Chemical BondingEasy
Question
For the following equilibrium
H2(g) ⇋ 2H(g)
There is 0.1% dissociation at equilibrium pressure of 1 atm. if pressure is 4 atm, % dissociation will be :-
H2(g) ⇋ 2H(g)
There is 0.1% dissociation at equilibrium pressure of 1 atm. if pressure is 4 atm, % dissociation will be :-
Options
A.0.4%
B.0.4%
C.0.2%
D.0.025%
Solution
Since dissociation is very less
Kp = 4x2P
4 × 1 × 10-6 × 1 = 4 × y24
y = 5 × 10-4 or 0.05 %
Kp = 4x2P
4 × 1 × 10-6 × 1 = 4 × y24
y = 5 × 10-4 or 0.05 %
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