Calculating Kp in two different ways


> 
> 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?


![enter image description here](https://i.stack.imgur.com/ojj82.png)




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


 3