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11th
Chemistry
Equilibrium
Relationship between Equilibrium Constant, Reaction Quotient and Gibbs Energy
Relationship between Equilibrium Constant, Reaction Quotient and Gibbs Energy
Equilibrium
Chemistry
Class 11
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Relation Between Gibbs Free Energy and Equillibrium Constant
9 mins
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Law of Chemical Equilibrium, Equilibrium Constant - Examples
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Relation Between Gibbs Free Energy and Equillibrium Constant
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Relating Equilibrium Constant, Reaction Quotient and Gibbs Free Energy
Example
Definitions
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Related questions
A vessel contains three gases A, B and C in the equilibrium:
A
⇌
2
B
+
C
At equilibrium, the concentration of A was
3
M and that of B was
4
M. On doubling the volume of the vessel, the new equilibrium concentration of B was
3
M. Calculate
K
c
and the initial equilibrium concentration of C.
View solution
A reaction mixture containing
H
2
,
N
2
and
N
H
3
has partial pressure 2 atm, 1atm and 3 atm respectively at 725 K. If the value of K for the reaction.
N
2
+
3
H
2
⇌
2
N
H
3
is
4
.
2
8
×
1
0
−
5
atm
−
2
at 725 K, in which direction the net reaction will go:
View solution
△
G
∘
for
2
1
N
2
+
2
3
H
2
⇌
N
H
3
is
−
1
6
.
5
k
J
m
o
l
−
1
at
2
5
∘
C
. Find out
K
p
for the reaction. Also report
K
p
and
△
G
∘
for:
N
2
+
3
H
2
→
2
N
H
3
at
2
5
∘
C
.
View solution
The equilibrium constant of a reaction doubles on increasing the temperature of the reaction from
2
5
o
C
to
3
5
o
C
. Calculate enthalpy change of the reaction, assuming it to be constant in this temperature range.
View solution
Density of equilibrium mixture of
N
2
O
and
N
O
2
at 1 atm and 384 K is
1
.
8
4
g
/
d
m
3
. Equilibrium constant of the following reaction is:
N
2
O
4
⇌
2
N
O
2
View solution
1 mole of
P
C
l
3
and 1 mole of
P
C
l
5
is taken in a vessel of 10 L capacity maintained at 400 K.At equilibrium, the moles of
C
l
2
is found to be
4
×
1
0
−
3
View solution
Steam undergoes decomposition at high temperature as per the reaction
H
2
O
(
g
)
⇌
H
2
(
g
)
+
2
1
O
2
(
g
)
,
Δ
H
∘
=
2
0
0
k
J
m
o
l
−
1
Δ
S
∘
=
4
0
J
m
o
l
−
1
The temperature at which equilibrium constant is unit is :
View solution
Calculate value of
′
l
n
(
K
e
q
)
' for the reaction at
250 K.
N
2
O
4
(
g
)
⇌
2
N
O
2
(
g
)
Given:
H
f
0
(
(
N
O
2
)
g
)
=
+
4
0
.
4
0
7
k
J
/
m
o
l
H
f
0
(
(
N
2
O
4
)
g
)
=
+
7
0
k
J
/
m
o
l
S
r
0
=
1
0
J
K
−
1
View solution
At 473 K,
K
c
for the reaction:
P
C
l
5
(
g
)
⇌
P
C
l
3
(
g
)
+
C
l
2
(
g
)
is
8
.
3
×
1
0
−
3
.
What will be the value of
K
c
, for the formation of
P
C
l
5
(
g
)
at the same temperature?
View solution
1
mole of
N
O
and
1
mole of
O
3
are taken in a
1
0
L
vessel and heated. At equilibrium,
5
0
%
of
N
O
(by mass) reacts with
O
3
according to the equation:
N
O
(
g
)
+
O
3
(
g
)
⇌
N
O
2
(
g
)
+
O
2
(
g
)
What will be the equilibrium constant for this reaction?
View solution
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More topics
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Homogeneous and Heterogeneous Equilibria
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Relationship between Equilibrium Constant, Reaction Quotient and Gibbs Energy
Factors Affecting Equilibria
Ionic Equilibrium in Solution
Acids, Bases and Salts
Ionization of Acids and Bases
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Buffer Solutions
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