1. Write the equilibrium-constant expression, in terms of concentrations, for each of the following reactions.
(a) TiCl3(s) W TiCl(s) + Cl2(g)
Kc = [Cl2]
(b) CuCl42-(aq) W Cu2+(aq) + 4 Cl – (aq)
Kc = ([Cu2+ ][Cl – ]4)/[CuCl42- ]
(c) CO(g) + H2O(g) W CO2(g) + H2(g)
Kc = ([CO2][H2])/([CO][H2O])
(d) 4 H3O+(aq) + 2 Cl – (aq) + MnO2(s) W Mn2+ (aq) + 6 H2O(l) + Cl2(g)
Kc = ([Mn2+ ][Cl2])/([H3O+ ]4[Cl – ]2)
2. At a given temperature, the equilibrium constant Kc for the reaction
2 SO2(g) + O2(g) W 2 SO3(g)
is 2.4 x 10-3. What is the value of the equilibrium constant for each of the following reactions at that temperature?
(a) SO2(g) + 1/2 O2(g) W SO3(g)
This chemical equation is obtained by dividing the stoichiometric coefficients found in the original chemical equation through by two. Hence, the value of Kc is the square root of the Kc value for the original chemical equation.
Kc = (2.4 x 10-3)1/2 = 0.049
(b) 2 SO3(g) W 2 SO2(g) + O2(g)
This chemical equation is the reverse of the original chemical equation. Thus, the value of Kc is the reciprocal of the Kc value for the original equation.
Kc = 1/(2.4 x 10-3) = 420 (to 2 significant figures)
(c) SO3(g) W SO2(g) + 1/2 O2(g)
This chemical equation is the reverse of the chemical equation found in part (a). Hence, the value of Kc is the reciprocal of the Kc value determined in part (a).
Kc = 1/0.049 = 20.
Also notice that this chemical equation can be obtained by dividing the stoichiometric coefficients found in the chemical equation in part (b) through by two. Thus, the value of Kc is also the square root of the Kc value determined in part (b).
Kc = (420)1/2 = 20.
3. At 500°C, the equilibrium constant Kp for the synthesis of ammonia
N2(g) + 3 H2(g) W 2 NH3(g)
is 1.45 x 10-5. Calculate the value of Kc at this temperature.
The pressure-based equilibrium constant (Kp) and the concentration-based equilibrium constant (Kc) are related by the following equation
Kp = Kc(RT)Dn
where Dn is the number of moles of gaseous products – the number of moles of gaseous reactants. For this reaction we obtain for Dn
Dn = 2 – 4 = – 2
Kc = Kp/(RT)-2 = Kp(RT)2
where R = 0.08314 L·bar/mol·K and T = 500°C + 273 = 773 K
Kc = (1.45 x 10-5) x [(0.08314 L·bar/mol·K)(773 K)]2 = 0.0599
4. If the equilibrium constant Kc for the reaction
N2(g) + O2(g) W 2 NO(g)
is 1.5 x 10-3, in which direction will the reaction proceed if the concentrations of the three gases are all 1.00 x 10-3 M ?
We need to calculate Qc and compare it to the value of Kc.
Qc = [NO]2/{[N2][O2]} = (1.00 x 10-3)2/{(1.00 x 10-3)(1.00 x 10-3)} = 1.00
Since Qc > Kc (1.00 > 1.5 x 10-3) the reaction will proceed to the reactants side (or shift to the left).
5. The reaction
2 NO2(g) W N2O4(g)
has an equilibrium constant, Kc, of 170 at 25EC. If 2.0 x 10-3 mol of NO2 is present in a 10.0-L flask along with
1.5 x 10-3 mol of N2O4, is the system at equilibrium? If it is not at equilibrium, does the concentration of NO2 increase or
decrease as the system proceeds to equilibrium? Justify your reasoning in both instances.
Once again we need to calculate Qc and compare it to the value of Kc.
[N2O4] = 1.5 x 10-3 mol/10.0 L = 1.5 x 10-4 M
[NO2] = 2.0 x 10-3 mol/10.0 L = 2.0 x 10-4 M
Qc = [N2O4]/ [NO2]2 = (1.5 x 10-4)/(2.0 x 10-4)2 = 3800
Qc Kc so the reaction is not at equilibrium.
Since Qc > Kc the reaction mixture shifts to the left (the reactants side) which increases the concentration of NO2.
6. An aqueous solution of ethanol and acetic acid, each at an initial concentration of 0.810 M, is heated to 100EC. At
equilibrium, the acetic acid concentration is 0.748 M. Calculate Kc for the reaction
CH3CH2OH(aq) + CH3COOH(aq) W CH3COOCH2CH3(aq) + H2O(l)
ethanol acetic acid ethyl acetate
The amount of acetic acid consumed is 0.810 M – 0.748 M = 0.062 M. From the stoichiometry of the reaction this is the same as the amount of ethanol consumed and the amount of ethyl acetate produced.
[CH3CH2OH] [CH3COOH] [CH3COOCH2CH3]
Initial 0.810 0.810 0
Change – 0.062 – 0.062 +0.062
Equilibrium 0.748 0.748 0.062
Kc = [CH3COOCH2CH3]eq /([ CH3CH2OH]eqC[ CH3COOH]eq)
Notice that water does not appear in the equilibrium-constant expression since it is the solvent.
Kc = (0.062)/[(0.748)(0.748)] = 0.11
7. You place 3.00 mol of pure SO3 in an 8.00-L flask at 1150 K. At equilibrium, 0.580 mol of O2 has been formed. Calculate
Kc for the following reaction at 1150 K.
2 SO3(g) W 2 SO2(g) + O2(g)
[SO3]initial = 3.00 mol/8.00 L = 0.375 M [O2]eq = 0.580 mol/8.00 L = 0.0725 M
[SO3] [SO2] [O2]
Initial 0.375 0 0
Change – 2x +2x +x
Equilibrium 0.375 – 2x 2x x
We know that [O2]eq = 0.0725 M = x
This allows us to calculate the equilibrium concentrations of both SO2 and SO3.
[SO2]eq = 2x = 0.145 M
[SO3]eq = 0.375 – 2x = 0.375 M – 0.145 M = 0.230 M
Kc = ([SO2]2eqC[O2]eq)/ [SO3]2eq = [(0.145)2(0.0725)]/(0.230)2
Kc = 0.0288
8. Kc for the interconversion of butane and isobutane is 2.5 at 25EC.
butane W isobutane
(a) If you place 0.017 mol of butane in a 0.50-L flask at 25EC and allow equilibrium to be established, what will be the
equilibrium concentrations of butane and isobutane?
[butane]initial = 0.017 mol/0.50 L = 0.034 M
[Butane] [Isobutane]
Initial 0.034 0
Change – x +x
Equilibrium 0.034 – x x
Kc = [Isobutane]/[Butane] = 2.5
x / (0.034 – x) = 2.5
x = 2.5(0.034 – x) = 0.085 – 2.5x
3.5x = 0.085
x = 0.085/3.5 = 0.024
[Isobutane]eq = x = 0.024 M
[Butane]eq = 0.034 M – 0.024 M = 0.010 M
(b) Calculate the percentage of molecules that exist as isobutane in an equilibrium mixture at 25EC.
We need to recognize that initially all of the molecules are butane.
% isobutane = ([Isobutane]eq/[butane]initial) x 100%
% isobutane = (0.024/0.034) x 100% = 71%
The percentage of molecules that are isobutane can also be calculated as follows.
x Þ fraction of molecules that are isobutane
1 – x Þ fraction of molecules that are butane
Kc = x/(1 – x) = 2.5
x = 2.5(1 – x) Þ x = 2.5 – 2.5x Þ 3.5x = 2.5
x = 2.5/3.5 = 0.71
0.71 x 100% = 71%
(c) For each of the following reaction mixtures calculate the reaction quotient Q and determine whether the reaction
mixture is at equilibrium. If a reaction mixture is not at equilibrium does [isobutane] increase or decrease as the reaction mixture proceeds to equilibrium?
Mixture 1: Mixture 2: [butane] = 0.025 M [butane] = 0.012 M
isobutane] = 0.035M [isobutane] = 0.035M
Mixture 1:
Qc = [Isobutane]/[Butane] = 0.035/0.025 = 1.4
Qc Kc so the reaction mixture is not at equilibrium.
Since Qc < Kc the reaction mixture shifts to the right (the products side) which increases the concentration of isobutane.
Mixture 2:
Qc = [Isobutane]/[Butane] = 0.035/0.012 = 2.9
Qc Kc so the reaction mixture is not at equilibrium.
Since Qc > Kc the reaction mixture shifts to the left (the reactants side) which decreases the concentration of isobutane.
9. The reaction
N2(g) + O2(g) W 2 NO(g)
contributes to air pollution whenever a fuel is burned in air at high temperature, as in a gasoline engine. At 1500 K,
Kc = 1.0 x 10-5. Suppose a sample of air has [N2] = 0.80 M and [O2] = 0.20 M before any reaction occurs. Calculate the
equilibrium concentrations of reactants and products after the sample has been heated to 1500 K.
Kc = ([NO]2)/([N2][O2]) = 1.0 x 10-5
[N2] [ O2] [NO]
Initial 0.80 0.20 0
Change – x – x + 2x
Equilibrium 0.80 – x 0.20 – x 2x
Since 100(Ka) < [O2]initial Þ [0.0010 < 0.20] Þ we can assume that 0.80 – x . 0.80 and 0.20 – x . 0.20. This simplifies our problem to
(2x)2/[(0.80)(0.20)] = 1.0 x 10-5
4x2 = 1.6 x 10-6
x = 6.3 x 10-4
[N2]eq . 0.80 M – 6.3 x 10-4 M . 0.80 M
[O2]eq . 0.20 M – 6.3 x 10-4 M . 0.20 M
[NO]eq = 2x = 1.3 x 10-3 M
10. Dinitrogen trioxide decomposes to NO and NO2 with DH = 40.5 kJ/mol.
N2O3(g) W NO(g) + NO2(g)
Predict the effect the following changes will have on the position of the equilibrium, that is, state which way the equilibrium
will shift (left, right, no change) when each of the following changes is made.
(a) Adding more N2O3(g)
Add reactant Þ Shift right
(b) Adding more NO2(g)
Add product Þ Shift left
(c) Increasing the volume of the reaction flask
Increased volume results in decreased pressure so the reaction mixture shifts in the direction with more moles of gas.
Shift right
(d) Lowering the temperature
The forward reaction is endothermic (DH > 0). Lowering the temperature causes the reaction mixture to shift in the exothermic direction.
Shift left
