An evacuated 1.000 L flask was filled at room temperature with 4.920E-2 mol NO (
ID: 881483 • Letter: A
Question
An evacuated 1.000 L flask was filled at room temperature with 4.920E-2 mol NO (nitrogen monoxide) and 7.74E-3 mol Br2 (bromine). The temperature was then raised to 50.5°C, and these substances reacted to give nitrosyl bromide, NOBr:
2NO(g) + Br(g) <--> 2NOBr(g)
When equilibrium was achieved, the pressure of the gases in the flask was found to be 1.315 atm.
n(Br2) = mol
n(NOBr) = mol (c) What are the partial pressures of the different gases in this equilibrium mixture? p(NO) = atm
p(Br2) = atm
p(NOBr) = atm (d) What is Kp at this temperature? Kp =
Explanation / Answer
a) we know that according to ideal gas equation
PV = nRT
Pressure = 1.315
Temperature= 50.5 = 273+50. K = 323.5 K
R = 0.0821
n= ?
V = 1
so
Total number of moles= PV / RT = 1.315 X 1 / 0.0821X 323.5 = 0.0495 moles
b)
2NO(g) + Br2(g) <--> 2NOBr(g)
Intial 0.04920 0.00774 0
change - 2x -x 2x
eqilibrium 0.0492-2x 0.00774-x 2x
As calcuated
0.0492 - 2x + 0.00774 -x +2x = 0.0495
0.05694 - 0.0495 = x
x = 0.00744
so mole of [NO] = 0.0492-2x = 0.0492-2(0.00744 ) = 0.03432
[Br2]= 0.00774- 0.00744 =0.0003
[NOBr] = 2 X 0.00744= 0.01488
c) Partial pressure of gas = mole fraction X totalpressure
Mole fraction = moles total moles
1) pNO = 0.03432 X 1.315 / 0.0495= 0.911
2) pBr2 = 0.0003 X 1.315 / 0.0495 = 0.00796
3) pNOBr = 0.01488 X 1.315 / 0.0495 = 0.395
d )
Kp = (pNOBr)2 / pNO)2X pBr2v =0.395 / 0.911 X 0.911 X0.00796 = 59.79