E04
Exp 04 - Equilibrium
‣ Reversible Reactions
‣ Steady State
‣ Part A: Reaction Time Data
‣ Dynamic Equilibrium
‣ Part B: Calibration Data
‣ Representing Equilibrium
‣ Chemical Equations
‣ Equilibrium Constant K
‣ Equilibrium Expression
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‣ Experiment
‣ Beer’s Law
‣ Analysis
‣ Next Meeting
Arrow Conventions
‣ Chemists commonly use two kinds of arrows in reactions to indicate the degree of
completion of the reactions.
‣ A single arrow indicates all the reactant molecules are converted to product
molecules at the end.
H2(g) + I2(g) ➞
‣ A double arrow indicates the reaction stops when only some of the reactant
molecules have been converted into products.
H2(g) + I2(g) ⇄ 2 HI(g)
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Reaction Dynamics
‣ When a reaction starts, the reactants are consumed and products are made.
‣ The reactant concentrations decrease and the product concentrations increase.
‣ As reactant concentration decreases, the forward reaction rate decreases.
‣ Eventually, the products can react to re-form some of the reactants, assuming the
products are not allowed to escape.
‣ As product concentration increases, the reverse reaction rate increases.
‣ Processes that proceed in both the forward and reverse direction are said to be
reversible.
H2(g) + I2(g) ⇄ 2 HI(g)
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Dynamic Equilibrium
‣ As the forward reaction slows and the reverse reaction accelerates, eventually
they reach the same rate.
‣ Dynamic equilibrium is the condition wherein the rates of the forward and reverse
reactions are equal.
‣ Once the reaction reaches equilibrium, the concentrations of all the chemicals
remain constant because the chemicals are being consumed and made at the same rate.
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[H2] = 8,
[I2] = 8,
[HI] = 0
At time 0, there are only reactants in the mixture, so only the
forward reaction can take place.
[H2] = 6,
[I2] = 6,
[HI] = 4
At time 16, there are both reactants and products in the mixture, so both the
forward reaction and reverse reaction can take place.
H2(g) + I2(g) ⇄ 2 HI(g)
At time 32, there are now more products than reactants in the
mixture, the forward reaction has slowed down as the
reactants run out, and the reverse reaction accelerated.
[H2] = 4,
[I2] = 4,
[HI] = 8
At time 48, the amounts of products and reactants in the mixture
haven’t changed; the forward and reverse reactions are
proceeding at the same rate. It has reached equilibrium.
[H2] = 4,
[I2] = 4,
[HI] = 8
H2(g) + I2(g) ⇄ 2 HI(g)
As the concentration of product increases and the concentrations
of reactants decrease, the rate of the forward reaction slows down,
and the rate of the reverse reaction speeds up.
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H2(g) + I2(g) ⇄ 2 HI(g)
At dynamic equilibrium, the rate of the forward reaction is equal to
the rate of the reverse reaction.
The concentrations of reactants and products no longer change.
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Equilibrium ≠ Equal
‣ The rates of the forward and reverse reactions are equal at equilibrium.
‣ But that does not mean the concentrations of reactants and products are equal.
‣ Some reactions reach equilibrium only after almost all the reactant molecules are
consumed; we say the position of equilibrium favors the products.
‣ Other reactions reach equilibrium when only a small percentage of the reactant
molecules are consumed; we say the position of equilibrium favors the reactants.
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When Country A citizens feel overcrowded, some
will emigrate to Country B .
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However, after a time, emigration will occur in both directions at the same rate, leading
to populations in Country A and Country B that are constant, but not necessarily equal.
Equilibrium Constant
‣ Even though the concentrations of reactants and products are not equal at
equilibrium, there is a relationship between them.
‣ The relationship between the chemical equation and the concentrations of
reactants and products is called the law of mass action.
‣ For the general equation, the law of mass action gives the relationship below.
aA + bB ⇄ cC + dD
‣ The lowercase letters represent the coefficients of the balanced chemical
equation.
‣ Always put products over reactants.
‣ K is called the equilibrium constant.
‣ K is unitless
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Writing Equilibrium Constant Expressions
So, for the reaction
2 N2O5(g) ⇄ 4 NO2(g) + O2(g)
the equilibrium constant expression is as follows:
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What Does the Value of Keq Imply?
‣ When the value of Keq >> 1, when
the reaction reaches equilibrium
there will be many more product
molecules present than reactant
molecules.
‣ The position of equilibrium favors
products.
‣ When the value of Keq << 1, when
the reaction reaches equilibrium
there will be many more reactant
molecules present than product
molecules.
‣ The position of equilibrium favors
reactants.
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Relationships between K and Chemical Equations
‣ When the reaction is written backward, the equilibrium constant is inverted.
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For the reaction aA + bB ⇄ cC + dD
For the reaction cC + dD ⇄ aA + bB
the equilibrium constant
expression is as follows:
the equilibrium constant
expression is as follows:
Relationships between K and Chemical Equations
‣ When the coefficients of an equation are multiplied by a factor, the equilibrium
constant is raised to that factor.
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For the reaction aA + bB ⇄ cC
For the reaction 2aA + 2bB ⇄ 2cC
the equilibrium constant
expression is as follows:
the equilibrium constant
expression is as follows:
Relationships between K and Chemical Equations
‣ When you add equations to get a new equation, the equilibrium constant of the
new equation is the product of the equilibrium constants of the old equations.
For the reactions (1) aA ⇄ bB and (2) bB ⇄ cC the equilibrium
constant expressions are as
follows:
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For the reaction aA ⇔ cC
the equilibrium constant
expression is as follows:
Heterogeneous Equilibria
‣ The concentrations of pure solids and pure liquids do not change during the course
of a reaction.
‣ Because their concentration doesn’t change, solids and liquids are not included in
the equilibrium constant expression.
‣ For the reaction the equilibrium constant expression is as follows:
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Heterogeneous Equilibria
The amount of C is different, but the amounts of CO
and CO2 remain the same. Therefore, the amount
of C has no effect on the position of equilibrium.
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E04
Exp 04 - Equilibrium
‣ Reversible Reactions
‣ Steady State
‣ Part A: Reaction Time Data
‣ Dynamic Equilibrium
‣ Part B: Calibration Data
‣ Representing Equilibrium
‣ Chemical Equations
‣ Equilibrium Constant K
‣ Equilibrium Expression
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‣ Experiment
‣ Beer’s Law
‣ Analysis
‣ Next Meeting
Part A
‣ Part A : Preparation of Beer’s Law Plot (calibration of absorbance to concentration)
‣ 1. Turn on the Spectronic 20 spectrophotometer, set the λ = ? nm, and allow it to
warm-up for 15 minutes.
‣ 2. Calibrate the Spectronic 20 using the procedure outlined in Appendix A. Always
remember the following when using a spectrophotometer cuvette:
All cuvettes should be wiped clean and dry on the outside with a tissue
Use a tissue when handling cuvettes or hold it at the very top
All solutions should be free of bubbles
Always position the cuvette with its reference mark aligned with the reference mark in the chamber
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Part A
‣ 3. Transfer approximately 30 mL of 0.200 M Fe(NO3)3 solution in a clean, dry
labeled 50 or 100 mL beaker. Note that there are two different stock solutions of Fe(NO3)3 at different
concentrations. Make sure you choose the correct solution for Part A or your
experiment will not work. ‣ 4. Transfer approximately 20 mL of 0.0020 M SCN- solution into a second clean, dry
labeled 50 or100 mL beaker. ‣ 5. Clean and dry five 100 mL beakers (or cups from stockroom) and label 1-5.
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Part A
‣ 6. Pipet 5.00 mL of the Fe(NO3)3 solution into beakers 1-5 using a 5 mL
volumetric pipet. Pipet 1, 2, 3, 4, and 5 mL of 0.0020 M SCN- solution into the
corresponding labeled beaker using a graduated pipet. Then add the correct number of mL of water to each beaker according
to the table below using a graduated cylinder. Use a clean glass stirring
rod to mix the solutions.
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Part A
‣ 7. Immediately determine the absorbance for each of the five solutions using the
Spectronic 20 set at λmax. Clean your beakers from Part A. Dispose of the waste
in the waste bottle.
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Part B
‣ Part B: Determining the Equilibrium Constant, Keq
‣ 1. Recalibrate the Spectronic 20 spectrophotometer.
‣ 2. Transfer approximately 30 mL of 0.0020 M Fe(NO3)3 solution to a clean, dry,
labeled 50 or 100 mL beaker.
‣ 3. Transfer approximately 20 mL of 0.0020 M KSCN solution to a clean, dry, labeled
50 or 100 mL beaker.
‣ 4. Clean and dry five test tubes and label 1-5.
‣ 5. Pipet 3.00 mL of the 0.0020 M Fe(NO3)3 solution into test tubes 1 – 5 using a
graduated pipet. Rinse the pipet with water, condition it, and then pipet 1, 2, 3, 4
and 5 mL’s of 0.0020 M KSCN solution into the corresponding test tube. Add the
correct mL of water to each tube so that the total volume is 10.00 mL (see table
below). Use a clean glass stirring rod to mix the solutions.
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Part B
‣ 6. Immediately determine the absorbance for each of the five solutions using the
Spectronic 20. Dispose of the solutions in the waste bottle when complete.
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Analysis
‣ 1. In order to calculate the Keq for each sample, you will need to determine
[Fe3+]eq , [SCN-]eq and [FeSCN2+]eq. First, calculate [FeSCN2+]eq, for each
sample in Part B using the Beer’s Law plot from Part A. Enter results in the data
summary table for Part B on page 6.
‣ 2. Determine [Fe3+]eq and [SCN-]eq) using the following equations:
[Fe3+]eq = [Fe3+]init – [FeSCN2+]eq [SCN-]eq = [SCN-]init –[FeSCN2+]eq
‣ First determine [Fe3+]init and [SCN-]init for each of the five samples in Part B.
Enter results in the data summary table on page 6.
‣ calculations for samples 2 – 5. Enter results in the data summary table on page 7.
Calculate the equilibrium constant, Keq, for each of the five samples.
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Analysis
‣ Example - determine [Fe]init, for sample #1. For this sample, 3.0 mL ferric nitrate
and 1 mL potassium thiocyanate were combined with 6 mL water. First, calculate
the mmols (or mols) of Fe3+ ion in the test tube.
‣ mmols Fe3+ ion = 0.0020 M * 3mL = 0.0020 mmols/mL * 3mL = 0.0060 mmols Fe3+
ion
‣ Calculate [Fe3+]init ion after dilution to 10 mL.
‣ [Fe3+]init = mmols Fe3+ ion/final volume = 0.0060 mmols/10mL = 0.00060 mmols/mL = 0.00060 M
‣ Calculate [Fe3+]eq using the equation [Fe3+]eq = [Fe3+]init – [FeSCN2+]eq substituting
values for [FeSCN2+]eq, and [Fe3+]init from the previous calculations for sample #1.
Enter your result in the data summary table on page 6.
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Analysis
‣ Repeat these calculations to determine [SCN1-]eq. For sample #1:
‣ mmols SCN1- ion = 0.0020 M * 1mL = 0.0020 mmols/mL * 1mL = 0.0020 mmols SCN1- ion
‣ [SCN- ]init = mmols SCN1- ion/final volume = 0.0020 mmols/10mL = 0.00020 mmols/mL = 0.00020 M
‣ Calculate [SCN-]eq using the equation [SCN-1]eq = [SCN-1]init –[FeSCN2+]eq
substituting values for [FeSCN2+]eq and [SCN-1]init from the previous calculations.
Enter your result in the data summary table.
‣ You now have values for [Fe3+]eq, [SCN-1]eq and [FeSCN2+]eq for sample #1. Write the equilibrium expression for this reaction and solve for Keq.
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Next Week
‣ Before next Meeting:
‣ Bring to class:
‣ Notebook
‣ You will not be turning in notebooks, but this
permanent record of your preparations,
observations and notes will be essential to
your success in this class.
‣ Textbook, calculator, pencils
(yes, you can use pen)
‣ Safety Glasses (you cannot participate in the next class without them)
‣ Read and bring a copy of the next experiment
Le Chatelier’s Principle
‣ Produce and bring to class:
‣ Your pre-lab for exp 05
‣ Your procedure summary for exp 05
‣ Review from your lecture text:
‣ Le Chatelier’s Principle
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We will start
with a quiz about
the experiment and
reading.
Questions?