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Byju's Answer
Standard XII
Physics
Tension in a String
The total pre...
Question
The total pressure at which
A
(
g
)
is
50
%
dissociated according to equation
A
(
g
)
⇌
B
(
g
)
+
C
(
g
)
is numerically equal to ______ times of
K
p
.
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Solution
The given reaction is :-
A
(
g
)
⇌
B
(
b
)
+
C
(
g
)
Initial :
P
0
0
(let)
pressure
At eqm :
P
(
1
−
α
)
α
P
α
P
(
α
is the degree of dissociation)
=
0.5
P
0.5
P
0.5
P
(
α
=
50
% given)
At eqm,
P
T
o
t
a
l
=
1.5
P
→
(
I
)
Now,
K
P
=
p
C
.
p
B
p
A
K
P
=
0.5
P
×
0.5
P
0.5
P
=
0.5
P
→
(
I
I
)
(
I
)
&
(
I
I
)
⇒
3
K
P
=
P
T
o
t
a
l
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Similar questions
Q.
Show that for reaction AB(g)
⇌
A(g) + B(g), the total pressure at which AB is 50% dissociated is numerically equal to three times of
K
p
.
Q.
The total pressure at which A
(
)
is
50
dissociated according to the equation A
(
)
⇔
B
(
)
+
C
(
)
is numerically equal to...........times of Kp.
Q.
For the reaction:
A
B
(
g
)
⇌
A
(
g
)
+
B
(
g
)
A
B
is
33
% dissociated at a total pressure of
p
. Then
p
K
p
will be equal to :
Q.
For the reaction
A
B
(
g
)
⇌
A
(
g
)
+
B
(
g
)
, AB is
33
%
dissociated at a total pressure of P. Therefore P is related to
K
p
is :
Q.
The pressure necessary to obtain
50
%
dissociation of
P
C
l
5
at
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is numerically equal to how many times the value of the equilibrium constant
K
p
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