At 35°C, K = 1.6 10-5 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) 3.8 mol of pure NOCl in a 2.0-L flask
NOCl ___ M
NO ___M
Cl2 __ M

(b) 2.4 mol of NO and 1.2 mol of Cl2 in a 1.0-L flask
NOCl___ M
NO ___ M
Cl2 ___ M

To solve this problem, we'll use the concept of the equilibrium constant (K) and the stoichiometry of the reaction.

First, let's write the balanced equation for the reaction:
2 NOCl(g) -> 2 NO(g) + Cl2(g)

For part (a), we have 3.8 mol of pure NOCl in a 2.0-L flask. To find the initial concentration of NOCl, we'll divide the number of moles by the volume:

Initial concentration of NOCl (a) = 3.8 mol / 2.0 L = 1.9 M

Since the initial concentration of NO and Cl2 is zero, their concentrations will be zero at the beginning.

Now, let's assume the final concentrations of NOCl, NO, and Cl2 at equilibrium to be x, y, and z, respectively.

According to the balanced equation, 2 moles of NOCl will produce 2 moles of NO and 1 mole of Cl2. This means that at equilibrium, the concentration of NO and Cl2 will be twice the concentration of NOCl.

Equilibrium concentration of NOCl (a) = Initial concentration of NOCl (a) - x = 1.9 - x M
Equilibrium concentration of NO (a) = 2x M
Equilibrium concentration of Cl2 (a) = x M

Now, we can use the given equilibrium constant (K = 1.6 x 10^-5 mol/L) to set up an equation:

K = [NO]^2 * [Cl2] / [NOCl]^2

Plugging in the equilibrium concentrations, we get:
K = (2x)^2 * x / (1.9 - x)^2

Now, we can solve this equation to find the value of x, which corresponds to the equilibrium concentration of NOCl (a).

For part (b), where we have 2.4 mol of NO and 1.2 mol of Cl2 in a 1.0-L flask, we can follow the same steps as before to find the equilibrium concentrations. The initial concentration of NO is 2.4 M, and the initial concentration of Cl2 is 1.2 M. The initial concentration of NOCl is zero.

Equilibrium concentration of NOCl (b) = Initial concentration of NOCl (b) + x = x M
Equilibrium concentration of NO (b) = Initial concentration of NO (b) - 2x = 2.4 - 2x M
Equilibrium concentration of Cl2 (b) = Initial concentration of Cl2 (b) - x = 1.2 - x M

Using the same equilibrium constant equation, we get:
K = [NO]^2 * [Cl2] / [NOCl]^2

Plugging in the equilibrium concentrations, we get:
K = (2.4 - 2x)^2 * (1.2 - x) / x^2

Now, we can solve this equation to find the values of x, which correspond to the equilibrium concentrations of NOCl (b), NO (b), and Cl2 (b).

I hope this helps! Let me know if you have any further questions.