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H 2 +I 2 ---> 2HI. If the system is equiliberated at [H 2 ] = 0.5 and [I 2 ] = 0.5 and [HI] = 1.23 in a one litre flask and suddenly 0.6 mole of HI is removed, what will be the near equilibrium composition of [H 2 ] when the equilibrium is re established??

H2+I---> 2HI. If the system is equiliberated at [H2] = 0.5 and [I2] = 0.5 and [HI] = 1.23 in a one litre flask and suddenly 0.6 mole of HI is removed, what will be the near equilibrium composition of [H2] when the equilibrium is re established??
 
 

Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
                H2    +    I2     --->   2HI
t=eq.          0.5             0.5               1.23
t = req.      0.5 – x/2    0.5 – x/2        1.23 – x
x = 0.6 (given)
[H2] after re-eqm. we get, => 0.5 – 0.6/2 = > 0.5  –  0.3  ===> 0.2 mol/l

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