At 35°C, Kc = 1.6 X 10-5 M for the reaction 2NOCI(g) 2NO(g) + Cl2 (g)' Calculate the

concentrations of all species at equilibrium for each of the following systems. Note any
simplifying assumptions made in the calculations.

3.0 mol NO(g) and 1.0 mol ofCl2(g) in a 1.0 L flask

To calculate the concentrations of all species at equilibrium, we need to use the given equilibrium constant (Kc) and the initial concentrations of the reactants.

The balanced equation for the reaction is:
2NOCI(g) ⇌ 2NO(g) + Cl2(g)

Given:
Initial concentration of NO(g) = 3.0 mol/L
Initial concentration of Cl2(g) = 1.0 mol/L

Let's assume x mol/liter is the amount of NOCI(g) that dissociates at equilibrium. This will result in the formation of x mol/L of NO(g) and Cl2(g).

At equilibrium, the concentrations will be:
[NOCI] = [initial NOCI] - x
[NO] = [initial NO] + x
[Cl2] = [initial Cl2] + x

Now, we can use the equilibrium constant expression to relate the concentrations of the species at equilibrium:

Kc = [NO]^2[Cl2] / [NOCI]^2

Substituting the equilibrium concentrations, we have:
Kc = ([initial NO] + x)^2 * ([initial Cl2] + x) / ([initial NOCI] - x)^2

Plugging in the given values:
Kc = (3.0 + x)^2 * (1.0 + x) / (1.6 * 10^-5)^2

Now, we solve for x:

1.6 * 10^-5 = (3.0 + x)^2 * (1.0 + x) / (1.6 * 10^-5)^2

Simplifying the equation and solving for x using algebra, you will find the value of x. Once you have the value of x, you can calculate the concentrations of all species at equilibrium by substituting x into the equilibrium expressions we defined earlier:

[NOCI] = [initial NOCI] - x
[NO] = [initial NO] + x
[Cl2] = [initial Cl2] + x

By plugging in the values, you will get the concentrations of all species at equilibrium for the given system.