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For rxn : H2+I2 gives 2HI value of equilibrium constant is 9 then what is the degree of dissociation of HI?

For rxn : H2+I2 gives 2HI value of equilibrium constant is 9 then what is the degree of dissociation of HI?

Grade:12th pass

2 Answers

Dilpreetkour Bali
28 Points
5 years ago
For this rxn alpha =2.k^1/2÷1+2k^1/2.substitute the value of k and get the ans
Alpha =2.9^1/2÷1+2.9^1/2
=6÷7
=.8
Vikas TU
14149 Points
5 years ago
In the given reaction he told the equilibrium constant for the formation of HI is 9 and he is asking the degree of dissociation of HI for which the equlibrium constant would be the reciprocal that is 1/9=0.111….
HI is in equilibrium with (0.5)H2 + (0.5)I2
            C                                          0         0
         C-Ca                                      Ca/2    Ca/2
0.1111=(Ca/2)/C-Ca
As C gets cancelled the value of a is 2/11=0.18.

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