Consider the formation of nitric oxide (NO) from oxygen and nitrogen, by the rea

Question

Consider the formation of nitric oxide (NO) from oxygen and nitrogen, by the reaction: O_2 (g) + N_2 (g) 2 NO (g) Find the equilibrium constant for the above reaction at 298 K. Find the equilibrium constant for the above reaction at 1200 K and lbar. You may neglect the effect of differences in heat capacity between the reactants and products, i.e., k_2 can be considered equal to 1. If the reaction starts with a mixture of 79 mol % N_2 and 21 mol O_2, find the extent of the reaction at 1200 K and l bar. There is no need to calculate the final value of the extent of the reaction but your final equation should only contain E as the unknown. How will the extent of the reaction change if you were to double the pressure? Now, consider the following reaction occurring simultaneously with the one in Part 1 2 NO (g) + O_2 (g) 2 NO_2 (g) Starting again from 79 mol % N_2 and 21 mol O_2 find the extent of both reactions at equilibrium at 1200 K and 1 bar. You may neglect the effect of differences in heat capacity between the reactants and products, i.e., k_2 can be considered equal to 1 for both reactions. There is no need to calculate the final values of the extent of the reactions, but your final equations should only contain epsilon_l and epsilon_2 as the unknowns.

Explanation / Answer

Gibbs free energy change= 2* standard gibbs free energy of NO- { 1* standard gibbs free energy of N2+ 1* standars gibbs free energy of O2} = 2* 86.57- { 1*0+1*0}=173.14 KJ

where 2,1,1 are coefficients of NO, N2 and O2 respectively, deltaG of NO= 86.57 Kj/mole and that of N2 and O2=0
but deltaG=-RT lnK R=8.314 J/mole.K, K is equilibrium constant

lnK= -deltaG/RT= -173.14*1000/(8.314* 298), K at 298K is = 4.5*10-31

let T1= 298 K, K1=4.5*10-31 , T2= 1200K, deltaH= 2*90.25-{1*0+1*0}= 180.50 Kj

from Van't Hoff equation

ln (K2/K1)= (deltaH/R)*(1/T1-1/T2)

ln (K2/K1)= (180.50*1000/8.314)*(1/298-1/1200) , K2= 2.73*107

for the reaction N2+O2--------->2NO is carried out with 0.79 moles N2 and 0.21 mole O2

let x= drop in moles of O2 to reach equilibrium ( extent of conversion)

hence at equilibrium,   moles : N2= 0.79-x. , O2= 0.21-x , NO=2x

K= [NO]2/ [N2][O2] = (2x)2/ (0.79-x)*(0.21-x) =4x2/(0.79-x)*(0.21-x)

when pressure is 2 atm, the partial pressures of N2 and O2 are 0.158 and 0.42, let P= change in partial pressure of O2 to reach equilibirum, at equilibrium, PNO= 2P, PN2= 0.158-P and PO2=0.42-P

K= 4P2/ (0.158-P)*(0.42-P), P is the extent of reaction. Since the no of moles of N2 and O2 increases, with an increase in pressure. . more no of moles of NO forms as per lechatlier principle.

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