You visited us 19 times! Enjoying our articles? Unlock Full Access!

Elevation in Boiling Point

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)

Open in App

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, $k = \frac{2.303}{2} l o g \frac{p^{0}}{p^{0} - 2 p} = \frac{2.303}{2} l o g \frac{320}{200} = \frac{2.303}{2} l o g 1.6 = \frac{2.303}{2} (4 \times l o g 2 - 1) = 0.2303$

Suggest Corrections

Similar questions

N_{2} O_{5}

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

N_{2} O_{5}

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

280 m m H g

580 m m H g .

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

N_{2} O_{5}

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

30

284.5

H g

584.5

H g

N_{2} O_{5}

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

N_{2} O_{5}

N e

N_{2} O_{5}

N_{2} O_{5},

N_{2} O_{5}

N_{2} O_{5}

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

30

284.5

m m

H g

584.5

m m

H g

Join BYJU'S Learning Program