Equil'm Law Expression Problems( continued)
9. A 1.00 L vessel has initially 7.00 g of CO_{(}_{g)} and 4.50 g of steam. If K = 3.59, how many grams of CO_{2} were found at equilibrium?
CO_{(}_{g)} + H_{2}O_{(g)} = H_{2(g)} + CO_{2(g)}

CO_{(g)} 
H_{2}O_{(g)} 
H_{2(g)} 
CO_{2(g)} 
Initial number of Moles /L 
7/(12+16 g/mole)/1.0 L= 0.25moles/L 
4.5/(2+16 g/mole)/1.0 L= 0.25 
0 
0 
Moles /L reacting/forming 
x 
x 
x 
x 
Moles /L at equil'm 
0.25  x 
0.25 x 
x 
x 
x^{2}/(0.25x)^{ }(0.25x) = 3.59
0 = 0.2243750000 1.795000000x + 2.59x^{2}
x = 0.1636361455, or 0.5294140475
But x cannot be greater than original number of
moles, so x = 0.1636361455 = 0.16 moles/L.
in one liter:
0.16 moles (44g/mole) = 7.0 g.
10. Very sneaky problem!
[HI]^{2}/([H_{2}][I_{2}])
= Ka
[HI]^{2}/[0.0007/1][0.00413/1] = 55.6
[HI] = 0.0127 moles/L (don't round off too much!)
V = 1.00 L but it contains three gases! So
for n you have to add all the moles
n = 0.0127 + 0.0007 +
0.00413 = 0.01753moles
T = PV/(nR) = 101(1)/(0.01753*8.31) =
693 K
11. At equilibrium there are only 1.7 X 10^{2} moles of Pb+2 present for every liter of solution. What K is the K for the following reaction?
PbCl_{2(s)} = Pb^{+2}_{(aq) +} 2 Cl^{}_{(aq)}
K = [Pb^{+2}][ Cl^{}_{(aq)}]^{2}
= [1.7 X 10^{2}][2 X 1.7 X 10^{2}]^{2}
=0.000020
12. Given: CO_{(}_{g)} + H_{2}O_{(g)} = H_{2(g)} + CO_{2(g)}
a. A student introduces 0.25 moles of CO and 0.25 moles of steam (H_{2}O_{(}_{g)}) in a 1.00 L flask. If K_{eq} at 800 C is 3.59 and temperature is constant, how much hydrogen will there be at equilibrium?

CO_{(g)} 
H_{2}O_{(g)} 
H_{2(g)} 
CO_{2(g)} 
Initial number of moles/L 
0.25 
0.25 
0 
0 
moles/L reacting/forming 
x 
x 
x 
x 
moles/L at equil'm 
0.25 x 
0.25 x 
x 
x 
x^{2}/(0.25x)^{ }(0.25x) = 3.59
0 = 0.2243750000 1.795000000x + 2.59x^{2}
x = 0.1636361455, or 0.5294140475
But x cannot be greater than original number of
moles, so x = 0.1636361455 = 0.16 moles/L.
Concentration [H_{2}] = 0.16 moles/L
b. Repeat for K = 6.00.

CO_{(g)} 
H_{2}O_{(g)} 
H_{2(g)} 
CO_{2(g)} 
Initial number of moles/L 
0.25 
0.25 
0 
0 
moles/L reacting/forming 
x 
x 
x 
x 
moles/L at equil'm 
0.25 x 
0.25 x 
x 
x 
x^{2}/(0.25x)^{2}= 6
0 = 0.375  3x + 5x^{2}
x = 0.1775255129, 0.4224744871 But x cannot be greater than original number of moles, so x = 0.1775255129= 0.18 moles/L
Concentration [H_{2}] = 0.18 M.
13. Given : N_{2(g)} + 3H_{2(g)} = 2 NH_{3(g)}
At 400 C, [N_{2(}_{g)}]_{eq}= 0.45 M
[H_{2(}_{g)}]_{eq}= 1.10 M
If K = 1.7 X 10^{2}, find the equilibrium concentration of NH_{3}.
N_{2(}_{g)} + 3H_{2(g)} = 2 NH_{3(g)}

N_{2(g)} 
H_{2(g)} 
NH_{3(g)} 
Initial number of moles 
Doesn't matter here. 
Doesn't matter 
Doesn't matter 
Moles reacting/forming 
Doesn't matter 
Doesn't matter 
Doesn't matter 
Moles at equil'm 
0.45 
1.10 
x 
x^{2}/[(0.45)(1.10)^{3}]= 1.7 X 10^{2} ( note volume = 1 L; they gave us actual concentrations in
moles/L)
x = 0.100 M
Concentration [NH_{3(}_{g)}]
= 0.100 M.
14. Given : H_{2(g)} + I_{2(g)} = 2 HI_{(g)}
[H_{2(}_{g)}]_{eq}= 0.200 M
[I_{2(}_{g)}]_{eq}= 0.200 M
If K = 55.6, find the equilibrium concentration of [HI]_{eq}_{}
H_{2(}_{g)} + I_{2(g)} = 2 HI_{(g)}

H_{2(g)} 
I_{2(g)} 
HI_{(g)} 
Initial number of moles 
Doesn't matter here. 
Doesn't matter 
Doesn't matter 
Moles reacting/forming 
Doesn't matter 
Doesn't matter 
Doesn't matter 
Moles at equil'm 
0.200 
0.200 
x 
x^{2}/[(0.200)(0.200)]= 55.6 (
note volume = 1 L; they gave us actual concentrations in moles/L)
x = 1.49
Concentration [HI] = 1.49 M.
15. Given : W_{(g)} + X_{(g)} = U_{(g)} + V_{(g) }+_{ }T_{(g)}
a. A student introduces W and X into a flask. The initial concentration of each reactant is 0.80 mole/L. Slowly they reach equil'm and [U]_{eq} = 0.60 mole/L. Find K.
W_{(}_{g)} + X_{(g)} = U_{(g)} + V_{(g) }+_{ }T_{(g)}

W_{(g)} 
X_{(g)} 
U_{(g)} 
V_{(g)} 
T_{(g)} 
Initial number of moles 
0.80 
0.80 
0 
0 
0 
Moles reacting/forming 
W/U=1:1, so W= 0.60 
X/U=1:1, so W= 0.60 
0.60 
V/U = 1:1,so V= 0.60 
T/U = 1:1,so T= 0.60 
Moles at equil'm 
0.800.60 =0.20 
0.800.60 =0.20 
0.60 
0.60 
0.60 
(0.60) (0.60) (0.60) /[(0.20)(0.20)]=
K( note volume = 1 L; they gave us actual concentrations in moles/L)
K = 5.4.
b. The experiment is repeated at a different temperature in a 2.0 L flask. If K = 0.5, and if there were initially 0.40 moles of W_{2} and 0.40 moles of X_{2}, what expression could be used to solve for the number of moles of U found at equil'm? ( Don't attempt to solve the expression.)
W_{(}_{g)} + X_{(g)} = U_{(g)} + V_{(g) }+_{ }T_{(g)}

W_{(g)} 
X_{(g)} 
U_{(g)} 
V_{(g)} 
T_{(g)} 
Initial number of moles 
0.40 
0.40 
0 
0 
0 
Moles reacting/forming 
x 
x 
x 
x 
x 
Moles at equil'm 
0.40  x 
0.40  x 
x 
x 
x 
(x/2) (x/2) (x/2) /[(0.40  x)/2( 0.40  x)/2]= 0.5( note volume = 2 L)
16. At 2000 C, K for the following reaction = 1.6 X 10^{3}.
2 NO_{(}_{g)} = N_{2(g)} + O_{2(g)}
If there was only NO introduced initially and we end up with an equilibrium concentration of 0.13 M for NO, what will be the concentration of both nitrogen and oxygen at equilibrium?
2 NO_{(}_{g)} = N_{2(g)} + O_{2(g)}

NO_{(g)} 
N_{2(g)} 
O_{2(g)} 
Initial moles/L 
Doesn't matter here. 
0 
0 
moles/L reacting/forming 
x 
x 
x 
moles/L at equil'm 
0.13 
x 
x 
x^{2}/(0.13^{2})= 1.6 X 10^{3}.
x = 0.0052 moles
[N_{2(}_{g)}] = [O_{2(}_{g)}] = 0.0052 moles/L
17. a. When is K_{eq} = 1 ?
For aB + cD = eF+ gH
Since, if K=1, then [F]^{e}[H]^{g
}= [B]^{a}[D]^{c}.
b. When is K_{eq}_{
}a_{ }large number?
If K= large number, then [F]^{e}[H]^{g}
>> [B]^{a}[D]^{c}.