# 15.4: Using Equilibrium Constants in Calculations

## Overview

In the last section we used the initial concentrations and the equilibrium concentration of one species to calculate the equilibrium constant. In this section we use the equilibrium constant to calculate the equilibrium concentration of all species. If K is a very large number, we have a product favored reaction and we do not need to do an equilibrium calculation, we simply base the theoretical yield on the complete consumption of the limiting reagent. But if K is a small number, or if you mixed "products" instead of "reactants", the reaction may not go to completion in the general Chemistry 1 sense, and you need to do an equilibrium calulation to calculate the final reactant and product.

• If the reaction proceeds as written (forming products and consuming reactants) the extent of reaction (x) is positive.
• If the reverse reaction procees (forming reactants and consuming products) the extent of reaction (x) is negative.

The overall strategy for these problems is to set up an RICE diagram and solve for x, the extent of reaction. Once you know x, you can calculate the equilibrium (final) concentration of each chemical species. This process is similiar to that in the last section (15-3), but you are now using the equilibrium constant expression to solve for the extent of reaction and not K, and then you use that value to calculate the equilbrium concentration of each species.

## Generic RICE Diagram

For the generic reaction

$aA +bB \rightleftharpoons cC + dD$

the RICE diagram is:

Reactants aA + bB cC dD
Initial [A]Initial [B]Initial [C]Initial [D]Initial
Change -ax -bx +cx +dx
Equilibrium [A]Eq [B]Eq [C]Eq [D]Eq

noting that:

$\left [ A\right ]_{Eq}=\left [ A\right ]_{Initial}-ax \\ \left [B\right ]_{Eq}=\left [ B\right ]_{Initial}-bx \\ \left [ C\right ]_{Eq}=\left [C\right ]_{Initial}+cx \\ \left [D\right ]_{Eq}=\left [D\right ]_{Initial}+dx \label{15.4.3}$

We know the equilbrium constant is:

$K=\frac{\left [C\right ]_{eq}^{c}\left [D\right ]_{eq}^{d}}{\left [A\right ]_{eq}^{a}\left [B\right ]_{eq}^{b}}$

substituting the above relationship into eq. \ref{15.4.3} gives us a single equation with just one unknown, x, which means we can algebraically solve for x.

$K=\frac{\left ( \left [C\right ]_{Initial}+cx \right )^{c}\left ( \left [D\right ]_{Initial}+dx \right )^d }{\left ( \left [A\right ]_{Initial}-ax \right )^{a}\left ( \left [B\right ]_{Initial}-bx \right )^b} \label{15-4.4}$

## Solving Equilibrium Problems

Lets take a closer look at eq \ref{15-4.4} and solve for x, first we realize this is a polynomial equation where the degree (maximum power of x) is determined by the sum of the stoichiometric coefficients of either the numerator (products) or denominator (reactants), depending on which is larger.

$a_nx^n + ... + a_2x^2 + a_1x + a_0$

This will make sense as you work through the problems, but the bottom line is these can become very complicated to solve and in this class we are going to look at several simple cases that are easy to solve.

Quadratic equations are polynomial equations of degree 2.

$ax^2 + bx + c = 0$

These occur for the following types of reactions (note, the highest sum of stoichiometric coefficients on either the reactant or product side is 2, and so the polynomial is quadratic)

\begin{align}A +B \nonumber & \rightleftharpoons C + D \\ A +B \nonumber & \rightleftharpoons C \\ 2A \nonumber & \rightleftharpoons C + D \\ A \nonumber & \rightleftharpoons C + D \end{align}

There will be two approaches we can use, the quadratic formula, and complete the square.

From algebra there are two roots to the quadratic equation, $$ax^2 + bx + c = 0$$ that can be determined from the quadratic formula.
$x= \dfrac{−b \pm \sqrt{b^2−4ac}}{2a}$

These two roots occur because there are two solutions to a square root. One of the answers will make no sense, typically giving a negative concentration for one species, meaning you consumed more of it than you have, which is impossible.

Note, if your reaction is product loaded, that is, you start with more products than you would have at equilibrium, the correct solution would be the one that has a negative x, if you set up the RICE diagram so that reactants are consumed and products are produced, that is, you are consuming the things on the right side of the equation, not producing them.

Consider the following problem: Calculate the equilibrium concentration of all species if 1 mol of COCl2 is placed in a 10 L container and undergoes the following reaction.

Problem $$\PageIndex{1}$$: $CO(g) + Cl_2(g) \rightleftharpoons COCl_2$In the following video this problem is solved, and it was noted that the COCl2 must be a reactant and so the equation was written backwards and the reciprocal of K was used.

Video $$\PageIndex{1}$$: Solution for Problem$$\PageIndex{1}$$, with the equation rewritten so the COCl2 is a reactant. Note how the extent of reaction (x) is a positive number.

After watching the above video, you should try and solve it as written in equation 15-4.9, that is, while treating COCl2 as a product, even though it is obviously a reactant. In this case, the solution that makes sense will have a negative value for x, the extent of reaction. This has been done with the following

Video $$\PageIndex{2}$$: Solution for Problem$$\PageIndex{1}$$, with the equation written with the COCl2 as product, even though it is obviously a reactant. Note how the extent of reaction (x) is a negative number

In the above two videos we knew the COCl2 was a product because there was no CO or Cl2 and therefor they could not be reactants, and thus the system had to be product loaded (as written n eq. 15-4.9). As K = 26.7 we know that at equilibrium there can be substantial amounts of both products and reactants coexisting at equilibrium, and if we had mixed similar quantities of all three species we would not have known ahead of time if the reaction would go to the right (reactant loaded - forming products), or to the left (product loaded - forming reactants). So in solving these problems you need to be prepared to check both roots of the square root term in the quadratic formula. Since your calculator gives the positive root, it is probably best to use that one first, and then if the answer makes no sense, go back and recalculate, using the other root.

#### Complete the Square

If two reactants create two products, and the species mixed are in stoichiometric proportions, you can simplify the problem where the equilibrium constant equals a single expression to the power of two.

$K=(expression\; of\; x)^2$

This is a much easier mathematical problem to solve than the quadratic equation, as you simply take the square of both sides and then algebraically solve for x. The technique works if you start with just reactants, or just products, or a mixture of the two, as long as they are in stoichiometric proportions.

Consider the following reaction

$H_2(g) + Br_2(g) \leftrightharpoons 2HBr(g)$

There are three types of problems we can use complete the square and avoid using the quadratic formula

1. Mix equal amounts of bromine and hydrogen
2. Mix any amount of hydrogen bromide gas
3. Mix equal amounts of bromine and hydrogen, along with any amount of hydrogen bromide gas.

(Note: from general chemistry1, HBr(aq) is hydrobromic acid, but HBr(g) is hydrogen bromide (gas).

Problem $$\PageIndex{1}$$: Calculate the equilibrium concentration of all species if the initial concentrations of hydrogen and bromide react 0.6 M and that of HBr is 1.25M and K = 50 for the reaction as written in equation 15-4.10.

$K=\frac{\left [HBr\right ]_{eq}^{2}}{\left [H_2\right ]_{eq}\left [Br_2\right ]_{eq}}= \frac{\left ( \left [HBr\right ]_{Initial}+2x \right )^{2}}{\left ( \left [H_2\right ]_{Initial}-x \right) \left ( \left [Br_2\right ]_{Initial}-x \right)}$

since $$H_2=Br_2$$, everything in the right becomes a single expression, as can be seen by substituting in actual values.

$K= \frac{\left ( \left [1.25\right ]_{Initial}+2x \right )^{2}}{\left ( \left [0.6\right ]_{Initial}-x \right )^2}=\left ( \frac{\left ( \left [1.25\right ]_{Initial}+2x \right )}{\left ( \left [0.6\right ]_{Initial}-x \right )}\right )^2$

Now you simply take the square root of both sides:\

$\sqrt{K}= \frac{\left [1.25\right ]_{Initial}+2x }{ \left [0.6\right ]_{Initial}-x }$

Video 15-4.3 shows how to solve this problem

Video $$\PageIndex{3}$$: Solution to a complete the square problem$$\PageIndex{1}$$:

#### Complete the Cube (or higher degree polynomials)

If n reactants create n products, and the species mixed are in stoichiometric proportions, you can simplify the problem where the equilibrium constant equals a single expression to the power of n.
$K=(expression\; of\; x)^n$

Lets consider a problem where three reactants form three products and they were mixed in stoichiometric proportions. This results in a cubic equation, and we will solve by completing the cube.s.

Consider the following reaction

$Cl_2(g) + 2HBr(g) \leftrightharpoons Br_2(g) + 2HCl(g)$

Problem $$\PageIndex{2}$$: Calculate the equilibrium concentration of all species if the initial concentrations of chlorine if 2.00 moles of chlorine and 4.00 moles of hydrogen bromide are added to a 10.0 liter container and K = 15.

Since we start with no products we know this is a reactant loaded system and will proceed to the right, meaning the extent of reaction will be a positive number. We also know that since K = 15, which is greater than 0.001 and less than 1,000, that there will be substantial amounts of reactants and products coexisitng at equlibrium, and so we have an equilibrium problem. That is for example, if K = 1015 we could treat this as a general chemistry 1 problem and say the amount of product produced is based on the complete consumption of the limiting reagent, and if K = 1015 we could say that no reaction occurs.

$K=\frac{\left [ Br_2 \right ]\left [HCl\right ]_{eq}^{2}}{\left [Cl_2\right ]_{eq}\left [HBr\right ]^2_{eq}}= \frac{\left ( \left [ Br_2 \right ]+x \right )\left ( \left [HCl\right ]_{Initial}+2x \right )^{2}}{\left ( \left [Cl_2\right ]_{Initial}-x \right) \left ( \left [HBr\right ]_{Initial}-2x \right)^2}$

substituting the initial concentrations:

$K= \frac{\left ( \left [ 0 \right ]+x \right )\left ( \left [0\right ]+2x \right )^{2}}{\left ( \left [0.200\right ]-x \right) \left ( \left [0.400\right ]-2x \right)^2}= \frac{\left ( x \right )\left (2x \right )^{2}}{\left ( \left [0.200\right ]-x \right) \left ( \left [0.400\right ]-2x \right)^2}$

Bringing the right hand side to a common power

$K= \frac{x^3}{\left ( \left [0.200\right ]-x \right)^3}=\left ( \frac{x}{0.200-x} \right )^3$

Now you simply take the cube root of both sides, and then solve for x:

$\sqrt[3]{K}=\frac{x}{0.200-x}$

Please watch the following YouTube if you need help solving this problem

Video $$\PageIndex{4}$$:

#### Method of Successive Approximations

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We typically use a RICE diagram to describe the chance in concentration as a reaction proceeds from initial to final (equilibrium) concentrations.  For the above generic reaction, the RICE diagram is:

Table $$\PageIndex{2}$$: Generic RICE Diagram
Reactants aA + bB cC dD
Initial [A]Initial [B]Initial   [C]Initial [D]Initial
Change -ax -bx   +cx +dx
Equilibrium [A]Eq [B]Eq   [C]Eq [D]Eq

noting that:

$\left [ A\right ]_{Eq}=\left [ A\right ]_{Initial}-ax \\ \left [B\right ]_{Eq}=\left [ B\right ]_{Initial}-bx \\ \left [ C\right ]_{Eq}=\left [C\right ]_{Initial}+cx \\ \left [D\right ]_{Eq}=\left [D\right ]_{Initial}+dx \label{15.3.3}$

where,

x=extent of reaction

and

$K=\frac{\left [C\right ]_{eq}^{c}\left [D\right ]_{eq}^{d}}{\left [A\right ]_{eq}^{a}\left [B\right ]_{eq}^{b}}$

There are two basic types of equilibria calculations we need to perform. In both cases you know what you mixed (initial concentrations), but in the first you calculate K from the concentration of one species after the reaction is over (equilibrium concentration), while in the other you know K and calculate the equilibrium concentration of all species once the reaction is over.

1. Determine K and equilibrium concentration of all species but one, given initial concentrations and the equilibrium concentration of that one species
2. Determine equilibrium concentration of all species given K and initial concentrations of all species.

We will use the equilibria of a weak acid as our example because we can measure the hydronium ion concentration with a pH meter.

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## WorkSheets

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: