Calculating Reactant and Product Quantities from Keq — Examples

When we started learning equilibrium calculations, we studied ICE tables. ICE stands for initial, change, and equilibrium. We used ICE to find the equilibrium concentrations. Unfortunately, your teacher might try to trick you! For example, you may have to work backward to find initial concentrations. So, today, we'll examine a couple of unique ICE examples.

Before we start, I want to remind you of what we've learned so far. First, we investigated reversible reactions. A reversible reaction reaches equilibrium when the forward and reverse rates are equal. So, we used the rate constants to generate a new relationship. That was the equilibrium constant! Then, we started calculating equilibrium constants. We learned we can use either concentration or pressure. When we have solutions, we use molarity. If we have gases, we need pressure units. So, we use mole fractions to compare partial pressures. Also, remember that pure solids and liquids we exclude from $K_{eq}$ calculations.

ICE Example Problems

1. Let's start with the synthesis reaction between hydrogen gas and, the halogen, iodine. Remember, both are diatomic elements. Let's say we start with equal pressures of both. But, we don't know the starting pressure. The final pressure of hydrogen iodide gas is 0.32 atmospheres. If the equilibrium constant is 5.0, calculate the initial concentrations.

• Strategy: We'll need to use an ICE table. But, our initial concentration for hydrogen and iodine will be "y." The initial value for hydrogen iodide will be zero.

First, we need to write our balanced chemical equation and our equilibrium constant expression.

• $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

• $K_{eq}=\frac{P_{HI}^{2}}{P_{H_{2}} \times P_{I_{2}}}=5$

Now we can create our ICE table.

$$\begin{array}{|r|c|c|c|} \hline \text{}&H_{2}&I_{2}&HI\\ \hline \text{Initial:}&y&y&0.0\\ \hline \text{Change:}&-x&-x&+2x\\ \hline \text{Equilibrium:}&y-x&y-x&0.0+2x=0.32atm\\ \hline \end{array}$$

First, we solve for $x$.

• $2x=0.32atm\rightarrow x=0.16atm$

Now we can use our equilibrium constant expression to solve for y.

• $K_{eq}=\frac{P_{HI}^{2}}{P_{H_{2}} \times P_{I_{2}}}\rightarrow 5=\frac{0.32^{2}}{(y-0.16)^2}\rightarrow (y-0.16)^{2}=\frac{0.32^{2}}{5} \rightarrow y-0.16= \sqrt{0.0205}\rightarrow y=0.30 atm.$

1. For our second problem, let's look at the decomposition reaction of chlorine gas. Chlorine's bond can spontaneously break. It's a critical reaction in organic chemistry. We use it to add chlorine to double and triple bonds. It's catalyzed by UV light.

Let's say our reaction is at equilibrium in a 1.0-liter container. We have 0.75 moles diatomic chlorine and 0.50 moles of chlorine atoms. Now, we add 0.25 moles of diatomic chlorine. Let's say we shine a lot of UV light. So, our new $K_{eq}=0.5$. What are the new equilibrium concentrations?

• Strategy: Again, we need an ICE table. We are in a 1.0-liter container. So, calculating concentrations is easy. In this case, our initial concentration of diatomic chlorine is $\frac{0.75mols+0.25mols}{1.0L}=1.00\frac{mols}{L}$. Our initial concentration of chlorine atoms is $0.50\frac{mols}{L}$. From Le' Chatelier's principle, we know the equilibrium will move towards the products.

First, we need to write the balanced equation and $K_{eq}$ expression.

• $Cl_{2(g)}\rightleftharpoons 2Cl_{(g)}$

• $K_{eq}=\frac{[Cl]^{2}}{[Cl_{2}]}$

So our ICE table will look like this.

$$\begin{array}{|r|c|c|} \hline \text{}&Cl_{2}&Cl\\ \hline \text{Initial:}&1.00&0.50\\ \hline \text{Change:}&-x&+2x\\ \hline \text{Equilibrium:}&1.00-x&0.50+2x\\ \hline \end{array}$$

Let's solve for x.

• $K_{eq}=0.5=\frac{(0.50+2x)^{2}}{(1.00-x)}\rightarrow .5-.5x=4x^{2}+2x+.25 \rightarrow 4x^{2}+2.5x-.25=0$

At this point, we have to use the quadratic formula.

• $x=\frac{-2.5\pm \sqrt{-2.5^{2}-4(4)(-..25)}}{2\times 4}\rightarrow \frac{-2.5\pm3.2}{8}\rightarrow x=0.087\frac{mol}{L}$.

Now, we can calculate the equilibrium concentrations.

• $[Cl_{2(g)}]=1.00-0.087=0.91\frac{mols}{L}$

• $[Cl_{g}]=0.50+ 2(.087)=0.67 \frac{mols}{L}$

• This is a very high concentration for this reaction. It's difficult to have a bond spontaneously break. So, we can assume we are at very high temperatures and using a UV light catalyst.

We've been studying equilibrium. You know how to calculate equilibrium constants with either concentration or pressure. For solution concentration we use units of molarity. For the gas pressures, we use partial pressures. So, we're working with mole fractions. Finally, we learned that pure solids and liquids get excluded from the equilibrium constant. Now, we can talk about how to find equilibrium constants experimentally. For general chemistry, we have two methods. Let's dive in!

Beer's Law Experimental Data

Refresher on Beer-Lambert Law

Our first method for finding the equilibrium constant uses the Beer-lambert law. Do you remember it? We studied it way back when we looked at light absorption. It states that absorbance is proportional to concentration.

• The formula is $A=\epsilon\times L \times c$.

• $A$ is absorbance. It's the amount of light that a sample absorbs. The opposite of absorbance is transmittance. That's the amount of light that went through the solution.
• $\epsilon$ is the molar absorptivity. It's a constant that depends on the solute.
• $L$ is the path length. It's the distance the light had to travel through our sample. Typically, the cuvettes we use have 1.0 cm path lengths.
• $c$ is the solution concentration in molarity.

If a sample obeys the Beer-Lambert law, it will produce a linear graph. The y-axis is absorbance. The x-axis is concentration. So, the slope of the line will be the molar absorptivity multiplied by the path length.

How do we perform a Beer-Lambert law experiment? Generally, you'll use it in an analytical chemistry unit or class. Analytical chemists measure and calculate the type and quantity of matter. They do that by using instruments. In the case of Beer's law, we would use a spectrophotometer. A spectrophotometer is a simple instrument. One side is a light source. On the other side is a sensor. We place the solution between the light and the detector.

The light source can shine many different wavelengths of light. We choose the wavelength that is our sample's maximum absorbance. According to Beer's law, a higher sample concentration equals more absorbance. So, less light gets to the sensor. It measures the number of photons that escape the sample.

All instrument-based techniques need calibration. So, we create a calibration curve. But, how? We have to use known concentrations of our solution. So, we know our x-axis values. The instrument tells us the absorbance. So, we know our y-axis values. Remember, the slope of the graph is molar absorptivity multiplied by path length. Once we have a calibration curve, we measure an unknown sample. We would use its absorbance to calculate concentration.

Beer-Lambert and $K_{eq}$

So, how does Beer's law relate to the equilibrium constant? If we know concentrations, we can calculate $K_{eq}$. Let's study the reaction between iron and thiocyanate ions.

$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$ $K_{eq}=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$

• Iron (III) thiocyanate absorbs light at 447 nanometers. Iron (III) and the thiocyanate do not.

We can use Beer's law to solve for the concentration of $FeSCN^{2+}$.

• $A=\epsilon\times L\times [FeSCN^{2+}]\rightarrow [FeSCN^{2+}]=\frac{A}{\epsilon\times L}$

The challenge is finding the equilibrium concentration of $Fe^{3+}$ and $SCN^{-}$.

• $[Fe^{3+}]_{eq}=[Fe^{3+}]_{initial}-[FeSCN^{2+}]$

• $[SCN^{-}]_{eq}=[SCN^{-}]_{initial}-[FeSCN^{2+}]$

Now we can plug everything back into the Equilibrium constant expression.

• $K_{eq}=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]} \rightarrow K_{eq}=\frac{\frac{A}{\epsilon \times L}}{([Fe^{3+}]_{initial}-[FeSCN^{2+}])\times([SCN^{-}]_{initial}-[FeSCN^{2+}] )}$

The previous equation has two unknowns $K_{eq}$ and $\frac{A}{\epsilon\times L}$. So, we can write it in the form $y=mx+b$. The math is complicated, so I'll only give the solution. Two chemists developed the formula in 1947!

• $\frac{A}{[Fe^{3+}][SCN^{-}]}=-K\times\frac{A([Fe^{3+}]+[SCN^{-}])}{[Fe^{3+}][SCN^{-}]}+L\times\epsilon$.

All the concentrations are our initial concentrations. So, we know them. We measure absorbance. Epsilon multiplied by path length we already knew from our calibration plot. So, if you had the data, you could use excel to generate a graph. The slope of the line is $K_{eq}$!

Potentiometry

We'll learn more about oxidation-reduction (redox) reactions later. But, you can also use them to calculate the equilibrium constant. You can split a redox reaction into two connected cells. Together, they create a battery! The equation that controls batteries is the Nernst equation.

• The formula is $E=E^{o}-\frac{RT}{nF} \times \ln{(Q)} \rightarrow E=E^{o}-\frac{0.0592}{n}ln{(Q)}$

• $E$ is the electric potential (voltage).
• $E^{o}$ is the standard electrode potential.
• $R$ is the gas constant.
• $T$ is the temperature.
• $n$ is the number of moles.
• $F$ is Faraday's constant.
• $Q$ is the reaction quotient.

• $Q=K_{eq}$ at equilibrium.

We can also use potentiometry for acid-base reactions. We rearrange the equation to solve for the pH. The Nernst equation becomes $pH=\frac{nF}{RT}(E^{o}-E)$. Experimentally, this equation becomes similar to a titration.

Using the Nernst Equation Experimentally

For the Nernst equation, start by making a calibration curve. Measure the voltage of a series of known concentrations. The voltage at equilibrium is zero. So, the x-intercept of our calibration curve will be the equilibrium concentration.

(Video) Equilibrium Calculations (ICE Tables) | General Chemistry

By Chad's Prep

In this video, Chad explains equilibrium calculations. He starts with an easy example. You're given the concentrations and have to calculate the constant. But, the following two examples are more challenging. For those, you have to use an ICE table. Chad walks you step-by-step through two ICE table problems. This is a helpful video for learning equilibrium calculations.