When solid ammonium chloride dissociates at a certain temperature in a $\pu{0.500 dm3}$ container, ammonia and hydrogen chloride are formed.

$$\ce{NH4Cl <=> NH3 + HCl}$$

The initial amount of ammonium chloride was $\pu{1.00 mol}$, and when the system had reached equilibrium there was a $\pu{0.300 mol}$ of ammonium chloride.

What is the numerical value of $K_\mathrm{c}$ for this reaction under these conditions?

A 0.490
B 1.63
C 1.96
D 3.27

  • 1
    $\begingroup$ If my cat asked me that, I would say: "Meteor, why in the world would you want to know about equilibrium constants? You're a cat!" $\endgroup$ – Ivan Neretin Nov 12 '17 at 7:10
  • $\begingroup$ @IvanNeretin So why in the world is your cat called Meteor? $\endgroup$ – Jan Nov 12 '17 at 14:42
  • 1
    $\begingroup$ @Jan Well, he has a shape resembling the letter M on his forehead, and he tends to run around wildly and wreak havoc. So the name kinda fits. $\endgroup$ – Ivan Neretin Nov 12 '17 at 15:04

You must note that $[\ce{NH4Cl}]$ does not change as the density of solids almost remains constant throughout reaction.

At equilibrium, number of moles of $\ce{NH3} $ and $\ce{HCl}$ will be same and equal to $0.7$ from stoichiometry of reaction.

So the expression of $\ce{K_c}$ will be

$$\ce{K_c = [NH3][HCl] = \left(\frac{0.7 mol}{0.5 L}\right)^2 } $$

which is $1.96 \, \ce{\frac{mol^2}{L^2}}$

| improve this answer | |

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