For a reaction in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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Q.
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

1 7/4 times of initial pressure

2 5/2 times of initial pressure

3 5 times of initial pressure

4 7/2 times of initial pressure

Ans.
(1)

$N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$

$\begin{matrix} t = 0 & p & 0 & 0 \\ t = \infty & p - 0.5 p & 2 \times 0.5 p & \frac{1}{2} \times 0.5 p \\ = 0.5 p & = p & = 0.25 p \end{matrix} \begin{matrix} Total = 1.75 p \\ = \frac{7}{4} p \end{matrix}$

Solution

Q.
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

1 7/4 times of initial pressure

2 5/2 times of initial pressure

3 5 times of initial pressure

4 7/2 times of initial pressure

Ans.
(1)

$N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$

$\begin{matrix} t = 0 & p & 0 & 0 \\ t = \infty & p - 0.5 p & 2 \times 0.5 p & \frac{1}{2} \times 0.5 p \\ = 0.5 p & = p & = 0.25 p \end{matrix} \begin{matrix} Total = 1.75 p \\ = \frac{7}{4} p \end{matrix}$

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Solution

Q. For a reaction N2O5(g) ⟶ 2NO2(g)+12O2(g) in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

1 7/4 times of initial pressure

2 5/2 times of initial pressure

3 5 times of initial pressure

4 7/2 times of initial pressure

Ans. (1) N2O5(g) ⟶2NO2(g)+12O2(g) t=0p00t=∞p-0.5p2×0.5p12×0.5p=0.5p=p=0.25pTotal=1.75p=74p

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Q.
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

Ans.
(1)

$N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$

$\begin{matrix} t = 0 & p & 0 & 0 \\ t = \infty & p - 0.5 p & 2 \times 0.5 p & \frac{1}{2} \times 0.5 p \\ = 0.5 p & = p & = 0.25 p \end{matrix} \begin{matrix} Total = 1.75 p \\ = \frac{7}{4} p \end{matrix}$