A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10^(–1) mol L–1. If value of Kc is 8.3 × 10^(–3), what are the concentrations of PCl3 and Cl2 at equilibrium?
Answers
Answered by
0
Answer:
hi friends
Explanation:
here is your answer
Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:
P
C
l
5
(
g
)
↔
P
C
l
3
(
g
)
+
C
l
2
(
g
)
PCl5(g) ↔ PCl3(g) + Cl2(g)
At equilibrium 0.5xx10^(-1) mol L^(-1)
x
m
o
l
L
−
1
xmolL-1
x
m
o
l
L
−
1
xmolL-1
It is given that the value of equilibrium constant,
K
C
KC is
8.3
×
10
−
3
8.3×10-3
Now we can write the expression for equilibrium as:
[
P
C
l
2
]
[
C
l
2
]
P
C
l
5
=
K
C
[PCl2][Cl2]PCl5=KC
⇒
x
×
x
0.5
×
10
−
1
=
8.3
×
10
−
1
⇒x×x0.5×10-1=8.3×10-1
⇒
x
2
=
4.15
×
10
−
4
⇒x2=4.15×10-4
⇒
x
=
2.04
×
10
−
2
⇒x=2.04×10-2
= 0.0204
= 0.02 (approximately)
Therefore at equilibrium.
[
P
C
l
3
]
=
[
C
l
2
]
=
0.02
m
o
l
L
−
1
Similar questions