For the reaction, $N_{2} O_{5} (g) \to 2 N O_{2} (g) + 0.5 O_{2} (g)$ , calculate the mole fraction of $N_{2} O_{5} (g)$ decomposed at a constant volume and temperature if the initial pressure is 600 mm Hg and the pressure at any time is 960 mm Hg. Assume ideal gas behaviour.
Write the numerical value (upto 2 digit only) after decimal point only as for 0.19, write 19).

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Solution

$- Initial pressure Final pressure N_{2} O_{5} (g) 600 660 - P \to 2 N O_{2} (g) 0 2 P + \frac{1}{2} O_{2} (g) 0 P / 2$
$P \propto m o l e$ if V, T are constant
(where mole equivalent to pressure P are decomposed)
Thus, $600 - P + 2 P + P / 2 = 960$
$∴ P = 240 m m H g$
Thus, mole of $N_{2} O_{5}$ decomposed $= \frac{240}{600} = 0.4$
So, answer is 40.

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Q. For the reaction :

N_{2} O_{5} (g) \to 2 N O_{2} (g) + 0.5 O_{2} (g)

The mole fraction of

N_{2} O_{5} (g)

decomposed at constant volume and temperature is

4 \times 10^{- x}

,then calculate the value of x. if the initial pressure is 600 mm Hg and the pressure at any time is 960 mm Hg is.
Assume ideal behaviour.

Assume ideal gas behaviour. For the reaction, $N_{2} O_{5} (g) ⇌ 2 N O_{2} (g) + 0.5 O_{2} (g)$ , the initial pressure is 600 mmHg and the pressure at any time is 960 mmHg. The fraction of $N_{2} O_{5} (g)$ decomposed at constant volume and temperature is: