3 detail [731() |consider the dissociation reaction of dinitrogen tetroxide: N2O

Question

3 detail

[731() |consider the dissociation reaction of dinitrogen tetroxide: N2O4(g)2 With thermodynamic data shown in Table 2. Calculate the value of G' at 298 K. 2)Calculate the value of Kp at 298 K. NO2(g) 74] o produce more hydrogen fro 1)Calculate K at 1000 K mixture of CO and H0 2)What percentage wou (3) What would be the stainless steel ves initrogen ssure of a tetroxide and that is confined in a fixed volume of 245 L at 298 K. Substance 1 ArkJimole) | (J Kmole) | C." (JKnole) | N2 Table 2. Thermodynamic data at 298 K (4)What would be th G, (kJ/mole) Given: At 1000 9.179 305.376 240.034 205.14 191.609 77.28 37.20 29.355 29.125 (l)AH.--(2)( 33 0951-9 179 (1)At 1000 -57 01

Explanation / Answer

The reaction is N2O4(g)<->2NO2(g)

deltaGo= deltaH0- T*deltaSO

for the given reaction

deltaHo= sum of standard enthalpy of products- sum of standard enthalpy of reactants

deltaHo= 2* deltaHo of NO2- 1* deltaH of N2O4

where 2, 1 are coefficient of NO2 and N2O4 in the reaction

=2*33.095-1*9.179 =57.011 Kj /mole

Similarly, deltSo= 2*240.034-1*305.376 J/K=174.7 J/K= 0.175 Kj/K.mole

deltaGo= deltaHo-T*deltaSO= 57.011- 298*0.175 KJ=4.861 KJ/mole

since deltaGO= -RT lnK, K= equilibrium constant

lnK= - 4.861*1000/(8.314*298), K=0.141

from moles of N2O4, n=1, V= 24.5L, T= 298K, R=0.0821L.atm/mole.K

P= nRT/V= 1*0.0821*298/24.5 =0.998 atm

Let x= drop in pressure of N2O4

At Equilibrium, PN2O4=0.998-x, PNO2=2x, P represents partial pressure of gas

KP= (PNO2)2/PN2O4= 4x2/(0.998-x)= 0.141, x2/(0.998-x)=0.141/4= 0.03525

When solved using excel, x=0.1715 atm

At Equilibrium, PN2O4=0.998-0.1715 =0.8265 atm and NO2= 2*0.1715=0.3430 atm

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