0
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… read this try to apply here. @NehalSamee $\endgroup$ – Karmanya GB Mar 20 '18 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.