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182 6 CHEMICAL EQUILIBRIUM
E6B.5(b) Treating all species as perfect gases so that aJ = pJ/p−○ , the equilibrium constant
for the reaction CH3OH(g) +NOCl(g)⇌ HCl(g) +CH3NO2(g) is
K = aCH3NO2aHCl
aCH3OHaNOCl
= (pCH3NO2/p−○)(pHCl/p−○)
(pCH3OH/p−○)(pNOCl/p−○)
= pCH3NO2 pHCl
pCH3OHpNOCl
= (xCH3NO2 p)(xHClp)
(xCH3OHp)(xNOClp)
= xCH3NO2xHCl
xCH3OHxNOCl
= Kx
where Kx is the part of the equilibrium constant expression that contains the
equilibrium mole fractions of reactants and products. Because K is indepen-
dent of pressure, and K = Kx in this case, it follows that Kx does not change
when the pressure is changed. Hence the percentage change in Kx is zero .
E6B.6(b) �e following table is drawn up for the N2(g) + O2(g) ⇌ 2NO(g) reaction,
supposing that in order to reach equilibrium the reaction proceeds to the right
by an amount z moles.
N2 + O2 ⇌ 2NO
Initial amount nN2 ,0 nO2 ,0 0
Change to reach equilibrium −z −z +2z
Amount at equilibrium nN2 ,0 − z nO2 ,0 − z 2z
Mole fraction, xJ
nN2 ,0 − z
ntot
nO2 ,0 − z
ntot
2z
ntot
Partial pressure, pJ
(nN2 ,0 − z)p
ntot
(nO2 ,0 − z)p
ntot
2zp
ntot
�e total amount inmoles is ntot = nN2 ,0+nO2 ,0 at all times. Treating all species
as perfect gases so that aJ = pJ/p−○ the equilibrium constant is
K =
a2NO
aN2aO2
= (pNO/p−○)2
(pN2/p−○)(pO2/p−○)
=
p2NO
pN2 pO2
= (2zp/n)2
[(nN2 ,0 − z)p/ntot] [(nO2 ,0 − z)p/ntot]
= 4z2
(nN2 ,0 − z)(nO2 ,0 − z)
Rearranging gives
K(nN2 ,0 − z)(nO2 ,0 − z) = 4z2
Hence (4 − K)z2 + K(nN2 ,0 + nO2 ,0)z − nN2 ,0nO2 ,0K = 0
�e initial amounts are calculated as
nN2 ,0 =
5.0 g
28.02 gmol−1
= 0.178... mol nO2 ,0 =
2.0 g
32.00 gmol−1
= 0.0625... mol