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Tardigrade
Question
Chemistry
The Δ G° for the reaction =+57 kJ M X(s) leftharpoons M(a q)++X(a q)- Hence Ks p for the reaction is
Q. The
Δ
G
∘
for the reaction
=
+
57
k
J
M
X
(
s
)
⇌
M
(
a
q
)
+
+
X
(
a
q
)
−
Hence
K
s
p
for the reaction is
1631
196
Equilibrium
Report Error
A
1
0
−
5
B
1
0
−
6
C
1
0
−
10
D
1
0
−
16
Solution:
Δ
G
∘
=
−
2.303
×
8.31
×
298
lo
g
K
s
p
⇒
lo
g
K
s
p
=
2.303
×
8.31
×
298
−
57000
On solving we get
K
s
p
≃
1
0
−
10