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 CO2 were found at equilibrium?

CO(g) + H2O(g) = H2(g) + CO2(g)

 CO(g) H2O(g) H2(g) CO2(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

x2/(0.25-x) (0.25-x) = 3.59

0 = 0.2243750000 -1.795000000x + 2.59x2

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/([H2][I2]) = 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?

PbCl2(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) + H2O(g) = H2(g) + CO2(g)

a. A student introduces 0.25 moles of CO and 0.25 moles of steam (H2O(g)) in a 1.00 L flask. If Keq at 800 C is 3.59 and temperature is constant, how much hydrogen will there be at equilibrium?

 CO(g) H2O(g) H2(g) CO2(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

x2/(0.25-x) (0.25-x) = 3.59

0 = 0.2243750000 -1.795000000x + 2.59x2

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 [H2] = 0.16 moles/L

b. Repeat for K = 6.00.

 CO(g) H2O(g) H2(g) CO2(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

x2/(0.25-x)2= 6

0 = 0.375 - 3x + 5x2

x = 0.1775255129, 0.4224744871 But x cannot be greater than original number of moles, so x = 0.1775255129= 0.18 moles/L

Concentration [H2] = 0.18 M.

13. Given : N2(g) + 3H2(g) = 2 NH3(g)

At 400 C, [N2(g)]eq= 0.45 M

[H2(g)]eq= 1.10 M

If K = 1.7 X 10-2, find the equilibrium concentration of NH3.

N2(g) + 3H2(g) = 2 NH3(g)

 N2(g) H2(g) NH3(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

x2/[(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 [NH3(g)] = 0.100 M.

14. Given : H2(g) + I2(g) = 2 HI(g)

[H2(g)]eq= 0.200 M

[I2(g)]eq= 0.200 M

If K = 55.6, find the equilibrium concentration of [HI]eq

H2(g) + I2(g) = 2 HI(g)

 H2(g) I2(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

x2/[(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 0.80-0.60 =0.20 0.80-0.60 =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 W2 and 0.40 moles of X2, 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) = N2(g) + O2(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) = N2(g) + O2(g)

 NO(g) N2(g) O2(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

x2/(0.132)= 1.6 X 10-3.

x = 0.0052 moles /L

[N2(g)] = [O2(g)] = 0.0052 moles/L

17. a. When is Keq = 1 ?

For aB + cD = eF+ gH

Since, if K=1, then [F]e[H]g = [B]a[D]c.

b. When is Keq a large number?

If K= large number, then [F]e[H]g >> [B]a[D]c.