Question

5) Gaseous iodine pentafluoride, IFs, can be prepared by the reaction of solid iodine and gaseous fluorine: unbalanced equation A 5.00 L, flask containing 10.0 g of I2 is charged with 10.0 g of F2, and the reaction proceeds until one of the reactants is used up (completely consumed). After the reaction is completed, the temperature in the flask is 125. Calculate the pressure inside the flask 5 points

Answer 1

Given

Intial conditions

Volume = 5 L

T = 273 K

Mass of I2 = 10 g

Molar mass of I2 = 253.8 g/mol

No. of moles of I2 = mass / molar mass = 10 g/ 253.8 g/mol = 0.0394 moles

No. of moles of F2 = 10g / 38 g/mol = 0.263 moles

I2 (g) + 5 F2 (g) ---> 2 IF5 (g)

Intial no. of moles of F2 present = 0.263 moles will contribute for the intital pressure inside the flask

PV = nRT

P * 5 L = 0.263 moles * 0.08206 L.atm/mol.K * 273 K

P1 = 1.18 atm

T1 = 273 K

By reaction

according to stoichiometry

1 mole of I2 will react with 5 moles of F2

so 0.0394 moles of I2 will require 0.197 moles of F2

as F2 is in excess I2 is limiting reactant

so there will be 0.197 moles of IF5 and (0.263 moles - 0.197) = 0.066 moles of F2 in gaseous form

so total no of moles of gas = 0.197 moles + 0.066 moles = 0.263 moles of gas

T = 125 C = 398 K

P * 5 L = 0.263 moles * 0.08206 L.atm/mol.K * 398 K

P = 1.72 atm Answer

Answer 2

iodine . I2,has a higher melting point than fluorine F2 , because its