# CHM 1046L Chemical Equilibrium Lab Report

CHM 1046LSpring 2023

Adil Alhinai

PID: 6258288

Chem 1046 U23

Junior Jimenez

Pre-Lab for Module 6 (Chemical Equilibrium Part 2)

Equilibrium Worksheet – Finding equilibrium concentrations and finding Kc

Keep in mind that you can only calculate the equilibrium constant from

equilibrium concentrations and vice-versa.

So if the problem only has initial (or non-equilibrium) concentrations, you first

need to calculate the equilibrium concentrations before finding the equilibrium

constant.

Alternatively you can also use the initial (or non-equilibrium) concentrations

and the equilibrium constant to determine equilibrium concentrations.

We can do this because stoichiometry relates the reactants to the products and

we can use this to determine how much products is formed and how much

reactant is lost.

Calculate equilibrium concentrations and the equilibrium constant from given initial

concentrations and the equilibrium concentration of a single reactant or product.

A flask is filled with 2.00 M CO2, which decomposes according to the following reaction

2CO2 (g) 2CO (g) + O2 (g)

At the very beginning of the reaction how much CO2 will be present in the flask initially?

2.00 M CO2

At the very beginning of the reaction how much CO and O2 will be present in the flask initially?

0 M CO and 0 M O2

Is the reaction currently at equilibrium? Why?

No, the reaction is not currently at equilibrium because it has not had enough time to reach

equilibrium.

Which direction will the reaction shift to reach equilibrium?

The reaction will shift in the direction of producing CO and O2 in order to reach equilibrium.

Will reactants be consumed or created? Would that be a + or a -?

Reactants will be consumed, so that would be a -.

Will products be consumed or created? Would that be a + or a -?

Products will be created, so that would be a +.

In a sentence describe what is happening in the reaction in terms of stoichiometry.

In the reaction, CO2 is being consumed and CO and O2 are being created according to the

stoichiometric ratios given in the equation.

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If a change of one mole was represented by x what would be the change of each of the reactants

and products (do not forget the signs).

Change in CO2 = -2x

Change in CO = +2x

Change in O2 = +x

In the table below fill in the blanks based on what you already know

2CO2 (g) 2CO (g) + O2 (g)

CO2

CO

O2

Initial concentration (I)

2.00M

0M

0M

Change in concentration (C)

-x

+x

+x

Equilibrium concentration (E)

(Add rows I and C)

2.00M -x

+x

+x

Problem

A flask is filled with 2.00 M CO2, which decomposes according to the following reaction

2CO2 (g) 2CO (g) + O2 (g)

Find the equilibrium constant of the reaction of the flask contains 0.90 M CO2 at equilibrium.

SHOW ALL YOUR WORK!

Since the equilibrium concentration of CO2 is 0.90 M, we can calculate x as follows: 2.00 M

CO2 – x = 0.90 M CO2

x = 1.10 M CO2

Therefore, the equilibrium concentrations are:

CO2: 0.90 M

CO: 1.10 M

O2: 1.10 M

The equilibrium constant (K) can be calculated using the equilibrium concentrations.

K = ([CO]^2 x [O2]) / [CO2] ^2

K = (1.10^2 x 1.10) / 0.90^2

K = 1.44

Part 1: Phase 1

The chemical reaction for the formation of the red-orange iron(III) thiocyanate complex ion is

Fe3+ (aq) + SCN- (aq) [FeSCN2+] (aq)

What is the source of the Fe3+ ions?

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The source of the Fe3+ ions is typically a metal salt, such as ferric chloride or ferrous sulfate.

What is the source of the SCN- ions?

The source of the SCN- ions is typically potassium thiocyanate.

The equilibrium constant for this reaction is given by

The stock solution in the stock beaker contains 10.00 mL of 0.1000 M iron(III) nitrate solution,

10.00 mL of 0.1000 M potassium thiocyanate solution, and 10.00 mL of 0.5000 M nitric acid

solution.

What is the total volume of the stock solution? SHOW YOUR WORK!

The total volume of the stock solution is 30.00 mL.

10.00 mL of 0.1000M Fe(NO3)3 + 10.00 mL of 0.1000M KSCN + 10.00 mL of 0.5000M HNO3

= 30.00 mL

How many moles of iron (III) nitrate are present in the stock solution? SHOW YOUR WORK!

The number of moles of Fe(NO3)3 in the stock solution is 0.3000 mol.

(10.00 mL x 0.1000 M) / (1000 mL/1L) = 0.01000 mol

0.01000 mol x 3 mol Fe (NO3)3/1 mol = 0.03000 mol Fe(NO3)3

How many moles of potassium thiocyanate are present in the stock solution? SHOW YOUR

WORK!

The number of moles of KSCN in the stock solution is 0.1000 mol.

(10.00 mL x 0.1000 M) / (1000 mL/1L) = 0.01000 mol

0.01000 mol x 1 mol KSCN/1 mol = 0.01000 mol KSCN

Based on the moles of iron (III) nitrate and moles of potassium thiocyanate, how many moles of

iron (III) thiocyanate do you expect to be formed (Hint: think limiting reagent)? We can do it this

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time because we have enough of one reagent to force the reaction to completion. SHOW YOUR

WORK!

The number of moles of FeSCN2+ expected to be formed is 0.03000 mol.

This is because Fe (NO3)3 is the limiting reagent, so the amount of FeSCN2+ formed will be

limited by the amount of Fe(NO3)3 present. Since there is 0.03000 mol of Fe (NO3)3 present

and the reaction ratio is 1 mol Fe(NO3)3 : 1 mol KSCN : 1 mol FeSCN2+, the amount of

FeSCN2+ formed will also be 0.03000 mol.

Once you determine the moles of iron (III) thiocyanate formed, determine the concentration of

iron (III) thiocyanate in the final solution. SHOW YOUR WORK!

The concentration of FeSCN2+ in the final solution is 0.1000 M.

Working:

(0.03000 mol FeSCN2+/1L) / (1000 mL/1L) = 0.03000M

0.03000M x (1L/1000 mL) = 0.1000M

Part 1: Phase 2

1. What is the concentration of iron (III) thiocyanate in solution 2 that contains 10.00 mL of

stock solution, labeled “Solution 1” (you just calculated the concentration of this solution in

phase 1) and 40.00 mL of nitric acid solution? SHOW YOUR WORK!

Concentration of iron (III) thiocyanate = 0.12 M

Volume of stock solution = 10.00 mL

Volume of nitric acid solution = 40.00 mL

Moles of iron (III) thiocyanate in stock solution = (0.12 M) (10.00 mL) / 1000 mL = 0.012 moles

Moles of nitric acid in nitric acid solution = (0.20 M) (40.00 mL) / 1000 mL = 0.008 moles

Total moles of iron (III) thiocyanate in solution 2 = 0.012 + 0.008 = 0.020 moles

Concentration of iron (III) thiocyanate in solution 2 = (0.020 moles) / (50.00 mL) × 1000 mL =

0.400 M

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2. What is the concentration of iron (III) thiocyanate in solution 3 containing 10.00 mL of

solution 2 (you just calculated the concentration of this solution in phase 2 question 1) and 40.00

mL of nitric acid solution? SHOW YOUR WORK!

Concentration of iron (III) thiocyanate = 0.400 M

Volume of solution 2 = 10.00 mL

Volume of nitric acid solution = 40.00 mL

Moles of iron (III) thiocyanate in solution 2 = (0.400 M) (10.00 mL) / 1000 mL = 0.040 moles

Moles of nitric acid in nitric acid solution = (0.20 M) (40.00 mL) / 1000 mL = 0.008 moles

Total moles of iron (III) thiocyanate in solution 3 = 0.040 + 0.008 = 0.048 moles

Concentration of iron (III) thiocyanate in solution 3 = (0.048 moles) / (50.00 mL) × 1000 mL =

0.960 M

3. What is the concentration of iron (III) thiocyanate in solution 4 containing 10.00 mL of

solution 3 (you just calculated the concentration of this solution in phase 2 question 2) and 40.00

mL of nitric acid solution? SHOW YOUR WORK!

Concentration of iron (III) thiocyanate = 0.960 M

Volume of solution 3 = 10.00 mL

Volume of nitric acid solution = 40.00 mL

Moles of iron (III) thiocyanate in solution 3 = (0.960 M) (10.00 mL) / 1000 mL = 0.096 moles

Moles of nitric acid in nitric acid solution = (0.20 M) (40.00 mL) / 1000 mL = 0.008 moles

Total moles of iron (III) thiocyanate in solution 4 = 0.096 + 0.008 = 0.104 moles

Concentration of iron (III) thiocyanate in solution 4 = (0.104 moles) / (50.00 mL) × 1000 mL =

2.080 M

Which substance did you use as the background blank to calibrate the spectrophotometer?

Explain your reasoning.

I used distilled water as the background blank to calibrate the spectrophotometer. This is

because distilled water has a neutral pH and does not contain any substances that could interfere

with the absorption of light. This allows for more accurate readings of the absorbance of the

solutions being tested.

Part 2

No calculations required.

Part 3

We will only be doing calculations for Solution 1 going forward. You will do the

calculations for Solutions 2 and 3 on your own. That should be relatively easier once

we go through the calculation for Solution 1 here.

Solution 1 calculations

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1) Solution 1 contains 10.00 mL of 0.002 M iron (III) nitrate solution, 4.00 mL of 0.002

M potassium thiocyanate solution, and 6.00 mL of 0.05 M nitric acid solution.

a) Calculate the concentration of iron (III) nitrate in the final solution in Solution 1. SHOW

ALL WORK!

The concentration of Fe(NO3)3 in the final solution of Solution 1 is 0.0014 M. (10.00 mL x

0.002 M) / (10.00 mL + 4.00 mL + 6.00 mL) = 0.008 M 0.008 M x (1L/1000 mL) = 0.0014 M

b) Calculate the concentration of potassium thiocyanate in the final solution in Solution 1. SHOW

ALL WORK!

The concentration of KSCN in the final solution of Solution 1 is 0.0008 M.

(4.00 mL x 0.002 M) / (10.00 mL + 4.00 mL + 6.00 mL) = 0.0016 M

0.0016 M x (1L/1000 mL) = 0.0008 M

Summarize the concentration of iron (III) nitrate and concentration of potassium thiocyanate in

Solution 1 from your calculations in questions 1a and 1b above. You can fill in the same for

Solutions 2-3 later after doing the calculations on your own time.

Solution

#

1#

2#

3#

mL of iron

(III) nitrate

mL of

potassium

thiocyanate

mL of

nitric

acid

10.00ml

10.00ml

6.00ml

4.00ml

6.00ml

10.00ml

6.00ml

4.00ml

4.00ml

M of iron

(III)

nitrate in

the final

solution

0.0014M

0.0014M

0.0008M

M of

potassium

thiocyanate

in the final

solution

0.0008M

0.0012M

0.0014M

Absorbance

of the

solution

0.062

0.092

0.098

You cannot do the limiting reagent thing here because you do not know that one reagent is in so

much excess like before that the reaction will go to completion. So, you must calculate the

concentration of the [FeSCN2+] from the graph you created before.

Is the absorbance value of Solution 1 within the range of values on the graph you created using

the data from Phase 2? So, can you use this absorbance value to determine the concentration of

[FeSCN2+] from the graph? (Think about the definition of the slope and what the x and y axis

are in this case and how it relates to the Beer-Lambert law).

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Consentrations vs Absorbances

0.12

y = 0.018x + 0.048

R² = 0.871

Absorbance

0.1

0.08

0.06

0.04

0.02

0

0.0008M

0.0012M

0.0014M

Concentration

No, they are not comparable test tube 1 is not within range so you cannot use it to

determine the concentration of [FeSCN2+].

Doing the calculation to find the equilibrium constant

Consider Solution 1 of Part 3,

What is the concentration of iron (III) nitrate aka iron (Fe3+) ions in this tube?

The concentration of Fe3+ ions in Solution 1 is 0.0014 M.

What is the concentration of potassium thiocyanate aka thiocyanate ions (SCN-) ions in this

tube?

The concentration of SCN- ions in Solution 1 is 0.0008 M.

Refer back to the graph produced for the Pre-lab Assignment to solve for [FeSCN2+]

According to Beer and Lambert’s law

A = l C

A = absorbance

= molar absorptivity

l = optical path length in cm (usually 1 cm is the length of the cuvettes used)

C = concentration

The equation of a line is y = mx + b

Where?

y = value for the Y axis

x = value of the X axis

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m = slope

b = y intercept

When the graph passes through the origin, the y intercept (b) is zero.

So your equation becomes y = mx

What is the y axis in this case?

Absorbance.

What is the x axis in this case?

Concentrations.

y = mx

A = l C

You can get the value of ( x l) from the slope of the graph and so can use this to find

concentration given the absorption and vice versa.

What is the value of the slope from the graph?

The slope of the graph is 0.0070.

What is the concentration of iron (III) thiocyanate in this tube calculated from the graph? SHOW

ALL WORK!

Therefore, the value of ( x l) is 0.0070.

Plugging this in the equation

A = lC A = 0.0070 C

Rearranging the equation

C = A/0.0070

Using the absorbance value of Solution 1, we can calculate the concentration of FeSCN2+ in

Solution 1.

The absorbance value of Solution 1 is 0.062.

Therefore, the concentration of FeSCN2+ in Solution 1 is 0.089 M.

Making an ICE table

A flask is filled with Fe3+ and SCN-, which decomposes according to the following reaction

Fe3+ (aq) + SCN (aq) [FeSCN2+] (aq)

At the very beginning of the reaction how much Fe3+ will be present in the flask initially? You

did this calculation and put down the value previously.

0.0014 mol.

At the very beginning of the reaction how much SCN- will be present in the flask initially? You

did this calculation and put down the value previously.

0.0008 mol.

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At the very beginning of the reaction how much [FeSCN2+] will be present in the flask initially?

Think about how much product will be present at the very beginning of the reaction.

0 mol

Was the reaction at equilibrium when the reaction started? Why?

No, the reaction was not at equilibrium when the reaction started because the amount of product

(FeSCN2+) present was 0 mol. This means that the reaction was just starting and had not yet

reached equilibrium.

Which direction will the reaction shift to reach equilibrium?

The reaction will shift in the direction of product to reach equilibrium. This means that more

FeSCN2+ will be formed as the reaction proceeds.

Will reactants be consumed or created? Would that be a + or a -?

Reactants will be consumed. This will be represented by a negative (-) sign.

Will products be consumed or created? Would that be a + or a -?

Products will be created. This will be represented by a positive (+) sign.

In a sentence describe what is happening in the reaction in terms of stoichiometry.

The reaction is proceeding with reactants being consumed and products being created in a 1:1:1

ratio.

If a change of one mole was represented by x what would be the change of each of the reactants

and products (do not forget the signs).

Change in Fe3+ = -x

Change in SCN- = -x

Change in [FeSCN2+] = +x

In the table below fill in the blanks based on what you already know

Fe3+ (aq) + SCN (aq) [FeSCN2+] (aq)

Initial concentration (I)

Change in concentration (C)

(in terms of x)

Equilibrium concentration (E)

(add rows I and C)

Fe3

0.0014

SCN 0.0008

[FeSCN2+]

0

-x

-x

+x

x

x

x

We know the value of x because this is the concentration of [FeSCN2+] which you calculated

from the graph. So, you can use this to calculate the value of x and therefore the equilibrium

concentrations of Fe3+ and SCN-. SHOW ALL YOUR WORK!

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[𝐹𝑒3+ ] = 0.0014 + (−0.089) = −0.0876

[𝑆𝐶𝑁 − ] = 0.0008 + (−0.089) = −0.0882

[𝐹𝑒𝑆𝐶𝑁 2+ ] = 0.089M.

Now that you have these equilibrium concentrations (values) plug them into the equilibrium

constant expression to calculate the value of Kc. SHOW ALL YOUR WORK!

Kc = 0.089/0.089 x 0.089 Kc = 1.00

This is the value of Kc that you found from Solution 1

Do you expect the value of Kc that you will find from Solutions 2-3 to be around this number or

be different from the number you just determined? Why or why not? Justify and explain your

reasoning.

I expect the value of Kc that I will find from Solutions 2 and 3 to be around this number,

because the concentration of the reactants and products are going to be similar in each solution

and so the equilibrium constant should also be similar.

Before you leave the lab, you need to do the same calculation that you did for solution 1

(beginning of Part 3 on page 5. Use this worksheet as a template to do those calculations on

solutions 2 and 3 and calculate an average value of Kc from all the 3 values.

Make sure you show the calculation showing how you determined the average. SHOW ALL

YOUR WORK!

ANS:

Solution 2: 10.00 mL of 0.002 M iron (III) nitrate, 6.00 mL of 0.002 M potassium thiocyanate,

and 4.00 mL of 0.05 M nitric acid.

M of iron (III) nitrate in the final solution: 0.0014 M

M of potassium thiocyanate in the final solution: 0.0012 M

absorbance of the solution: 0.092

Solution 3: 6.00 mL of 0.002 M iron (III) nitrate, 10.00 mL of 0.002 M potassium thiocyanate,

and 4.00 mL of 0.05 M nitric acid.

M of iron (III) nitrate in the final solution: 0.0008 M

M of potassium thiocyanate in the final solution: 0.0014 M

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absorbance of the solution: 0.098

Using the absorbance value of Solution 2, we can calculate the concentration of FeSCN2+ in

Solution 2. The absorbance value of Solution 2 is 0.092. Therefore, the concentration of

FeSCN2+ in Solution 2 is 0.132 M. Using the absorbance value of Solution 3, we can calculate

the concentration of FeSCN2+ in Solution 3.

The absorbance value of Solution 3 is 0.

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