When gaseous F2 and solid I2 are heated to high temperatures, the I2 sublimes and gaseous iodine heptafluoride forms. 280 torr of F2 and 4.00g of I2 are put into a 2.50L container at 250K and the container is heated to 550K.
a) what is the final pressure in atm?
b) what is the partial pressure of I2 gas in atm?
I2 + 7F2
2IF7
No of moles of I2 = given mass / molar mass = 4 g / 253.8 g mol-1 = 0.0158 mol
No of moles of F2 :
We know, PV = nRT n =
PV/RT
PF2 = 280 torr = 0.368 atm, V = 2.5 L, T = 250 K, R = 0.082 L atm K-1 mol-1
n = ( 0.368 atm X 2.5 L) / (0.082 L atm K-1 mol-1 X 250 K) = (0.92 / 20.5) mol = 0.0449 mol
1 mol of I2 reacts with 7 moles of
F2 0.0158 mol
of I2 will completely react with 7 X 0.0158 = 0.1106 mol
of F2
But this is lesser than the number of moles of F2 available. It means that I2 is present in excess and that F2 is the limiting reagent.
7 mol of F2 reacts with 1 moles of
I2 0.0449 mol
of F2 will completely react with (1/7) X 0.0449 = 0.0064
mol of I2
Remaining moles of I2 = 0.0158 mol - 0.0064 mol = 0.0094 mol
Also, 7 mol of F2 form 2 moles of
IF7 0.0449 mol
of F2 will form (2/7) X 0.0449 = 0.0128 mol = No of
moles of IF7 formed.
After the reaction is complete there are 0.0128 mol IF7 + 0.0094 mol I2
So, total no of moles after completion of rxn = 0.0222 mol
(a) Pfinal = (nTotal RT) /
V
Pfinal = (0.0222 mol X 0.082 L atm K-1
mol-1 X 550 K) / 2.5 L
Pfinal = 0.4005 atm
(b) pI2 = partial pressure of I2 , PT = Total pressure
xI2 = mole fraction of I2 = moles of I2 / total no of moles = 0.0094 mol / 0.0222 mol = 0.4234
pI2 = xI2 X PT = 0.4234 X 0.4005 atm = 0.1696 atm
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