chemical equilibrium MC

1.
conservation of mass.
as it's a close system, no material is lost. no matter what form does the reactant turn into, all matter is retained in the flask.


2.
should the last reaction be
2NOCl + O2 <======> 2NO2 + Cl2 ?

first, write the Kc for first 2 equations:
K1 = [NO]^2 [O2] / [NOCl]^2
K2 = [NO]^2 [O2] / [NO2]^2

Kc of unknown reaction = [NO2]^2 [Cl2] / [NOCl]^2 [O2]

now, look at the terms. do they look familiar from K1 and K2 ?
Arrange the terms, you'll see
Kc = { [NO]^2 [Cl2] / [NOCl]^2 } x { [NO2]^2 / [NO]^2 [O2] }
= K1 / K2

do the simple division yields answer.

When one encounter this type of question, always write down Kc of all given equations. then, arrange the terms to fit the target Kc.
sometimes, you'll need using square and square-root or things similar.
or, just randomly produce an expression to see if it fits.


3.
Kc = [Cl2]^2 / [CCl4]
no. of mole of Cl2 produced = 0.06M x 1dm^3 = 0.06 mole
no. of mole of CCl4 used = 0.06/2 = 0.03 mole (look at the equation and mole ratio!).
so final no. of mole of CCl4 = 0.31 - 0.03 = 0.28 mole
conc. = 0.28 / 1 = 0.28 M

Kc = (0.06)^2 / 0.28 = 0.0128 M^-1


4.
similarly, but notice the unit of the answers.
" M^3 "
from the equation, product side has 3 mole of gas, so it has already contribute M^3 ;
that means, the reactant does NOT contribute to the unit of Kc; it's probably not a gas and need not be included in the expression of Kc.

now write the Kc.
Kc = [NH3]^2 [CO2]
no. of mole of CO2 = 0.18 M x 1 dm^3 = 0.18 mole
form equation, mole ratio of CO2 : NH3 = 1:2
so no. of mole of NH3 = 0.36 mole
conc. of NH3 = 0.36 M

Kc = 0.36^2 x 0.18 = 0.023 M^3

this question is a bit more tricky than Q.3, but still do-able if you're careful enough.