Chemical Equilibrium?

Question

At 35°C, K = 1.6 x10^-5 mol/L for the following reaction.

2 NOCl(g) <--> 2 NO(g) + Cl2(g)

Calculate the concentrations of all species at equilibrium for each of the following original mixtures.

(a) 2.4 mol of NO and 1.2 mol of Cl2 in a 1.0 L flask
(b) 1.4 mol of NOCl and 1.4 mol of NO in a 1.0 L flask
(c) 2.8 mol of NOCl, 2.8 mol of NO, and 1.4 mol of Cl2 in a 1.0 L flask
(d) 1.9 mol/L concentration of all three gases
(e) 3.2 mol of NO and 1.1 mol of Cl2 in a 1.0 L flask

Can anyone explain to me how to do these problems?

Answer 1

The K refers to the equlibrium concentrations of the 3 gases as follows
[NO]^2[Cl2]/[NOCl]^2 = 1.6x10-5
Equilibrum can be attained from either the reactants end (no products to start), the products end (no reactant to start), or 2 of the 3 gases. Your problems illustrate this. I'll show you (d), hopefully you can work out the others.

When you have all 3 gases, it helps to guess which way the reaction is going. If you "plug" in the initial reactants, the result is 1.9, which is much higher than K. So, suspect that the reaction will run backwards. So let x be the moles/L of chlorine lost at equilibrium, so at equilibrium, we have 1.9-x moles. Since all reactions are in a liter flask, thats where we get the per liter. Then the other concentrations at equilibrium are
1.9-2x for NO and 1.9+2x for NOCl.
Substituting, (1.9-x)(1.9-2x)^2 /(1.9+2x)^2 =K
You can solve this by algebra or by trial and error. For example, if x = 0.9, we have (1)(0.01)/3.6^2.
which is about 7x10-4. From the term for NO, x cant exceed 0.95, otherwise the expression goes to zero. So, the answer should be pretty close to 0.95 for Cl2, 3.7 for NOCl, and somewhere around 0.04 mole/L.

Answer 2

In area equilibrium mass of each area reaches on equilibrium point and keeps to be there... jointly as in chemical equilibrium composition of gadget does no longer substitute whilst that's remoted from surrounding.. Sajid Iran Khan (pupil of Wah Engineering college,Wah Cantt .Pakistan)

Answer 3

[NO}^2 x [CL] divided by reactants which are [NOCL]^2
Take the concentrations of the products and see if they have coefficents. If they do take the power of them ex: NO^2. Then umltiple all products together and divide them by the reactants (which you raise to any nessicary powers). The answer you get will equal K so basically:

[NO]2 x [CL] = 1.6 x 10^-5
[NOCL]2