Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl_2 (g) + 3

Question

Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl_2 (g) + 3F_2 (g) rightarrow 2 ClF_3 (g) A 2.00 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 855 mmHg. What is the pressure of CLF_3 in the reaction vessel after the reaction? Express your answer to three significant figures. What is the total pressure in the reaction vessel after the reaction? Express your answer to three significant figures.

Explanation / Answer

From gas law

PV=nRT, n= moles of gas = PV/RT

P= partial pressure of   the given gas in atom, V= 2L, R=0.0821 L.atm/mole.K, T= 298K

When chlorine is the gas, P= 337/760 atm =0.4434 atm, n= 0.4434*2/(0.0821*298)= 0.036 moles

Whenn Fluorine is the gas, P= 855/760 = 1.125 atm, n= 1.125*2/(0.0821*298)= 0.092

Theoretical moles ratio of of Cl2: F2 = 1:3

Actual ratio given =0.036:0.092 = 1:2.6

This suggests F2 is the limiting reactant. Hence all the F2 get consumed. Moles of ClF3 formed =(2/3)*0.092=0.061 moles

Hence partial pressure of ClF3= 0.061*0.0821*298/2 atm =0.746 atm, converting this into mm Hg, partial pressure of ClF3= 0.746*760 mm Hg =566.96 mm Hg

Moles of Cl2 reacted= 0.092/3= 0.031            , excess moles of Cl2 remaining=0.036-0.031=0.005

Partial pressure of Cl2= 0.005*0.0821*298/2 atm =0.0612 atm= 0.0612*760 mm Hg= 46.5 mm Hg

Total pressure after reaction = partial pressure of Cl2+ partial pressure of ClF3=46.5+566.96=613.46 mm Hg

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