 m m H g$
$\Rightarrow p^{0} = 320 m m H g$ and p = $60 m m H g$
So, $k = \frac{2.303}{2} l o g \frac{p^{0}}{p^{0} - 2 p} = \frac{2.303}{2} l o g \frac{320}{200} = \frac{2.303}{2} l o g 1.6 = \frac{2.303}{2} (4 \times l o g 2 - 1) = 0.2303$

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Similar questions

Q. The decomposition of

N_{2} O_{5}

according to the equation

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

is a first order reaction. After 30 min, from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm Hg. On complete decomposition, the total pressure is 584.5 mm Hg. The rate constant of the reaction is:

Q. The decomposition of

N_{2} O_{5}

according to equation:

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

, is a first-order reaction. After

30

minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be

284.5

mm of

H g

and on complete decomposition, the total pressure is

584.5

mm of

H g

The rate constant of a reaction is:

N_{2} O_{5}

decomposes according to the equation,

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

This is a first order reaction. A closed container has

N_{2} O_{5}

and

N e

gases at a total pressure of 333.8 mm of Hg. Thirty minutes after

N_{2} O_{5}

starts decomposing, the total pressure is found to be 384.5 mm Hg. On complete decomposition of

N_{2} O_{5},

the total pressure is 684.5 mm Hg. What is rate constant of decomposition of

N_{2} O_{5}

?
(Take ln(1.169) = 0.156)

Q. The decomposition of

N_{2} O_{5}

according to the equation,

2 N_{2} O_{5} (g) ⟶ 4 N O_{2} (g) + O_{2} (g)

, is a first order reaction. After

30

minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be

284.5

m m

H g

and on completion, the total pressure is

584.5

m m

H g

. Calculate the rate constant of the reaction.

The decomposition of a compound

P

, at temperature

T

according to the equation

2 P_{(g)} \to 4 Q_{(g)} + R_{(g)} + S_{(l)}

is the first order reaction. After

30

minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be

317

H g

and after a long period of time the total pressure observed to be

617

H g

The total pressure of the vessel after

75

minute, if the volume of liquid

S

is supposed to be negligible is:

(Given : Vapour pressure of

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at temperature

T = 32.5

H g

)

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The given reaction is 2N2O5(g)→4NO2(g)+O2(g) p000initiallyp0−2p4ppat 2 min02p0p02after long time Given: p0+3p=500 mmHg and 2.5p0=800 mmHg ⇒p0=320 mmHg and p = 60 mmHg So, k=2.3032logp0p0−2p=2.3032log320200=2.3032log 1.6=2.3032(4×log2−1)=0.2303