The decomposition reaction 2 N 2 O 5 g Δ 2 N 2 O 4 g + O 2 g is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm.. After Y × 10 3 s the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 × 10 − 4 s − 1 , assuming ideal gas behavior, the value of Y is

Solution:

$2 N_{2} O_{5 (g)} \overset{Δ}{\to} 2 N_{2} O_{4} (g) + O_{2} (g)$

Given rate constant of the reaction = $5 \times 10^{- 4} \sec^{- 1}$

$\begin{array}{l} \frac{1}{2} \frac{{dP}_{N_{2} O_{5}}}{dt} = K_{overall} P_{N_{2} O_{5}} \\ \Rightarrow \frac{{dp}_{N_{2} O_{5}}}{dt} = 2 K_{overall} P_{N_{2} O_{5}} \end{array}$

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

$1 - x x x / 2 \begin{array}{l} 1 + \frac{x}{2} = 1.45 \\ x = 0.90 atm \end{array}$

$\begin{array}{l} y \times 10^{- 3} = \frac{2.303}{2 \times 5 \times 10^{- 4}} \log \frac{1}{0.1} \\ y = \frac{23.03}{10} = 2.30 \end{array}$

Solution:

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