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??

Question

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??

Vikas TU · Answer

H 2 + I 2 ---> 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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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+I₂---> 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??

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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+I2 ---> 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??

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H₂+I₂---> 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??