In the reaction- 2 N 2 O 5- 4 NO 2- O 2- initial pressure and rate constant- k are 500 atm and 26-87 - 10-4 sec -1 respectively- After 10 minutes- the final pressure of N 2 O 5 take- log 5-0-7 A- 100 atmB- 250 atmC- 150 atmD- 200 atm

The correct option is A 100 atmIf P_t is the pressure after time t, then k = 2.303/t log P_0/P_t ∴ 26.87 × 10^ 4s^ 1 = 2.303/600 log 500/P_t ⇒ log500/P_t = ...

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Byju's Answer

Standard XII

Chemistry

Reaction Quotient

In the reacti...

Question

In the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ ; initial pressure and rate constant, (k) are $500 a t m$ and $26.87 \times 10^{- 4} s e c^{- 1}$ respectively. After 10 minutes, the final pressure of $N_{2} O_{5}$
(take: $l o g 5 = 0.7$ )

100 a t m

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250 a t m

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150 a t m

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200 a t m

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Solution

The correct option is A $100 a t m$
If $P_{t}$ is the pressure after time t, then

$k = \frac{2.303}{t} l o g \frac{P_{0}}{P_{t}}$

$∴ 26.87 \times 10^{- 4} s^{- 1} = \frac{2.303}{600} l o g \frac{500}{P_{t}}$
$\Rightarrow l o g \frac{500}{P_{t}} = \frac{16122 \times 10^{- 5}}{2.303}$

$\Rightarrow l o g \frac{500}{P_{t}} = 0.7$

$\Rightarrow l o g 500 - l o g P_{t} = 0.7$

$\Rightarrow l o g P_{t} = 2.7 - 0.7 = 2$

$\Rightarrow P_{t} = 10^{2} = 100$ atm

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In the reaction, 2N2O5→4NO2+O2; initial pressure and rate constant, (k) are 500 atm and 26.87×10−4 sec−1 respectively. After 10 minutes, the final pressure of N2O5 (take: log5=0.7)

The correct option is A 100 atmIf Pt is the pressure after time t, then k=2.303tlogP0Pt ∴26.87×10−4s−1=2.303600log500Pt ⇒log500Pt=16122×10−52.303 ⇒log500Pt=0.7 ⇒log500−logPt=0.7 ⇒logPt=2.7−0.7=2 ⇒Pt=102=100 atm

In the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ ; initial pressure and rate constant, (k) are $500 a t m$ and $26.87 \times 10^{- 4} s e c^{- 1}$ respectively. After 10 minutes, the final pressure of $N_{2} O_{5}$
(take: $l o g 5 = 0.7$ )