Consider the following reaction: N_2O_4 (g) doubleheadarrow 2 NO_2 (g) 0.1638 mo

Question

Consider the following reaction: N_2O_4 (g) doubleheadarrow 2 NO_2 (g) 0.1638 moles of N_2O_4 and 0.0116 moles of NO_2 are placed in an empty 5.66 L vessel at 292 degree K. At equilibrium, the total pressure in the vessel is found to be 0.775 bar. a. Assuming that the mixture behaves ideally, compute the total number of moles of gas present at equilibrium. b. Use an ICE table to compute the extent of the reaction at equilibrium at 292 degree K. c. Compute the partial pressures of N_2O_4 and NO_2 at equilibrium at 292 degree K. d. Use the data in Table C.3 in your textbook to compute Delta_r G degree and Delta_r H degree at 1000 degree K. e. Compute the equilibrium constant at 1000 degree K. f. Assuming that the Delta_r H degree is constant over the temperature range of interest, compute the equilibrium constant at 1186 degre K. g. Compute Q and Delta_r G (non-standard) for this reaction at 1000 degree K when the partial pressure of NO_2 is 2.44 bar and that of N_2O_4 is 1.96 bar.

Explanation / Answer

The reaction is N2O4ß-> 2NO2

                      N2O4                             NO2

Initial           0.1638                           0.0116

Change           -x                                     2x

Equilibrium   0.1638-x                     0.0116+2x

Total moles at equilibrium = 0.1638-x+0.0116+2x= 0.1754+x

Temperature , T = 292 K, P=0.775 bar =0.775*0.9869 atm = 0.765 atm, V= 5.66L

R=0.0821L.atm/mole.K

Number of moles from PV=nRT, n= PV/RT= 0.765*5.66/(0.0821*292) =0.1806

Hence 0.1754+x= 0.1806, x= 0.1806-0.1754=0.0052

Hence at equilibrium N2O4= 0.1638-0.0052 = 0.1586 and NO2= 0.0116+2*0.0052=0.022

b)Extent of reaction = 100*(moles reacted/ total moles )= 100*(0.0052/0.1638)= 3.17%

c) Partial pressure =moles fraction* total pressrue, mole fraction = moles/ total moles

partial pressures : N2O4= (0.1638-0.0052)/(0.1754+0.0052)*0.775 =0.6805 bar

NO2= 0.775-0.6805 = 0.0945

d) deltaH of reaction = deltaH of products –deltaH of reactants

enthalpy change : NO2= 33.18Kj/mole and N2O4= 9.16 KJ/mol

= 2* deltaH of NO2- deltaH of N2O4 = 2*33.18-9.16= 57.2 KJ

Simlarly entropy change =2*239.95 J/Kmole.-1*304.18 J/Kmole = 175.72J/K

Gibbs free energy change at 1000K= deltaH- T*deltaS= 57.2*1000-1000*175.72 J=-118520 Joules =

-118.52KJ

Equilibrium constant at 1000K, lnK= -deltaG/RT= 118520/(8.314*1000) =1552648

From ln (K1186/K1000)= (deltaH/R)*(1/1000-1/1186)

K1186= equilibrium constant at 1186K, K1000= equilibrium constnat at 1000K

deltaG= -118520 joules

ln(K1186/K1000)= (57.2*1000/8.314)*(1/1000-1/1186)

K1186= 4567401

g)

Q = [PNO2]2/ [PN2O4] = (2.44)2/ 1.96 = 3.04

deltaG= deltaG0+RTlnQ=

deltaGo = 57.2*1000-298.15*175.72 +8.314*1000*ln(3.04)=14053 joules

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