I think it's
I2 → 2I isn't it ? If it's then i can solve the sum
The Degree of dissociation of I2 molecule of 1000*C and under atmospheric pressure is 40% by volume.Find the total equilibrium pressure on the gas if the dissociation is reduced to 20% at the same temp.
I think it's
I2 → 2I isn't it ? If it's then i can solve the sum
I2(g)→2I(g)
P 0
P-p' 2p'
total pressure=P+p'=1
p'/P=0.4
so P=1/1.4=5/7
p'=2/7
K=(4/7)2/(3/7)=16/21
I2(g)→2I(g)
P 0
0.8P 0.4P
total pressure=1.2P
K=16/21=0.16P2/0.8P=P/5
P=80/21
total pressure=1.2P=1.2*80/21=96/21=32/7=4.57 atm