Problem 15.28 Consider the equilibrium N2( g )+O2( g )+Br2( g )2NOBr( g ) Part A
ID: 541131 • Letter: P
Question
Problem 15.28
Consider the equilibrium
N2(g)+O2(g)+Br2(g)2NOBr(g)
Part A
Calculate the equilibrium constant Kp for this reaction, given the following information (at 296 K ):
2NO(g)+Br2(g)2NOBr(g)Kc=2.0
2NO(g)N2(g)+O2(g)Kc=2.3×1030
Express your answer using two significant figures.
Problem 15.28
Consider the equilibrium
N2(g)+O2(g)+Br2(g)2NOBr(g)
Part A
Calculate the equilibrium constant Kp for this reaction, given the following information (at 296 K ):
2NO(g)+Br2(g)2NOBr(g)Kc=2.0
2NO(g)N2(g)+O2(g)Kc=2.3×1030
Express your answer using two significant figures.
Explanation / Answer
for the given reaction, N2(g)+ O2(g)+ Br2(g)<----> 2NOBr, KP= [PNOBr]2/ [PN2] [PO2] [PBr2]
for the given reaction, 2NO(g)+ Br2(g) <---> 2NOBr(g), Kc= 2, Kp = KC*(RT)deltan, deltan= change in moles of gases during the reaction, =2-1-2=-1, Kp = 2*(0.0821*298)-1 = 0.082= [PNOBr]2/ [PNO]2 [PBr2] (1)
for the 2NO(g) <---> N2(g) + O2(g), KC= 2.3*1030, Kp = KC*(RT)deltan, deltan= 2-1-1= 0, Kp =KC= 2.3*1030
Kp = [PN2] [PO2]/ [PNO]2 (2), Eq.2/Eq.1 = [PN2] [PO2] [PBr2] /[NOBr]2= 2.3*1030/0.082 = 2.8048*1031
which is the required value of Kp. ( P stands for partial pressure)