1.25 mol of NOCl was placed in a 2.50 L reaction chamber at 427 celciciu degre . After equilibrium was reached, 1.1 mole of NOCl remained. Calculate the equilibrium constant Kc for the reaction

............2NOCl ==> 2NO + Cl2

initial.....1.25.......0.....0
change.......-2x.......2x....x
equil......1.25-2x.....2x.....x

From problem, 1.25-2x =1.1 mols
Therefore, x = 0.075 mols
Convert x, 2x, and 1.25 to moles/L. M = moles/2.5L.
You now know (NO), (Cl2), and (NOCl).
Substitute into the Kc expression for NOCl and solve for Kc.

where did you get this question from ?

0.000432

dr bobs response was correct, but for anyone who needs to go further, 2NOCl is 1.1, 2NO is 0.15, and Cl2 is 0.075 by x. To get Kc, it would be (0.15^2)(0.075)/(1.1^2), which equals 1.4x10^-3.

To calculate the equilibrium constant, Kc, for the reaction, we need to use the formula:

Kc = ([Products]) / ([Reactants])

However, we need to first determine the initial concentrations of the reactants and the final concentration of the remaining reactant after equilibrium has been reached.

From the problem statement, we are given that the initial amount of NOCl was 1.25 mol, and 1.1 mol remained after equilibrium. To find the initial concentration of NOCl, we divide the initial amount by the volume of the reaction chamber:

Initial concentration of NOCl = 1.25 mol / 2.50 L = 0.5 M

The concentration of NOCl at equilibrium is 1.1 mol / 2.50 L = 0.44 M.

Now we can use these concentrations to calculate the equilibrium constant Kc:

Kc = ([NOCl]^n) / ([NO]^m[Cl2]^p)

In this case, the reaction is given as:

2NOCl → 2NO + Cl2

So, the concentrations used in the equation will be:

[NOCl] = 0.44 M
[NO] = 0 M (since NOCl was the limiting reactant and was not present in the products)
[Cl2] = 0 M (since Cl2 was not present in the reactants)

Therefore, the equation becomes:

Kc = (0 M)^2 / (0.5 M)^2

Since both [NO] and [Cl2] are 0, the equation becomes Kc = 0 / (0.5)^2 = 0.

Therefore, the equilibrium constant Kc for the reaction is 0.

5.6×10^-10