Calculate K_{eq} with Reactants/Products: Units & Coefficients

[crybllrd]

We just started learning this in chem, and I thought it was quite simple.
I missed the particular lecture, and am now paying the price:

The following chemical reaction has reached equilibrium:
2NO + O$$_{2}$$ $$\rightarrow$$2NO$$_{2}$$
Calculate the equilibrium constant (K$$_{eq}$$) if the concentrations of reactants and products are 1.26 × 10$$^{-10}$$ M for NO, 2.28 × 10$$^{-3}$$ M for O$$_{2}$$, and 9.32 × 10$$^{-6}$$ M for NO$$_{2}$$.

K$$_{eq}$$=$$\frac{products}{reactants}$$

This is the equation I found, but what units do I use? Molar, mol, kg, ...
And how do coefficients play a role?

How do I proceed?

BTW I see my subscripts aren't correctly formatted, LATEX says it's correct, so I don't know how to fix it...

 

[Bohrok]

The molarities of the products go on top and molarities of the reactants on the bottom.

If you try to use LaTeX along with regular text, it won't be formatted very well; just stick to one or the other:
2NO + O₂ → 2NO₂
$2NO + O_2 \rightarrow 2NO_2$

 

[crybllrd]

Thats what I thought.

However, the answer is multiple choice, and I don't see my calculation there. That's why I was wondering if coefficients come into play.

$$K_{eq}=\frac{1.26e-10+2.28e-3}{9.32e-6}\approx244$$



So I thought maybe multiply each by its coefficient:

$$K_{eq}=\frac{(2)(1.26e-10)+(2.28e-3)}{(2)(9.32e-6)}\approx1.06e-8$$

My choices are:

$$2.40e12, 3.28e-7, 2.61e-17, or 2.40e-12$$

Am I making a mistake somewhere?

 

[Yanick]

The stoichiometric coefficients become exponents in Keq expressions. For example

aA + bB ---> cC + dD

Keq = ([C]^c*[D]^d)/([A]^a*^b)

 

[Borek]

http://en.wikipedia.org/wiki/Reaction_quotient

 

[crybllrd]

Still not getting a correct answer.
So, products over reactants, and each to the power of it's coefficient, right?

$$K_{eq}=\frac{(9.32e-6)^{2}}{(1.26e-10)^{2}+(2.28e-3)}=3.8e-8$$


My options are:

$$(A) 2.40e12 (B) 3.28e-7 (C) 2.61e-17 (D) 2.40e-12$$

I'm doing something wrong here, and I really have no idea what. This is due by 11pm, but more importantly I have an exam next Thursday so I need to figure it out.
Thanks again for the help so far.

 

[crybllrd]

Oh, after posting that I see my mistake.
I shouldn't add products or reactants, but multiply them. I had the correct equation, but for some reason I wanted to add them.
I got the correct answer now, 2.40e12.
Thanks again for all the help!

 

[hawkingfan]

OK, I'll explain how they derived the equation for the equilibrium constant and why it has that format. I don't know if you have studied the rates of reactions yet in your chem class but I'll still include that part of the explanation anyway.

As we all know, the rate of most chemical reactions depends on the concentration of its reactants. Suppose that we have a chemical reaction of the form:

aA + bB ----> cC + dD

The rate of a chemical reaction at any time "t" can be considered as the rate of disappearance of any particular reactant. (does not matter which one since they all disappear in a given ratio. The disappearance of all others can be determined if we know just how fast one of them is consumed)

The rate of any chemical reaction at a specified temperature by the equation:

rate = k([A]^m)(^n)

where m and n are the orders of reaction with respect to each reactant. These "orders" of reaction depend on the pathway by which the reaction proceeds. The parameter K is called a "rate constant" which must be experimentally determined and is dependent on the temperature of the system.

For a lot of simple reactions, the orders of reaction are equal to the value of the stoichiometric coefficients. (with the exception of zero order reactions or reactions with rate-determining steps) Therefore the rate equation for the forward reaction becomes

rate = k([A]^a)(^b)

***choose your equilibrium reaction appropriately and be sure of the mechanism. There are cases where this does derivation for equilibrium does not work.

Since this is a reversible reaction, we know that the products C and D can reacts again to give A and B at a particular rate. The rate equation for this reverse reaction is:

rate = K([C]^c)([D]^d)

I chose an uppercase K since the rate constants for the forward and reverse reactions are different. The definition of equilibrium requires that the forward and reverse reactions be equal. Therefore:

rate of forward reaction= k([A]^a)(^b) = K([C]^c)([D]^d) =rate of reverse reaction

If we solve for the ratio of rate constants on the left hand side, we get.

(k/K) = ([C]^c)([D]^d) / ([A]^a)(^b)

Since both rate constants are arbitrary constants, we can easily define the "equilibrium constant" as the ratio of these constants and we see that it is equal to the term on the right hand side. Once we know the equilibrium constant, we can calculate the concentrations of each chemical species based on certain initial conditions.