CHEMISTRY 1AO3
TUTORIAL #6
Oct. 1923, 2009
Solutions – Chapters 15 and 5
______________________________________________________________________________
Page 1 of 7
The following problems (17) are based on material from Chapters 15 and class notes.
1.
Write the expression for the equilibrium constant,
K
c
for each of the following reactions:
SOLUTION:
(i)
2 SO
2
(g) + O
2
(g)
2 SO
3
(g)
K
c
= [SO
3
]
2
/[SO
2
]
2
[O
2
]
(ii)
CH
3
COOH(aq) + H
2
O(l)
CH
3
COO
(aq) + H
3
O
+
(aq)
K
c
= [CH
3
COO
][H
3
O
+
]/[CH
3
COOH]
(iii)
CH
4
(g) + H
2
O (g)
CO (g) + 3H
2
(g)
Kc
= [CO][H2]3/[CH4][H2O]
(iv)
4 NO
2
(g)
2 N
2
O (g) + 3 O
2
(g)
K
c
= [N
2
O]
2
[O
2
]
3
/[NO
2
]
4
(v)
PCl
3
(l) + Cl
2
(g)
PCl
5
(s)
K
c
= 1/[Cl
2
]
2.
For the gas phase reaction H
2
+ I
2
2HI at 490
o
C, the equilibrium
concentrations (in mol litre
1
) are:
[H
2
] = 8.62
10
4
M,
[I
2
] = 2.63
10
3
M, [HI] = 1.02
10
2
M.
(a)
Calculate
K
c
for the equilibrium
K
c
= [HI]
2
/[H
2
][I
2
] = (1.02
10
2
)
2
/(8.62
10
4
)(2.63
10
3
)
= 45.9 (no units)
(b)
What is
K
c
for the reverse reaction, such that:
HI
1/2 H
2
+ 1/2I
2
?
For the reverse process,
K
c
'
= 1/
K
c
and also the stoichiometry is divided in half, so take
the square root:
K
c
'
= (1/
K
c
)
1/2
=
(1/45.9)
1/2
=
0.148
(c)
When one mole of H
2
and one mole of I
2
in a 1.00 litre container are heated to 490
o
C,
what are the equilibrium concentrations of H
2
, I
2
and HI ?
(mol/L)
[H
2
]
[I
2
]
[HI]
initial
1.00
1.00
0
change
 x
 x
+ 2x
at equilibrium
(1.00  x)
(1.00  x)
(2x)
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View Full DocumentCHEMISTRY 1AO3
TUTORIAL #6
Oct. 1923, 2009
Solutions – Chapters 15 and 5
______________________________________________________________________________
Page 2 of 7
K
c
= (2x)
2
/(1  x)
2
= 45.9
(this expression is a perfect square – take square root of both sides)
therefore, (2x)/(1  x) =
√
45.9 = 6.77, and solving for x gives
x = 0.772 mol/L
[H
2
] = [I
2
] = (1.00  x) = 0.23 mol/L
[HI] = 2x = 1.54 mol/L
(d)
Under the same conditions as for (c), an extra 0.20 mole of H
2
are added; what are the
concentrations of H
2
, I
2
and HI when equilibrium has been reached?
One could start with the equilibrium conditions for (c), but it is easier to assume the
additional moles were added at the beginning. The final equilibrium will be the same
either way.
(mol/L)
[H
2
]
[I
2
]
[HI]
initial
1.20
1.00
0
change
 x
 x
+ 2x
at equilibrium
(1.20  x)
(1.00  x)
(2x)
K
c
= (2x)
2
/(1.20  x)(1.00  x) = 45.9
which gives a quadratic: 41.9x
2
– 100.98x + 55.08 = 0
Find a solution, with the quadratic formula:
x
= [  b
(b
2
 4ac) ] / 2a, in this case a = 41.9, b = 100.98 and c = 55.08
Solve the quadratic to see that:
x = 1.58 or 0.834
(the value of 1.58 is not possible, since it would give a negative value for [H
2
] and [I
2
] )
therefore,
x = 0.834
is the correct solution
[H
2
] = (1.20  x) = 0.37 mol/L
[I
2
] = (1.00  x) = 0.17 mol/L;
[HI] = 2x = 1.67 mol/L
(e)
Explain the impact of performing the original reaction in (a) at higher pressure in terms
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 Spring '08
 Landry
 Equilibrium, Kc

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