Q)HI was heated in a closed tube at 440 degrees Ctill equilibrium is reached . at this temperature 22% of HI was dissociated . The equilibrium constant for this dissociation is -a)0.282 b)0.0796 c)0.0199 d) 1.99.
Raghav Rao , 8 Years ago
Grade 12
2 Answers
Priya
Last Activity: 3 Years ago
Given reaction :
2HI is in equilibrium with H2 and I2
2HI H2 I2
Initial moles:. 2. -. -
At equilibrium. 2-2x. X. X
X=0.22 ( amount dissociated of HI)
So,
K= (x)(x)/(2-2x)^2
K=(0.22)^2/(1.56)^2
K=0.0199
Arvind
Last Activity: 3 Years ago
Given reaction :
2HI is in equilibrium with H2 and I2
2HI H2 I2
Initial moles:. 2. -. -
At equilibrium. 2-2x. X. X
X=0.22 ( amount dissociated of HI)
So,
K= (x)(x)/(2-2x)^2
K=(0.22)^2/(1.56)^2
K=0.0199
Given reaction :
2HI is in equilibrium with H2 and I2
2HI H2 I2
Initial moles:. 2. -. -
At equilibrium. 2-2x. X. X
X=0.22 ( amount dissociated of HI)
So,
K= (x)(x)/(2-2x)^2
K=(0.22)^2/(1.56)^2
K=0.0199
Given reaction :
2HI is in equilibrium with H2 and I2
2HI H2 I2
Initial moles:. 2. -. -
At equilibrium. 2-2x. X. X
X=0.22 ( amount dissociated of HI)
So,
K= (x)(x)/(2-2x)^2
K=(0.22)^2/(1.56)^2
K=0.0199
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