The rate of reaction at 23.03 sec after the reaction has been initiated at tempe | TestHub

Passage / Comprehension

Dinitrogen pentoxide decomposes on heating to nitrogen dioxide and oxygen following first-order kinetics:

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

Initially, $0.2$ mole of $N_{2} O_{5}$ was taken in a vessel of capacity 1 L and heated to a Certain temperature of $T$ (Kelvin), at which the reaction begins. The concentration of $N_{2} O_{5}$ is measured at different time intervals by withdrawing samples of the reaction mixture. A plot of $ln [N_{2} O_{5}]$ vs time is shown in Figure 1. The rate constant ( $k$ ) of the reaction is determined from the slope of this linear plot. The experiment was repeated at five different temperatures, and the rate constants were calculated for each temperature. A plot of $ln k$ versus $(t e m p er a t u re)^{- 1}$ is shown in Figure 2.

The rate of reaction at 23.03 sec after the reaction has been initiated at temperature T kelvin will be

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Answer:

Solution:

$k = \frac{2.303}{t} lo g \frac{a _{0}}{a _{t}}$

$1 \times 1 0^{- 1} = \frac{2.303}{23.03} lo g \frac{a _{0}}{a _{t}}$

$lo g \frac{a _{0}}{a _{t}} = 1$

$\frac{a _{0}}{a _{t}} = 10$

$a_{t} = \frac{a _{0}}{10} = \frac{0.2}{10} = 0.02$

Rate $= k [N_{2} O_{5}]_{t = 23.03 s e c}$

Rate $= 1 \times 1 0^{- 1} \times 0.02 = 0.002$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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