Unit 5, Lesson 06: Equilibrium Problems #1 Worksheet With Answers Page 3
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1
2
3
4
5↔
↔
↔ ↔
5)
For the reaction:
PCl
PCl
+
Cl
K
= 0.042
5 (g)
3 (g)
2 (g)
eq
How many moles of PCl
must be reacted (put into) in a 0.500 litre flask in order to give 0.10 moles of
5
chlorine at equilibrium? ( [PCl
] = 1.15 M; n = C x V = 0.58 moles of PCl
needed )
5
5
PCl
PCl
Cl
5 (g)
3 (g)
2 (g)
x
0
0
I
– 0.20
+ 0.20
+ 0.20
C
E
x – 0.20
0.20
0.20
Keq
= [PCl
] [Cl
]
From the ICE table, x represents the initial concentration of PCl
,
3
2
5
[PCl
]
but the question asked for the number of moles of PCl
5
5
0.042
=
[0.20] [0.20]
Convert concentration to moles:
[ x – 0.20 ]
n = C x V
x
= 1.15
= 1.15 M x 0.500L
= 0.58 mol of PCl
(2 sig digs)
5
↔
↔ ↔
↔
–4
6)
At 1530˚C, for the reaction: N
+ O
2 NO
, the value of K
is 1.20 x 10
.
2 (g)
2 (g)
(g)
eq
1.00 mole of each nitrogen and oxygen gas are combined in a 1.00 L reaction container and allowed to
reach equilibrium. Determine the equilibrium concentration of NO
at these conditions.
(g)
N
O
2 NO
2 (g)
2 (g)
(g)
C = n/V = 1.0 M
1.0 M
0
I
C
– x
– x
+ 2x
E
1.0 – x
1.0 – x
2x
2
Keq
=
[NO]
Can we ignore the –x?
[N
] [O
]
2
2
1.0
is 8333 which is more than 500,
–4
–4
2
1.20 x 10
so we can ignore the –x
1.20 x 10
=
[2x]
[1.0 ] [1.0 ]
–4
2
1.20 x 10
= 4x
(remember to square both terms)
– 3
x
= 5.48 x 10
Substitute x into the last row of the ICE table to find the equilibrium concentrations:
at eq’m,
[NO] = 0.0110 M (3 sig digs)
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