Chem31 Outreach Workshop, Monday Nov. 22nd
Happy Thanksgiving!!!
1) Determining Equilibrium Constants
a) Write the expression for the equilibrium constant of the following
reaction:
4 NO2 (g) +
O2 (g)
2 N2O5 (g) Æ
K = [ P(NO2)4 * P(O2)] / P(N2O5)2
b)The gas cis-2-butene is converted to its isomer, trans-2-butene, also in
the gas phase, when it is heated. A 7 mol sample of cis-2-butene is heated
to 400oC in an 8.00 flask, and when the system reaches equilibrium 3.92 mols
of trans-2 butene are present. (no other products are formed) Calculate
the equilibrium constant for this reaction.
Cis-2-butene(g) ÅÆ trans-2-butene (g)
3.92 mols of trans-2-butene created means that 3.92 mols of cis-2-butene
were consumed, leaving 3.08 mols of cis-2-butene
[cis] = 3.08mols / 8.00= .385 M
[trans] = 3.92mols/8.00= .490M
Kc = [trans] / [cis] = .490M / .385M = 1.27
c) An empty 2.00L flask is filled with .200 mols of HI gas and is heated to
453oC to produce H2 and I2 gases. The reaction is allowed to come to
equilibrium, where it is found that the concentration of HI is now 0.078M.
Calculate Kc.
Find the initial [HI] = .200 mol / 2.00L = 0.100M
Set up the table of changes:
Conc (M)
2HI (g)
Initial
Change
Equilibrium
0.100M
-2x
0.100-2x
ÅÆ
H2(g)
0
+x
x
+
I2(g)
0
+x
x
First solve for x, so that we can determine all concentrations at equilibrium:
[HI] = .078M = 0.100-2xso
x = 0.011M
Now solve for
[H2] = x = [I2] = x = 0.011M
Plug into expression for Kc
Kc = [H2]*[I2] / [HI]2 = (.011*.011) / .0782 = 0.020
d) Determine Kc for the following reaction.
Kp = 2.1*10-4 at exactly 1000K
CaCO3(s) ÅÆ CaO (s) + CO2 (g)
Kp = Kc(RT)∆n(gas) so Kc = Kp / [(RT)∆n(gas)]
∆n (gas) = 1-0 = 1
Kc = (2.1*10-4 ) * [.0821 (atm*L/mol*K) * 1000K ]-1 = 2.6*10-6
e) Determine the equilibrium constant for the following reaction, given that
the K’s for the dissolution of HClO2(aq) and HNO2(aq) are .0110 and
4.57*10-4, respectively.
HClO2 (aq)
Σ
NO2- (aq)
+
(1)
HClO2 (aq)
(2)
H+ (aq)
HClO2 (aq)
+
ClO2- (aq)
H+ (aq)
+
NO2- (aq)
-
NO2 (aq)
+
HNO2 (aq)
+
ClO2- (aq)
NO2- (aq)
ClO2- (aq)
K = (0.0110) * (1 / 4.57*10-4) = 24.1
+
HNO2 (aq)
2. At 125 °C, KP = 4.0 atm-2 for the reaction:
2 NaHCO3 (s)
Na2CO3 (s) + CO2 (g) + H2O (g)
A 1.00 L flask containing 10.0 g of NaHCO3 (s), 0.10 atm of CO2 (g), and 0.10
atm of H2O (g) was heated to 125 °C. (Waymouth, Winter 1998 Exam 2)
a) Calculate the partial pressures of CO2 (g) and H2O (g) at equilibrium.
b) Calculate the masses of Na2CO3 (s) and NaHCO3 (s) present at
equilibrium. (Molar mass of Na2CO3 = 106 g mol-1; molar mass of NaHCO3 =
84.0 g mol-1.)
a)
First calculate Q to determine which way the equilibrium will
shift:
2
NaHCO3 ]
[
(1)
1
Q=
=
=
= 100 > K P
[Na 2CO 3 ][CO 2 ][H2O ] (1)(PCO 2 )(PH 2O ) (0.10)(0.10)
Therefore the system will shift to form reactants.
Now set up the table:
• units in atm
• let x = amount of CO2 (g) formed
Na2CO2 +
CO2
+
H2O
initial
(s)
0.10
0.10
change
+x
+x
+x
equil
(s)
0.10 + x
0.10 + x
2
NaHCO 3 ]
[
(1)
1
KP =
=
=
[Na 2CO3 ][CO2 ][H 2O ] (1)(PCO 2 )(PH 2O ) (0.10 +– x )2
2 NaHCO3
(s)
–2x
(s)
= 4.0
1
= 2.0; x = 0.40 atm
(0.10 – x)
At equilibrium PCO2 = PH2O = 0.10 + x = 0.50 atm
b)
We can determine the new amounts of the solids because we know the
original masses present and we know the change that has occurred from our
table. We know x in atm, so convert to moles:
(0.40 atm )(1.00L)
PV
=
= 0.0122 mol
n=
L atm ⎞⎟
RT ⎛⎜
0.0821
(398 K )
⎝
mol K ⎠
There were no grams of Na2CO3 present at the start and a
change of +x to get to equilibrium so
⎛ 106 g ⎞
⎟ = 1.29 g Na 2CO3
0.0122 mol Na2CO3 ⎜
⎝ 1mol ⎠
For NaHCO3 there were 10g to begin with and a change of –2x to get to
equilibrium, so first determine how x for NaHCO3
⎛ 84.0 g ⎞
⎟ = 1.02 g NaHCO 3
0.0122 mol NaHCO3 ⎜
⎝ 1mol ⎠
So we have 1.29 g Na2CO3 present at equilibrium, and
10.0 g – 2(1.02 g) = 7.96 g NaHCO3 remain