Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Standard free energy change for an equilibrium is zero, the value of Kp is
Q. Standard free energy change for an equilibrium is zero, the value of
K
p
is
2468
159
Haryana PMT
Haryana PMT 2009
Equilibrium
Report Error
A
0
36%
B
1
58%
C
2
3%
D
100
3%
Solution:
Standard free energy change,
Δ
G
o
=
−
2.303
RT
lo
g
K
p
If
Δ
G
o
=
0
⇒
0
=
−
2.303
RT
lo
g
K
p
lo
g
K
p
=
0
∴
K
p
=
1