/15 KNOW
1.
/33 INQ
NAME:
EQUILIBRIUM QUEST A
For the equilibrium system below, use Le Châtelier’s Principle to determine the shift in
the equilibrium and the effect on the indicated variable. (12 K)
N2 (g) + 2 O2 (g) + energy ⇌ 2 NO2 (g)
Stress
SHIFT
Variable
EFFECT OF SHIFT
Increase [oxygen]
right
[N2 (g) ]
decrease
Decrease Volume
right
Keq
No effect
Decrease Temperature
left
Keq
decrease
Add a Catalyst
No effect
[O2 (g)]
No effect
Add Ne at Constant Volume
No effect
[NO2 (g)]
No effect
Decrease Pressure
left
[N2 (g) ]
increase
2.
For each reaction below predict whether the reactants or products will be favoured. (3 K)
⇌ N2O4 (g)
Keq = 6.5 x 10-5
___reactants______
H2 (g) + Cl2 (g) ⇌ HCl (g)
Keq = 7.60 x 109
____
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Keq = 3.5 x 10 -4
____
A)
2 NO2
B)
C)
(g)
products_______
reactants__________
PART TWO: PROBLEMS (33 marks INQ)
MAKE SURE YOU ORGANIZE YOUR SOLUTIONS. I WILL NOT MARK A MESS OF
NUMBERS!!!!!!!!!!
1.
Given the data below, calculate the value for the equilibrium constant. (2 marks)
2 NO (g) + Cl2(g) ⇌ 2 NOCl (g)
At equilibrium [NO (g)] = 1.20 M [Cl2(g)] = 0.75 M, [NOCl (g)] = 0.88 M
Keq
= [NOCl]2/[NO][Cl2]
= (0.88)2/(1.20)2(0.75)
= 0.72
Therefore Keq is 0.72
2.
Carbonyl bromide, COBr2, decomposes to form carbon monoxide and bromine gases. At
250°C, carbonyl bromide is 38 % molar decomposed. If a 2.0 L vessel originally contains
3.6 mol of carbonyl bromide. Calculate the value of the equilibrium constant at 250°C.
(4 marks)
COBr2(g) ⇌ CO(g) + Br2(g)
MR
I
C
E
1
1.8
-x
1.8-x
1
0
+x
x
1
0
+x
x
At equilibrium COBr2 is 38% decomposed
Therefore
x= (0.38)(1.8)
x= 0.68
Keq
3.
= [CO][Br2]/[COBr2]
=(0.68)(0.68)/(1.8-0.68)
=0.41
A 5.0 L vessel initially contains 2.5 mol of carbon dioxide and 1.5 mol of hydrogen. The
following equilibrium is established;
CO2 (g) + H2(g) ⇌ H2O (g) + CO (g)
a)
MR
I
C
E
Keq = 2.00 x 10-4
Find the equilibrium concentrations of all species. (4 marks)
CO2 (g) + H2(g) ⇌ H2O (g) + CO (g) Keq = 2.00 x 10-4
1
1
1
1
0.50
0.30
0
0
-x
-x
+x
+x
0.50-x 0.30-x
x
x
Keq= [H2O][CO]/[CO2][H2]
0.000200 = (x)(x)/(0.50-x)(0.30-x)
Approx works
0.000200(0.15)=x2
x=0.0055
[H2O]=0.0055 M, [CO]=0.0055 M, [CO2]=0.50 M, [H2]=0.30 M
b)
Calculate the % reaction for the equilibrium system above. (3 marks)
CO2 (g) + H2(g) ⇌ H2O (g) + CO (g)
Moles =2.50
1.50 therefore hydrogen is the limiting factor and a maximum of 1.50
mol (0.30 M) of CO can be made.
% rxn =equil/theor x 100%
= 0.0055/0.30 x 100%
=1.8
4. Hydrogen cyanide gas decomposes to form hydrogen gas and cyanogen gas (C2N2). At
240°C, the value of the equilibrium constant is 0.45. Determine the equilibrium concentrations
of all species if 1.80 mol of hydrogen and 1.80 mol of cyanogen gas are placed in a 3.0 L vessel
and allowed to come to equilibrium. (5 marks)
2 HCN(g) ⇌ H2(g) + C2N2(g) Keq=0.45
MR
2
1
1
I
0
0.60
0.60
C
+2x
-x
-x
E
2x
0.60-x 0.60-x
Keq=[H2][C2N2]/[CN]2
0.45=(0.60-x)2/(2x)2
0.67(2x)=0.60-x
2.34x=0.60
x=0.26
[HCN]=0.52 M [H2]=0.34 M [C2N2]=0.34 M
5.
A gaseous reaction contains 1.00 mol of SO2Cl2 gas, 0.60 mol of SO2 gas and 0.32 mol of
Cl2 gas in a 2.0 L vessel. If the system is allowed to come to equilibrium as shown
below, determine the concentration of all species at equilibrium. (5 marks)
SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g)
Keq = 0.011
MR
1
1
1
I
0.50
0.30
0.16
C
-x
+x
+x
E
0.50-x 0.30+x 0.16+x
Keq=[ SO2 ][Cl2]/[SO2Cl2]
0.011=(0.30+x)(0.16+x)/0.50-x
0.0055-0.011x=(0.048+0.46x+x2)
x2+0.471x+0.0425=0
a=1, b=0.471, c=0.0425
x1=-0.12, x2=-0.35
[SO2Cl2]=0.62 M, [SO2 ]=0.18 M, [Cl2]=0.04 M
6.
A 1.0 L vessel contains 3.0 mol of A and B and some amount of both C and D. The
amount of D initially present is twice the amount of C. The gases are allowed to come to
equilibrium as shown below. At equilibrium, the concentration of D is 4.0 mol/L. Given
that the value of the equilibrium constant is 81.0, calculate;
A)
the equilibrium concentrations of all species. (7 marks)
B)
the initial amounts of C and D (3 marks)
A (g) +
MR
I
C
E
1
3.0
-x
3.0-x
B(g) ⇌
1
3.0
-x
3.0-x
2 C (g) + 2 D (g) Keq = 81.0
2
2
y
2y
+2x
+2x
y+2x
2y+2x=4.0
=2.0-x+2x
y=2.0-x
= 2.0+x
Keq = [C]2[D]2/[A][B]
81 =(2.0+x)2(4.0)2/(3.0-x)2
(9)=(2.0+x)(4.0)/ (3.0-x)
27.0-9.0x = 8.0+4.0x
19=13x
x=1.46
a) [A]=1.56 M, [B]=1.56 M, [C]=3.46 M, [D]=4.0 M
b) y=2.0-1.46 = 0.54 M
initial amount of C=0.54 mol and initial amount of D=1.08 mol