In the reaction, $2N_{2}O_{5} ightarrow 4NO_{2} + O_{2}$ initial pressure is 500 atm and rate constant k is $3.38 imes 10^{- 5}s^{- 1}$ after 10 min the final pressure of $N_{2}O_{5}$ is

Question

In the reaction, $2N_{2}O_{5} ightarrow 4NO_{2} + O_{2}$ initial pressure is 500 atm and rate constant k is $3.38 	imes 10^{- 5}s^{- 1}$ after 10 min the final pressure of $N_{2}O_{5}$ is

Dr. Abubacker · Accepted Answer

(a)
From the unit of rate constant ($i.e.,s^{- 1}),$ it is clear that the reaction is of first order.
$2N_{2}O_{5} ightarrow 4NO_{2} + O_{2}$
Hence, for first order reaction ,
$k = \frac{2.303}{t}\log\frac{p_{0}}{p_{t}}$
$	herefore 3.38 	imes 10^{- 5} = \frac{2.303}{10 	imes 60}\log\frac{500}{p_{t}}$
Or $\log\frac{500}{p_{t}}$=0.00880
$	herefore\ \ \ \ \ \ \frac{500}{p_{t}} = anti\log{0.00880}$
=1.02
$p_{t} = \frac{500}{1.02} = 490\ atm$

Q. In the reaction, $2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}$ initial pressure is 500 atm and rate constant k is $3.38 \times 10^{- 5}s^{- 1}$ after 10 min the final pressure of $N_{2}O_{5}$ is

Explanation

Reviewed by Dr. Abubacker

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