Calculate the degrees of dissociation of N 2 O 4 into NO 2 at 25? C and 100? C a

Question

Calculate the degrees of dissociation of N2O4 into NO2 at 25? C and 100? C at 1 atm. Use the thermodynamic data table (page 988).

B. Thermodynamie Data Thermodynamic Data for Selected Elements and Inorganic Compounds at 1 bar and 298 K (continued) 188.7 -228.6 -237.2 -120.4 H20(9) 75.3 H:00) H 202(1) He(a) Hg(!) Hg (aq) HgO(red) -285.8 -187.8 109.6 0 20.79 27.98 164.4 -32.2 44.06 54.44 -90.8 116.13 55.19 26.48 29.16 64.18 29.58 K (aq) KOH(s) KCI(s) KCIOs(s) KNO1(s) Kr(g) -252.38 283.27 -424.8 82.59 142.97 132.9 164.08 28.03 14.23 50.21 32.68 -436.8 -289.9 393.1 0 100.3 -492.7 20.79 23.64 -293.8 -443.9 LiOH(s) Mg2+(aq) MgCl2(s) -487.2 Mg(s) 0 23.89 -466.9 601.8 -641.3 -138.1 MgO(s) -569.6 -591.8 26.78 89.62 37.41 N2(9) NH3(9) NHt (ag) NH CIs) NaH4U) NO(9) NO2(9) N204(9) N20(9) HNO,(I) HNo,(ag) -46.3 -16.6 192.5 35.66 -132.5 113.4 -202.87 149.4 50.63 139.3 29.86 210.6 33.9 240.5 98.29 103.6 -80.7 81.56 109.87 207.6-111.3 146.4 51.21 59.0 Na(s) 28.41 -240.12

Explanation / Answer

We know that

Delta G = Delta H - T Delta S

The dissociation of N2O4 is

N2O4 --> 2NO2

1) Delta H = 2XDelta H NO2 - Delta N2O4 = 2 X 33.9 - 9.7 = 58.1 KJ

Delta S = 2 X Delta S NO2 - Delta S N2O4 = 2X 240.5 - 304.3 = 176.7 J / mole K = 0.176 KJ / mol K

At temperature 25 0C = 298 K

Delta G = 58.1 - 298 X 0.176 = 5.65 KJ

Delta G = -RT lnK

lnK = -5.65 / 0.008314 X 298 = -2.28

K = 0.102

at temperature 100 0C = 373 K

Delta G = 58.1 - 373 X 0.176 = -7.55 KJ

Delta G = -RT lnK

lnK = -7.55 / 0.008314 X 373= -2.43

K =0.088

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