A sample of Hl (g) is placed in flask at pressure of 0.2 atm . At equilibrium the partial pressure of Hl is 0.04 atm . What is kp for given equilibrium?
2HI (g) ---> H2 (g) + I2 (g)
suneel10:
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kp=4
solution: kp= [H2]*[I2]/[HI]^2
=0.08*0.08/0.04*0.04
=4
DEGREE OF DISSOCIATION=0.2-0.04/2=0.08
nhi sama aaya
samaj*
DEKH PHLE TOTAL PRESSUE 0.2 THA JO SIRF HI KI VJH SE THA.. AB HI KA TOH 0.04 MTLB DECREASE HUYA 0.16..AGR HI 2 JAYEGE TOH EK EK H2 ND I2 BNEGE MTLB EQBM PE H2 ND I2=0.08.... AB VALUE PUT KRDE KP K FORMULE M
ho gya
thnx
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