At 1800 K, ethane gas decomposes to ethene and hydrogen. Rate constant for the reaction is $10^{- 3} {Pa}^{- 1} \sec^{- 1}$ If initial pressure of ethane is $3 \times 10^{5} Pa$ , how many sec would it take for the pressure to reach $5 \times 10^{5} Pa$

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$3.33 \times 10^{- 2} \sec$

$6.66 \times 10^{- 3} \sec$

$1000.4 \sec$

$1800.2 \sec$

answer is C.

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Detailed Solution

2nd order reaction

$\begin{matrix} \underset{3 \times 10^{5} - x}{C_{2} H_{6}} ⟶ \underset{x}{C_{2} H_{4}} + \underset{x}{H_{2}} \\ \begin{aligned} \Rightarrow 3 \times 10^{5} + x = 5 \times 10^{5} \\ \Rightarrow \frac{1}{10^{5}} = \frac{1}{3 \times 10^{5}} + 10^{- 3} \times t \\ \Rightarrow t = \frac{2}{3} \times 10^{- 2} \sec . \end{aligned} \end{matrix}$