JEE Advance - Chemistry (1992 - No. 13)
2NO + O2 $$\to$$ 2NO2 $$\to$$ N2O4
The dimer, N2O4, solidifies at 262 K. A 250 ml flask and a 100 ml flask are separated by a stop-cock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm. and the smaller one contains oxygen at 0.789 atm. The gases are mixed by the opening stopcock and after the end of raction the flasks are cooled at 220K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume the gases to behave ideally)
Explanation
2NO + O2 $$\to$$ 2NO2 $$\to$$ N2O4
Number of moles of NO = $${{PV} \over {RT}}$$
$$ = {{1.053 \times 250} \over {0.0821 \times 300 \times 1000}} = 0.0107$$
Number of moles of
$${O_2} = {{0.789 \times 100} \over {0.0821 \times 300 \times 1000}} = 0.0032$$
Now 2 moles of NO need 1 mole of O2 for conversation into NO2.
$$\therefore$$ 0.0032 moles of O2 react with NO
= 0.0064 moles
NO left unreacted
= 0.0107 $$-$$ 0.0064 = 0.0043 mol
Total volume of the vessels
= 250 + 100 = 350 ml
Oxygen will be completely converted into NO2 and NO2 will then be completely converted into N2O4 (dimer) which becomes solid at 262 K; hence at 220 K, N2O4 is in solid state and only NO is present in gaseous state. Thus the whole volume (250 + 100 = 350 ml) of 350 ml is occupied by NO that has been left unreacted.
Therefore the pressure, P of NO gas = $${{nRT} \over V}$$
$$ = {{0.0043 \times 0.082 \times 220} \over {0.350}} = 0.221$$ atm
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