The degree of dissociation of $\ce{NH3}$ at $\pu{1 atm}$ is 20 % as follows: $$\ce{2NH3 <=> N2 + 3H2} $$

I follow two ways but end up with two different answers.

Here, I assume that initial amount of reactant mole is 2 and that of product is 0. Then I take the amount of ammonia in equilibrium to be $2-2\alpha$, nitrogen to be $\alpha$ and hydrogen to be $3\alpha$. Then, I find $K_\mathrm{p}$ (where $\alpha=0.2$).

Here, as dissociation is 20 %, I assume that amount of ammonia in equilibrium will be 0.8 and that of nitrogen and hydrogen will be 0.2 and 0.6 respectively. Then also, I find $K_\mathrm{p}$, but it differs from the initial one.

Where am I mistaken?

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First solution is perfect.

The issue is in the second solution.

If we start with 1 mole of $\ce{NH3}$ then with 20% dissociation we are left with 0.8 moles as 0.2 moles react these 0.2 moles give o.1 mole of $\ce{N2}$ and 0.3 moles of $\ce{H2}$

Hence $$K_\mathrm p = \frac{[\ce{H2}]^3[\ce{N2}]}{[\ce{NH3}]^2} = \frac{[\ce{0.3}]^3[\ce{0.1}]}{[\ce{0.8}]^2}$$

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  • $\begingroup$ That means for $2 mol$ , I should say $1.6 mol$ $NH_{3}$ remains in equilibrium ...? $\endgroup$ – Nehal Samee Mar 19 '18 at 5:42
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    $\begingroup$ @NehalSamee yes that is perfectly correct $\endgroup$ – Karmanya GB Mar 19 '18 at 13:22
  • $\begingroup$ ...But, if we take that total mole at equilibrium be 100 , then equilibrium contains 80 mol $NH_{3}$ , 15 mol $H_2$ and 5 mol $N_2$ ...Then , the calculation doesn't match ... $\endgroup$ – Nehal Samee Mar 20 '18 at 14:20
  • $\begingroup$ that is not the definition of degree of dissociation degree of dissociation is % of reactant that undergoes reaction @NehalSamee $\endgroup$ – Karmanya GB Mar 20 '18 at 15:29
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    $\begingroup$ en.wikipedia.org/wiki/… read this try to apply here. @NehalSamee $\endgroup$ – Karmanya GB Mar 20 '18 at 15:44

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