# 13.4 Equilibrium calculations  (Page 4/11)

 Page 4 / 11

## Calculation of changes in concentration

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium , we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1. Determine the direction the reaction proceeds to come to equilibrium.
1. Write a balanced chemical equation for the reaction.
2. If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Q c from the initial concentrations and compare to K c to determine the direction of change.
2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
1. Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
2. Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
3. Solve for the change and the equilibrium concentrations.
1. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x , and check any assumptions used to find x .
2. Calculate the equilibrium concentrations.
4. Check the arithmetic.
1. Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.

## Calculation of concentration changes as a reaction goes to equilibrium

Under certain conditions, the equilibrium constant for the decomposition of PCl 5 ( g ) into PCl 3 ( g ) and Cl 2 ( g ) is 0.0211. What are the equilibrium concentrations of PCl 5 , PCl 3 , and Cl 2 if the initial concentration of PCl 5 was 1.00 M ?

## Solution

Use the stepwise process described earlier.

1. Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl 5 is

${\text{PCl}}_{5}\left(g\right)⇌{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)$

Because we have no products initially, Q c = 0 and the reaction will proceed to the right.

2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

Let us represent the increase in concentration of PCl 3 by the symbol x . The other changes may be written in terms of x by considering the coefficients in the chemical equation.

$\begin{array}{llll}{\text{PCl}}_{5}\left(g\right)\hfill & ⇌\hfill & {\text{PCl}}_{3}\left(g\right)+\hfill & {\text{Cl}}_{2}\left(g\right)\hfill \\ -x\hfill & & x\hfill & x\hfill \end{array}$

The changes in concentration and the expressions for the equilibrium concentrations are:

3. Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}\phantom{\rule{0.2em}{0ex}}=0.0211$
$=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}$

This equation contains only one variable, x , the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

$0.0211=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(1.00-x\right)}$
$0.0211\left(1.00-x\right)={x}^{2}$
${x}^{2}+0.0211x-0.0211=0$

Appendix B shows us an equation of the form ax 2 + bx + c = 0 can be rearranged to solve for x :

$x=\phantom{\rule{0.2em}{0ex}}\frac{-b\phantom{\rule{0.2em}{0ex}}±\phantom{\rule{0.2em}{0ex}}\sqrt{{b}^{2}-4ac}}{2a}$

In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a , b , and c yields:

$x=\phantom{\rule{0.2em}{0ex}}\frac{-0.0211\phantom{\rule{0.2em}{0ex}}±\phantom{\rule{0.2em}{0ex}}\sqrt{{\left(0.0211\right)}^{2}-4\left(1\right)\left(-0.0211\right)}}{2\left(1\right)}$
$=\phantom{\rule{0.2em}{0ex}}\frac{-0.0211\phantom{\rule{0.2em}{0ex}}±\phantom{\rule{0.2em}{0ex}}\sqrt{\left(4.45\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)+\left(8.44\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)}}{2}$
$=\phantom{\rule{0.2em}{0ex}}\frac{-0.0211\phantom{\rule{0.2em}{0ex}}±\phantom{\rule{0.2em}{0ex}}0.291}{2}$

Hence

$x=\phantom{\rule{0.2em}{0ex}}\frac{-0.0211+0.291}{2}\phantom{\rule{0.2em}{0ex}}=0.135$

or

$x=\phantom{\rule{0.2em}{0ex}}\frac{-0.0211-0.291}{2}\phantom{\rule{0.2em}{0ex}}=-0.156$

Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M .

The equilibrium concentrations are

$\left[{\text{PCl}}_{5}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.00-0.135=0.87\phantom{\rule{0.2em}{0ex}}M$
$\left[{\text{PCl}}_{3}\right]=x=0.135\phantom{\rule{0.2em}{0ex}}M$
$\left[{\text{Cl}}_{2}\right]=x=0.135\phantom{\rule{0.2em}{0ex}}M$
4. Check the arithmetic.

Substitution into the expression for K c (to check the calculation) gives

${K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.135\right)\left(0.135\right)}{0.87}\phantom{\rule{0.2em}{0ex}}=0.021$

The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.

What is Stoichiometry?
A radioactive subtance has a half life of 20hrs what fraction of the original radioactive nuclear will remain in 80hrs
1/16th
Ian
general properties of transition metal
pls how do i learn chemistry
am just tried of it😭🙏
Same here
Brianna
organic chem tutor, hell teach you everything.. take notes clearly in a notebook and read it before you go to sleep until you know every topic like the back of your hand
Professor
I find it very hard to understand and remember things I read please I need help. Is just as if chemistry is difficult to me
Precious
may God help us. .me writing jamb this year
pray it will favour me
get good or find something else. don't do it if you don't love it
Professor
what is dative bond?
A coordinate covalent bond, also known as a dative bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.
dharshika
how can i write iupac nunber
Emmanuel
In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a method of organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry. It is published in the Nomenclature of Organic Chemistry
Nike
How can I know the logic for writing the configuration of an element?
Brianna
s and p diagram
Papa
I dont understand that
Brianna
what is elemental composition of earth
what if we try copper and the hydrogen what happened
law of definite proportion
Law of Definite Proportion states that all pure samples of the same chemical compound contains the same elements in proportion by mass
Eunice
what is chemistry
It is that branch of Science which deals with the study of composition, structure and properties of matter .... Ok
ShAmy
The branch of natural science that deals with the constitution of substances and the changes that they undergo as a consequence of alterations in the constitution of their molecules
Young
what is the constitution of substance
ShAmy
picture of periodic table
what is matter
What are the classes of colloid
percentage composition
what is fermentation
what is distillation
Chisom
were is the topic of chrorides