Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L,

Question

Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 65.0 kPa, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 250 kPa e temperature at the beginning and the end of the experiment is 22 °C After the stopcock is opened, the gases mix and react to form nitrogen dioxide NO o2 Which gases are present at the end of the reaction? NO NO2 Calculate the partial pressures of the gases. If the gas was consumed completely, put 0 for the answer Number Number Number kPa kPa kPa NO o, NO

Explanation / Answer

from idealgas equation

PV = nRT

n = no of mol of NO = PV/RT

P = 65 kpa = 0.6415 atm

V = 6 L

R = 0.0821 l.atm.k-1.mol-1

T = 22 c = 295 k

= (0.6415*6)/(0.0821*(295))

= 0.16 mol

n = no of mol of O2 = PV/RT

P = 250 kpa = 2.46 atm

V = 1.5 L

R = 0.0821 l.atm.k-1.mol-1

T = 22 c = 295 k

= (2.46*1.5)/(0.0821*(295))

= 0.152 mol

2NO + O2 ---> 2NO2

2 mol NO = 1 mol O2

limiting reactant = NO

no of mol of NO2 produced = 0.16 mol

no of mol of NO remaining = 0 mol

no of mol of O2 remaining = 0.152-(0.16/2) = 0.072 mol

gases present,at the end of the reaction = only O2,NO2

pNO2 = nRT/V = 0.16*0.0821*295/(6+1.5) = 0.516 atm

     = 52.3 kpa

pNO = 0 kpa

pO2 = 0.072*0.0821*295/(6+1.5) = 0.232 atm

    = 23.5 kpa

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