Kc
[
[N
NO2 O2 ]
42]4.66 10 ^3
0.8(
02
0x)
2
x
0.804
0x^2
x3.73 10 ^3 4.66 10 ^3 x 4 x^2
4 x^2 4.66 10 ^3 x3.73 10 ^3 0Solving by the quadratic formula gives x3.00 10 ^2 and x3.11 10 ^2. We use
x3.00 10 ^2.
The original equilibrium concentrations are[NO 2 ] 2 x M6.00 10 ^2 M
[N 2 O 4 ](0.800x) M(0.8003.00 10 ^2 ) M0.770 M__? mol NO 2 1.00 L6.00 10 ^2 mol NO 2__? mol N 2 O 4 1.00 L0.770 mol N 2 O 4(b) When the volume of the reaction vessel is halved, the concentrations are doubled, so the
new initialconcentrations of N 2 O 4 and NO 2 are 2(0.770 M)1.54 Mand 2(6.00^10 ^2 M)
0.120 M,respectively.N 2 O 4 (g) 34 2NO 2 (g)
new initial 1.54 M 0.120 M
change due to rxn x M 2 x M
new equilibrium (1.54x) M (0.120 2 x) MKc4.66 10 ^3 Rearranging into the standard form of a quadratic equation givesx^2 0.121x1.81 10 ^3 0Solving as before gives x0.104 and x0.017.
The maximum value of xis 0.060 M,because 2xmay not exceed the concentration of NO 2
that was present after the volume was halved. Thus, x0.017 Mis the root with physical
significance. The new equilibrium concentrations in the 0.500-liter container are[NO 2 ](0.120^2 x) M(0.1200.034) M0.086 M
[N 2 O 4 ](1.54x) M (1.540.017) M 1.56 M__? mol NO 2 0.500 L0.043 mol NO 2__? mol N 2 O 4 0.500 L0.780 mol N 2 O 4In summary,First Equilibrium Stress New Equilibrium0.770 mol of N 2 O 4 Decrease volume from 0.780 mol of N 2 O 4
0.0600 mol of NO 2 1.00 L to 0.500 L 0.043 mol of NO 21.56 mol N 2 O 4
L0.086 mol NO 2
L(0.120 2 x)^2
1.54x[NO 2 ]^2
[N 2 O 4 ]0.770 mol N 2 O 4
L6.00 10 ^2 mol NO 2
LThe value of xis the number of moles
per liter of N 2 O 4 that react. So xmust
be positive and cannot be greater than
0.800 M.
0 x 0.800 MLeChatelier’s Principle tells us that
a decrease in volume (increase in
pressure) favors the production of
N 2 O 4.
Q9.35 10 ^3Q K
shift left
m8888888The root x0.104 would give a
negativeconcentration for NO 2 , which
is impossible.
(0.120)^2
1.54732 CHAPTER 17: Chemical Equilibrium