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Grade 12Physical Chemistry

Q)HI was heated in a closed tube at 440 degrees C
till 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.

Profile image of Raghav Rao
9 Years agoGrade 12
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2 Answers

Profile image of Priya
5 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 
Profile image of Arvind
5 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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