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Question

When 36.6gN2O4(g) is introduced into a 1.0litre flask at 27C. The following equilibrium reaction occurs: N2O4(g)2NO2(g);Kp=0.1642 atm.
(a) Calculate Kc of the equilibrium reaction?
(b) What are the number of moles of N2O4 and NO2 at equilibrium?
(c) What is the total gas pressure in the flask at equilibrium?
(d) What is the percent dissociation of N2O4?

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Solution

36.6gofN2O4No.ofmoles=36.6920.40
Pv=nRT
P=nvRT36.692×1×0.084×300=9.80atm.
Kp=(2x)29.80x=0.1642
4x2=1.610.1642x
4x2+0.1642x1.61=0
x=0.614
PNO2=1.228atmPN2O4=9.186
(a)KC=KP(RT)ΔngΔng=21=1
0.1642×(0.084×300)10.1642/24.630.0067
(c)Totalpressure=9.186+1.22810.414atm
(b)nN2O4=PVRT9.186×10.084×3000.373mol.
NNO2=PVRT=1.2280.084×3000.050mol.
(d)%dissociation0.400.3730.40×1000.0675×1006.75%

1220504_764595_ans_f63e66ec6b624f7daeb757c1b6e2b732.JPG

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