Chemistry Quesiton about Equilibrium (College)?

Question

A 2.5 mole NOCL in 1.5 L in a 400 degrees Celcius solution. After it reaches equilibrium, 28% of NOCl was dissacciated.

2NOCL(g)---->2NO(g)+Cl2(g) (Reversible Reaction)

Calculate Kc.

I am stumped. Anyone have an idea?

Answer 1

Gotta do the walk to show you can talk the talk.
The initial conc of NOCl is 2.5 M/1.5 L or 1.67 M/L (you don't need to have a liquid to have mole/liter problems). If 28% of this is gone, you have
0.72(1.67) or about 1.2 M/L left, and about 0.45 moles reacted. From the reaction equation, you would have 0.45 moles of nitrous oxide and 0.225 moles of Cl2.

Then, since Kc = [NO]^2 [Cl] / [NOCl]^2, just substitute in the equilibrium concs and solve.

Answer 2

Dissociation constant involves the follwing:
[product A]ex A times [product B]ex B divided by [reactant] ex Reactant. ex = exponent. [A]ex 2 means the concentration of A "[A]" to the second power.

The brackets [ ] means "concentration"

What this means is as follows:
[NO] ex2 times [Cl] divided by [NOCl] ex2 or this is read as:
"The concentration of NO to the 2nd power times the concentration of Cl2 to the first power divided by NOCl to its second power".

This is only a start. Generally, the products of the chemical reaction are placed in the numerator and the reactants are placed in the denominator.

I hope this helps as a start. You'll need a lot more time....spend it!