HS Chemistry - Reversible Reactions

More About Dynamic Equilibrium

Overview of The Page

This page will cover:

What is the equilibrium constant of a reversible reaction?
What are ICE tables? How are they used?
What is the pressure constant of a reversible reaction?

The previous subpage showed the position of equilibrium of a reversible reaction graphically:

Position of Equilibrium on a spectrum

This gives us the information we need it to give, but as noted on the previous page, they aren't formal diagrams to use - they're just drawn to help visualize what is meant by position of equilibrium. Wouldn't it be great to have something that could be formally and officially used to convey the same information? Preferably something that didn't require drawing an entire diagram.

As it turns out, there is - the equilibrium constant $K~c~$ of a reversible reaction.

The K_c of a reaction is a ratio of the concentrations of product molecules to the concentrations of reactant molecules. It shows the extent of the reaction. Essentially:

K_c = products ÷ reactants

When K_c = 1 for a reversible reaction, 50% of the molecules will be product molecules and 50% will be reactant molecules when the system is at equilibrium. The amount of products will equal the amount of reactants, and K_c will evaluate to 1.

When K_c > 1, more of the molecules will be product molecules than reactant molecules when the system is at equilibrium. Therefore, a reaction where K_c > 1 favors the products.

Conversely, when K_c < 1, more of the molecules will be reactant molecules than product molecules when the system is at equilibrium. Therefore, a reaction where K_c < 1 favors the reactants.

But say we're given the reaction:

H₂ + Cl₂ ⇌ 2HCl

To find the K_c for this reaction $or just about any reaction, really$ :

K_c = products ÷ reactants =

$\(concentration of product1$ × $concentration of product2$ × $concentration of product3$ × …)
÷ $\(concentration of reactant1$ × $concentration of reactant2$ × $concentration of reactant3$ × …)

The reaction given above can be rewritten as:

H₂ + Cl₂ ⇌ HCl + HCl

The concentration of a molecule is written as [molecule]. This is measured in mol/dm³ $moles per cubic decimeter$

So the concentration of HCl would be written as [HCl]

The K_c for the reaction H₂ + Cl₂ ⇌ HCl + HCl is:

K_c = $HCl\] × \[HCl$ ÷ $H~2~\] × \[Cl~2~$

Which simplifies to:

K_c = [HCl]² ÷ $H~2~\] × \[Cl~2~$

The original reaction was:

H₂ + Cl₂ ⇌ 2HCl

And the K_c of the reaction was:

K_c = [HCl]² ÷ $H~2~\] × \[Cl~2~$

Since there were 2HCl molecules, the HCl's concentration was just squared $as it was multiplied twice$ .

Thus, for any reversible chemical reaction that follows the format:

aM + bN + cO + … ⇌ dP + eQ + fR + …

Where the lowercase letters are variables representing coefficients and the uppercase letters are variables representing molecules $O is not Oxygen, and N is not Nitrogen, they are just variables$ , the K_c will be:

K_c = $\[P\]^d^ × \[Q\]^e^ × \[R\]^f^ × …$ ÷ $\[M\]^a^ × \[N\]^b^ × \[O\]^c^ × …$

Throughout this entire subpage, in calculating the K_c of a reaction, nowhere have the starting concentrations been mentioned. That's because the starting concentrations DO NOT affect the K_c of a reaction. Since the K_c is a ratio, the exact concentrations at the start don't factor in, and thus don't affect K_c.

If we don't know the concentration of a molecule, we can find it using an ICE table:

	Reactant1	Reactant2	Product1	Product2
Initial $what were the initial concentrations$ ?
Change $how much did the concentrations<p></p>change by over the reaction$ ?
Equilibrium $what are the concentrations<p></p>of the molecules at equilibrium$ ?

We could add more columsn for each reactant or each product if we needed to /- we're not just limited to two reactants and two products.

Let's take the following question, just to show how to fill out an ICE table:

5 moles of H₂ gas and 5 moles of F₂ gas react with each other in a closed system. When equilibrium is reached, there are 7.5 moles of HF. What is the equilibrium constant of the equation?

First, set everything up:

The equation: H₂ + F₂ ⇌ 2HF

K_c = [HF]² ÷ $H~2~\] × \[F~2~$

The table:

H₂ F₂ 2HF

Initial

Change

Equilibrium

Then fill the given information in the table, and add in the change values:

	H₂	F₂	2HF
Initial	5	5	0
Change	−0.5n	−0.5n	+n
Equilibrium			7.5

We know that the change for both reactants is −0.5n because there is one of each reactant molecule. The total product change is +n, so to counteract this, the total reactant change must be -n. Since there are two different reactant molecules, and there is one of each in the reactant molecules, the total reactant change is split evenly between them. So the change for each reactant is −0.5n.

Given that information, we can calculate that n = 7.5. Using that, we can calculate the equilibrium concentrations for H₂ and F₂ $5 \+ −0.5n, or 5 − 0.5n$ . Plugging in the values gives us:

	H₂	F₂	2HF
Initial	5	5	0
Change	−0.5n	−0.5n	+n
Equilibrium	1.25	1.25	7.5

Now we can calculate the equilibrium constant for this reaction:

K_c = [HF]² ÷ $H~2~\] × \[F~2~$ = 7.5² ÷ $1.25 × 1.25$ = 36

This reaction strongly favors the products.

Partial Pressure

Remember how, when dealing with gases, pressure also affects equilibrium? When dealing with gases, therefore, there is not just an equilibrium constant, but also a pressure constant, K_p

K_p follows much the same concept as K_c. The only difference is that instead of the molar concentrations being divided by one another, it's the pressures that are being divided by one another.

K_p = products ÷ reactants =

$\(pressure of product1$ × $pressure of product2$ × $pressure of product3$ × …)
÷ $\(pressure of reactant1$ × $pressure of reactant2$ × $pressure of reactant3$ × …)

Using the ideal gas law $PV = nRT$ , P $pressure$ = nRT ÷ V

The molar concentration, moles per volume $indicated by \[molecule\]$ , is equal to n ÷ V

So P = [molecule]RT

Since in K_p, the pressures of the products are divided by the pressures of the reactants, for any reversible chemical reaction that follows the format:

aM + bN + cO + … ⇌ dP + eQ + fZ + …

Where the lowercase letters are variables representing coefficients and the uppercase letters are variables representing molecules $O is not Oxygen, and N is not Nitrogen, they are just variables$ , and all the molecules are gas molecules, the K_p will be:

K_p = $\(\[P\]RT$ ^d × $\[Q\]RT$ ^e × $\[Z\]RT$ ^f × …) ÷ $\(\[M\]RT$ ^a × $\[N\]RT$ ^b × $\[O\]RT$ ^c × …)

If one of the reactant/product molecules isn't a gas, then that molecule is just removed when calculating Kp $not from the original reaction though\!$ and the K_p is then calculated as normal.

Simplifying K_p further $limiting the example to only six molecules for readability, although it can work with more$ :

K_p = $\(\[P\]RT$ ^d × $\[Q\]RT$ ^e × $\[Z\]RT$ ^f) ÷ $\(\[M\]RT$ ^a × $\[N\]RT$ ^b × $\[O\]RT$ ^c)

K_p = $\[P\]^d^ × \[Q\]^e^ × \[Z\]^f^$ ÷ $\[M\]^a^ × \[N\]^b^ × \[O\]^c^$
×
$RT^d^ × RT^e^ × RT^f^$ ÷ $RT^a^ × RT^b^ × RT^c^$

K_p = K_c × RT^{$def^ ^÷^ ^abc$}

K_p = K_c × RT^∆n
- ∆n = def ÷ abc

R is the universal gas constant, and T is the temperature of the system at equilibrium.

If we don't know the pressure for one of the molecules, we can use an ICE table to find the molar concentration, and then find the pressure of the gas from there, as P = [molecule]RT