The value of the equilibrium constant Kc is?

Question

Initially 0.500 mol/L NOCl(g) reaches equilibrium as shown in the equation below:

2 NOCl_(g) ↔ Cl_2(g)  + 2 NO_(g)

At equilibrium concentration of Cl₂(g) is 0.15 mol/L. 

The value of the equilibrium constant Kc is a.bc?

Write your answer for a.bcd in the blank, including the decimal.

Your Answer:

Answer 1

I'll put a disclaimer right up front that this may be a bit tricky to follow in a text format like this, since I won't be able to draw tables and equations as easily as I would like to, but I'll give it a shot. Hopefully this will be somewhat legible.

So to start, recall that for any reversible reaction (say, aA + bB <-> cC + dD), the equilibrium constant (K_c, in this case) for that reaction is defined by the equation [C]^c [D]^d / [A]^a [B]^b, where [A] is the molar concentration of chemical A at equilibrium and so on. We can make this equation more specific by plugging in the information given by the balanced reaction, so that K_c = [Cl₂][NO]² / [NOCl]². If we can find all of the equilibrium concentrations of the gasses involved, we can just plug them in to find the equilibrium constant.

So what do we know?

1. We know that we start with 0.500 mol/L of NOCl (and nothing else). We can assume that NOCl is the only chemical present at the start of the reaction, since no amounts are mentioned for any other chemicals.
2. We know that end with 0.15 mol/L of Cl₂, along with however much NO was created and however much NOCl we have left over. We generally assume that the container is closed (so all of the NO and Cl₂ had to come from only the 0.500 mol/L NOCl that we put in at the start) and rigid (so that we can treat concentration like moles, since the volume of the container is constant).
3. We know the stoichiometric ratio by which NOCl decomposes into NO and Cl₂. That is, we know how much of each chemical we have relative to each other chemical we have.

So lets start with the third point. We know that we have 0.15 mol/L of Cl₂ at equilibrium. Since all of this had to come from the decomposition of NOCl, and because the decomposition of NOCl creates two moles of NO for every mole of Cl₂, there must be 0.30 mol/L of NO in the container at equilibrium. Furthermore, since it takes two moles of NOCl to create one mole of Cl₂, we must have used 0.30 mol/L to make all of this Cl₂ and NO. So our equilibrium concentration of NOCl must be 0.500 mol/L - 0.30 mol/L (what we started with minus what we used).

Going back to our equation for the equilibrium constant, this gives us:

[Cl₂] = 0.15 mol/L

[NO] = 0.30 mol/L

[NOCl] = 0.20 mol/L

K_c = [Cl₂][NO]² / [NOCl]² = ((0.15 mol/L)(0.30 mol/L)²) / (0.20 mol/L)² = 0.3375

Which rounds off to 0.34 to two significant figures (since 0.15 only goes to the second decimal place).

Hope this helps! I'm happy to answer any questions/issues I may have missed with this response!

Answer 2

To determine the K_c of this reaction, you first need to use the equilibrium concentration of Cl₂ to stoichiometrically find the equilibrium concentrations of NOCL and NO:

1. NO eq. conc.=[(.15 mol/L Cl₂)(2 mol NO)]÷(1 mol Cl₂)= .3 mol NO produced at eq.
2. NOCl eq conc.=[(.15 mol/L Cl₂)××(2 mol NOCL)]÷(1 mol Cl₂)= .3 mol NOCl used (this is the amount of NOCl used in the reaction, so to get the amount left at equilibrium, you will have to subtract this from its initial concentration:
3. (.5 mol/L)–(.3 mol/L)= .2 mol/L NOCl at eq.
4. Now, to find K_c, set up the K expression (K= [products]÷[reactants]=[(eq. conc. Cl₂)(eq. conc. NO)]÷[(eq. conc. NOCL)] (make sure to raise the coefficients on each of these molecules to the exponents of their respective concentrations in the K expression):
5. K_c=[(.15)(.3)²]÷[(.2)²]= .3375 or 3.375x10^-1