For the reaction $N_{2} O_{5} (g) ⇌ 2 {N O}_{2} (g) + 1 / 2 O_{2} (g)$ , calculate the mole fraction of $N_{2} O_{2} (g)$ decomposed at a constant volume and temperature, if the initial pressure is $600 m m$ $H g$ and the pressure at any time is $960 m m$ $H g$ . Assume ideal gas behaviour. If answer is $x$ then report $10 x$

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Solution

If p is the partial pressure of $N_{2} O_{5}$ that has decomposed
$N_{2} O_{5} (g) ⇋ 2 N O_{2} (g) + \frac{1}{2} O_{2} (g)$
600-p 2p p/2
Pressure at any time $= (600 m m H g - p) + 2 p + \frac{p}{2} = 600 m m H g + \frac{3}{2} p$
Equating this to 960 mmHg ,we get
$p = (\frac{2}{3}) (960 - 600) = 244 m m H g$
The mole fraction of $N_{2} O_{5}$ decomposed would be
$x = \frac{244}{600} = 0.407$

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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:

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For the reaction N2O5(g)⇌2NO2(g)+1/2O2(g), calculate the mole fraction of N2O2(g) decomposed at a constant volume and temperature, if the initial pressure is 600mm Hg and the pressure at any time is 960mm Hg. Assume ideal gas behaviour. If answer is x then report 10x

If p is the partial pressure of N2O5 that has decomposed N2O5(g)⇋2NO2(g)+12O2(g) 600-p 2p p/2Pressure at any time=(600mmHg−p)+2p+p2=600mmHg+32pEquating this to 960 mmHg ,we getp=(23)(960−600)=244mmHgThe mole fraction of N2O5 decomposed would bex=244600=0.407