Grade 12th passPhysical Chemistry

please please explain in detail...........
the degree of dissociation (a)

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Yasir Rather

7 Years agoGrade 12th pass

1 Answer

aparazita sharma

7 Years ago

Hey Yasir!

Given equation is $2HI\rightleftharpoons H2+I2$ and we have to calculate relation between Kp and a.Now we know that Kp and Kc are related as Kp=Kc(RT)^∆ng........(1)

Where ∆ng is change in moles of products and reactants which is 0 in this case as 1mole H2 + 1 mole I2 = 2mole products= 2 mole reactant(HI)

Now eq (1) turns out to be Kp=Kc

That means if we write equation for Kc and eqaute it with Kp we will get relation b/w Kp and a.

Let's assume that we had 2 moles of HI initially.If a is degree of dissociation then no. of dissociated moles of HI will be 2a. So we are left with 2—2a moles at eqm.Simultaneously a moles of H2 and a moles of i2 form at eqm...

$2HI\rightleftharpoons H2+I2$

At eqm. 2-2a. a . a.

So Kc for this is (H2)(I2)/(HI)²

. a*a/(2-2a)²=a ²/(2-2a)² and this is equal to Kp acc to eqn 1..

Now taking square root both sides...we get √kp =a/2-2a

Solving it we get , 2√Kp / 1+2√Kp.....which is option d in ques (√2Kp is probably misprinted).

Hope it helps!!