Class 4: Intro to Equilibrium

Question 1

Define dynamic equilibrium and the law of mass action.

Answer 1

Dynamic Equilibrium:
* State where forward and reverse reactions occur at equal rates
* Concentrations of reactants and products remain constant
* However, molecules are continually being converted between reactants and products
* No net change in overall composition over time

Law of Mass Action:
* Relates equilibrium concentrations of reactants and products
* For a reaction: aA + bB ⇌ cC + dD
* Equilibrium constant (K) = [C]^c[D]^d / [A]^a[B]^b
* [X] represents molar concentration of species X
* K is constant at a given temperature
* Larger K means products are favored at equilibrium
* Smaller K means reactants are favored at equilibrium
* Allows calculating equilibrium concentrations from initial concentrations

So in summary, dynamic equilibrium describes the state, while the law of mass action quantifies the relationship between concentrations at equilibrium.

Question 2

Use the law of mass action to write an equilibrium constant expression for any reaction and calculate the value from provided data.

Answer 2

Writing the Equilibrium Constant (K) Expression:
* For a general reaction: aA + bB ⇌ cC + dD
* K = [C]^c[D]^d / [A]^a[B]^b
* [X] is the molar concentration of X at equilibrium
* Exponents are stoichiometric coefficients in balanced equation

Calculating K from Provided Data:
* Equilibrium concentrations of reactants/products are given
* Substitute the values into the K expression
* K = ([C]^c[D]^d) / ([A]^a[B]^b)
* Raise each concentration to its stoichiometric coefficient power
* Divide product concentrations by reactant concentrations

Example:
For the reaction: 2NO2 ⇌ N2O4
With [NO2] = 0.5 M and [N2O4] = 1.2 M at equilibrium
K = [N2O4] / [NO2]^2
= (1.2) / (0.5)^2
= 4.8

So the key steps are:
1) Write K expression from balanced equation
2) Substitute known equilibrium concentrations
3) Raise each concentration to stoichiometric coefficient power
4) Divide product over reactant concentrations

Question 3

Calculate the reaction quotient (Q) and use it to determine the direction of the reaction.

Answer 3

• The reaction quotient (Q) is similar to the equilibrium constant (K) expression, but uses initial/current concentrations instead of equilibrium concentrations.
• For a general reaction: aA + bB ⇌ cC + dD
  Q = [C]^c[D]^d / [A]^a[B]^b
• To calculate Q:
  
  • Substitute initial/current concentrations into the K expression
  • Raise each concentration to its stoichiometric coefficient power
  • Divide product concentrations by reactant concentrations
• Comparing Q to K reveals the direction the reaction will proceed:
  
  • If Q < K, the reaction will proceed forward (reactants → products)
  • If Q > K, the reaction will proceed reverse (products → reactants)
  • If Q = K, the reaction is already at equilibrium
• The reaction will shift in the direction that makes Q = K at equilibrium
  
  • Increasing Qproduct concentrations drives Q → K
  • Decreasing Qreactant concentrations drives Q → K

So in summary, calculate Q, compare it to K, and the reaction will shift in the direction that forces Q → K.

Question 4

Calculate equilibrium concentrations or the equilibrium constant depending on the information provided

Answer 4

Calculating Equilibrium Concentrations:
* If initial concentrations and K are provided:
- Substitute initial concentrations and K into reaction quotient (Q) expression
- Set Q = K and solve for the unknown equilibrium concentration(s)
- May require using approximations or quadratic formula

• If some equilibrium concentrations and K are provided:
  
  • Substitute known values into K expression
  • Rearrange to solve for unknown equilibrium concentration(s)

Calculating Equilibrium Constant (K):
* If all equilibrium concentrations are provided:
- Substitute equilibrium concentrations into K expression
- Evaluate K by raising concentrations to stoichiometric coefficients
- Divide products by reactants concentrations

• If initial concentrations and some equilibrium concentrations are provided:
  
  • Calculate reaction quotient Q using initial concentrations
  • Substitute known equilibrium concentrations into Q expression
  • Set Q = K to solve for K

So in summary:
- Use provided info and K expression
- Substitute known values
- Set Q = K if calculating equilibrium concentrations
- Isolate the unknown(s) and solve

Question 5

Define dynamic equilibrium and the law of mass action

Answer 5

Dynamic Equilibrium: if molecules can react to form products, then products can react to re-form reactants
Law of Mass action: the proposition that the rate of the chemical reaction is directly proportional to the product of the activities or concentrations of the reactants.

Question 6

Use the law of mass action to write an equilibrium constant expression for any reaction and calculate the value from provided data

Answer 6

K=([Concentration of product C]^stoich coefficient [Concentration of product D]^stoich coefficient) / ([conc. Of reactant A]^stoich coefficient [conc. Of reactant B]^stoich coefficient

Question 7

Calculate the reaction quotient (Q) an use it to determine the direction of the reaction

Answer 7

Q is the same equation as Kc
If Q<K: the numerator needs to get bigger, so it will go to the FORWARD REACTION
IfQ>K: the denominator needs to get bigger so it will go to the REVERSE REACTION

Question 8

Calculate equilibrium concentrations or the equilibrium constant depending on the info provided

Answer 8

ICE tables
I stands for initial concentration. This row contains the initial concentrations of products and reactants.
C stands for the change in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign and a variable; this is because this row represents an increase or decrease (or no change) in concentration.
E is for the concentration when the reaction is at equilibrium. This is the summation of the initial and change rows. Once this row is completed, its contents can be plugged into the equilibrium constant equation to solve for Kc

Question 9

N2(g) + 3 H2(g) ⇄ 2 NH3(g)
A 1 L vessel contains 1.00 mol of N2, 0.500 mol of H2, 0.0866 mol of NH3, and is at equilibrium.
Considering the simulation above, what can you say about the rate of formation and the rate of
consumption of NH3.

Answer 9

If it is at equilibrium thent he rates of the forward and reverse reactions are the same

Question 10

law of mass action for the general equilibrium reaction

Answer 10

Kc=([C]^c[D]^d)/([A]^a[B]^b)

lower case letter, exponents which are moles of that molecule
Upper case chemical species concentrations (conc shown by {})

Question 11

A. If Kc is very large then at equilibrium the beaker will contain (Circle one)

Answer 11

Mostly products

Question 12

B. If Kc is very small then at equilibrium the beaker will contain (Circle one)

Answer 12

Mostly reactants

Question 13

C. If Kc is very approximately 1 then at equilibrium the beaker will contain (Circle one)

Answer 13

Equal amount of each

Question 14

CO(g) + 2 H2(g) ⇄ CH3OH(g)
Calculate the equilibrium constant,
K, for each experiment. Measured
molar equilibrium concentration are
provided

Experiment
CO(g) (M) H2(g) (M) CH3OH(g) (M)
1 0.0911 0.0822 0.00892
2 0.0753 0.151 0.0247
3 0.138 0.176 0.0620

Answer 14

1 solved out -

1. 14.49
2. 14.38
3. 14.5

(0.00892)/(0.0911)(0.0822)^2
= 14.4911496798

What can you conclude about the starting position of the experiment compared to the ratio of
concentrations at equilibrium?

Does not matter!

d. Describe the direction of the reaction to reach equilibrium in each of the experiments.
1. Reactants ® Products
2. Products ® Reactants
3. Products ® Reactants

Question 15

What can you conclude about the absolute concentration and ratio of concentrations at
equilibrium?

Answer 15

Though equilibrium concentrations may be different, the ratio, K, is constant at a given K

Question 16

H2(g) + Cl2(g) ⇄ 2 HCl(g)

Keq = 1.2 x 10^-12 @ 473K

If we start with 0.37 M H2, 0.49 M Cl2, and 0.0 M HCl:

A reaction quotient, Q, is calculated using the Law of Mass action with concentrations that are not
necessarily at equilibrium. Will the reaction above proceed towards the reactants or the products and
how did Q help you decide?

Q= (HCl)^2/(H2)(Cl2) = 0

Q<K so the reaction must proceed to the right (products)

Answer 16

• Q is the reaction quotient
• K is the equilibrium constant
• Comparing Q and K:
• If Q = K, the reaction is already at equilibrium
• If Q < K, the reaction will proceed from reactants to products to reach equilibrium
• If Q > K, the reaction will proceed from products to reactants to reach equilibrium
• Driving forces:
• Q < K means products are favored (forward reaction driven)
• Q > K means reactants are favored (reverse reaction driven)
• System adjusts to make Q = K at equilibrium
• Calculating Q:
• Use the Law of Mass Action with current concentrations
• Q = [products]^x / [reactants]^y (based on balanced equation)
• Compare calculated Q to known K value
• Using Q vs K:
  1) Calculate Q from current concentrations
  2) Compare Q to K
  3) If Q ≠ K, predict direction of reaction to reach equilibrium
  4) Reactants favored if Q > K
  5) Products favored if Q < K

So in essence, comparing the reaction quotient Q to the equilibrium constant K allows you to determine which direction the reaction will proceed to reach equilibrium.

Question 17

Since Q = 0 < K = 1.2 x 10^-12, the reaction will proceed towards the products to reach equilibrium.

The ICE table setup is correct:
Copy code H2(g) + Cl2(g) ⇌ 2 HCl(g)

Initial 0.37 M 0.49 M 0 M
Change -x -x +2x
Equilibrium (0.37-x) (0.49-x) 2x
A few key points:

For reactants, change is -x (decreasing)
For products, change is +2x (increasing, based on balanced coefficients)
At equilibrium, concentrations are (initial - change) for reactants
At equilibrium, concentration is (change) for products

The stoichiometric 2x for HCl is because there are 2 moles of HCl produced per reaction, based on the balanced equation coefficients.
The decision to add or subtract x is based on the direction the reaction proceeds to reach equilibrium from the initial non-equilibrium concentrations.

a. How do you know to add or subtract the x for each reactant and product?

Based on the direction of the reaction to reach EQ

Answer 17

x = +/- 2.33 ^10-7
only the plus makes sense!

c. Using your value for x, find the equilibrium concentrations of H2, Cl2, and HCl.
[HCl] = 2x = 4.66 ´ 10-7 M
[H2] = 0.37 - 2.27 ´ 10-7 = 0.37 M
[Cl2] = 0.49 - 2.27 ´ 10-7 = 0.49 M

There were many steps to solving for the equilibrium concentrations of H2, Cl2, and HCl. How
might you check to verify you got the right answer?
Recalculate K: K = (4.66 ^ 10-7 M)2
/(0.37*0.49) = 1.197 ^ 10-12 Very close to 1.2 ^ 10-12 !

Question 18

1. Do a calculation to show the concentration of pure H2O in moles/L. Why is it that pure liquids and solids are not included in Kc expressions?

Answer 18

WE ARE LOOKING TO FIND THE CONCENTRATION OF H2O IN MOLES/LITER…

LETS START WITH THE DENSITY OF WATER, BECAUSE THAT IS A KNOWN VALUE WITH
SIMILAR UNITS (p = mass /volume).

ONCE WE KNOW GRAMS/LITER, WE CAN CONVERT TO MOLES/LITER USING THE MOLAR MASS OF H2O!

pwater: 1 g/ML

LETS CONVERT THIS TO g/L
1g | 1000mL. = 1000
1ml | 1 L g/L

TIME TO CONVERT THIS TO MOLES /L

1000g/L | 1 mol H2O = 55.5
1L | 18.02 g H20 mol/L

PURE H20(1) IS TYPICALLY NOT INCLUDED IN EQUILIBRIUM EXPRESSIONS BECAUSE THIS CONCENTRATION IS SO COMPARATIVELY HIGH. THROUGHOUT THE REACTION, THIS CONCENTRATION LIKELY WILL NOT CHANGE BY A MEASURABLE AMOUNT. THIS IS TRUE FOR OTHER PURE LIQUIDS AND SOLIDS AS WELL - SINCE THEIR CONCENTRATIONS WILL REMAIN FAIRLY CONSTANT, THEY CAN BE EXCLUDED FROM THE EQUILIBRIUM EXPRESSION.

Question 19

Q vs K

Answer 19

Q in any time of reaction
K only at equilibrium

at equilibrium Q=K

Question 20

Problem:

1.30 mol N2(g) and 1.65 mol H2(g) are added to a 2.00 L container
At equilibrium, 0.100 mol NH3(g) is formed
Need to find equilibrium concentrations of N2(g) and H2(g)

Steps:
a) Balanced equation: N2(g) + 3H2(g) ⇌ 2NH3(g)
b) Set up ICE table:
Initial: 1.30 mol N2, 1.65 mol H2
Change: -x mol N2, -3x mol H2, +2x mol NH3 (x = 0.050 mol)
Equilibrium: 1.30 - 0.050 = 1.25 mol N2, 1.65 - 0.150 = 1.50 mol H2, 0.100 mol NH3
c) Calculate equilibrium concentrations:
[N2]eq = 1.25 mol / 2.00 L = 0.63 M
[H2]eq = 1.50 mol / 2.00 L = 0.75 M
So, the equilibrium concentrations are [N2]eq = 0.63 M and [H2]eq = 0.75 M.

Answer 20

1. Writing Balanced Chemical Equations:
   
   • Identify reactants and products
   • Write the formulas of reactants on the left and products on the right
   • Balance atoms of each element by adjusting coefficients (whole numbers)
   • No fractional coefficients allowed
2. Setting up an ICE Table:
   
   • Under “Initial”: Write initial moles of reactants
   • Under “Change”: Use -x for reactants consumed, +y for products formed
     
     • x and y related by balanced equation (e.g. 2x = y)
   • Under “Equilibrium”: Initial + Change
3. Finding x (amount of reaction):
   
   • Equate amount of any product at equilibrium to the “Change” expression
   • Solve for x
   • Substitute x into ICE table expressions
4. Calculating Equilibrium Concentrations:
   
   • [X]eq = Equilibrium Amount of X / Volume
   • Convert from moles to molarity (M = mol/L)
5. Relating Amounts by Stoichiometry:
   
   • Use balanced coefficients
   • e.g. 2 mol A reacts with 3 mol B to make 2 mol C
     
     • If 0.1 mol C formed, 0.05 mol A reacted (0.1/2)
6. Algebraic Manipulation:
   
   • Rearrange equation to isolate unknown
   • Substitute known values and solve

The key things to practice are: setting up the ICE table, using stoichiometry to relate amounts, rearranging algebraic equations, and converting between moles and molarity. Work through practice problems methodically.

Question 21

To calculate Kc at 300 K using the given Kc value at 2000 K, we can use the van’t Hoff equation, which relates the equilibrium constant to temperature:
ln(K2/K1) = (-ΔH°/R) * [(1/T2) - (1/T1)]

Answer 21

Where:
K2 = Equilibrium constant at T2 (unknown, to be calculated)
K1 = Equilibrium constant at T1 (given as 2.9 x 10^8 at 2000 K)
ΔH° = Standard enthalpy change for the reaction (J/mol)
R = Ideal gas constant = 8.314 J/mol·K
T2 = 300 K (low temperature)
T1 = 2000 K (high temperature)
To use this equation, we need to find ΔH° for the reaction. The bond enthalpies given can be used to estimate ΔH°:
ΔH° = ΣD(bonds broken) - ΣD(bonds formed)
For the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)
Bonds broken:
1 N≡N bond (941 kJ/mol)
3 H-H bonds (3 x 436 kJ/mol) = 1308 kJ/mol
Total = 941 + 1308 = 2249 kJ/mol
Bonds formed:
2 N-H bonds (2 x 391 kJ/mol) = 782 kJ/mol
ΔH° = 2249 kJ/mol - 782 kJ/mol = -92.6 kJ/mol
Substituting the values in the van’t Hoff equation:
ln(K2/2.9x10^8) = (-(-92600))/(8.314) * [(1/300) - (1/2000)]
ln(K2/(2.9x10^8)) = 11160 * (0.00233)
ln(K2/(2.9x10^8)) = 26.0
Taking the exponential of both sides:
K2 = 2.9 x 10^8 * e^26.0
K2 = 4.64 x 10^26
Therefore, the equilibrium constant Kc at 300 K is approximately 4.64 x 10^26.

Question 22

van’t Hoff equation

Answer 22

The van’t Hoff Equation:
ln(K2/K1) = (-ΔH°/R) * [(1/T2) - (1/T1)]

Where:
K2 = Equilibrium constant at temperature T2
K1 = Equilibrium constant at temperature T1
ΔH° = Standard enthalpy change of the reaction (J/mol)
R = Universal gas constant = 8.314 J/mol·K
T2 = Temperature 2 (K)
T1 = Temperature 1 (K)

Using the van’t Hoff Equation:
1) You need to know the equilibrium constant (K) at one temperature. This is K1.
2) Calculate ΔH° for the reaction using bond energies or tabulated values.
3) Substitute the known values of K1, ΔH°, R, T1 into the equation.
4) Rearrange to solve for ln(K2/K1)
5) Take the exponential of both sides to get K2/K1
6) Multiply by K1 to get K2

Notes:
- ΔH° is assumed to be constant over the temperature range
- Useful when you know K at one temperature and want to find K at another
- Increasing temperature favors the endothermic direction (positive ΔH°)
- Decreasing temperature favors the exothermic direction (negative ΔH°)

To calculate ΔH°:
ΔH° = Σ(Bond Enthalpies of Bonds Broken) - Σ(Bond Enthalpies of Bonds Formed)

So sum the bond enthalpies being broken and subtract the sum of the bond enthalpies being formed in the balanced reaction.

Question 23

to find equilibrium conc. of a chemical species plug x back in

Question 24

Q>K

Answer 24

reverse reaction dominates

Question 25

which carbon atoms are asymmetric (chiral) in an organic molecule:

Answer 25

1) Draw the Lewis structure of the molecule, showing all atoms and bonds.

2) Identify the carbon atoms that are bonded to four different substituents (atoms or groups attached to it).

3) For each carbon identified in step 2, assign priorities to the four substituents based on the Sequence Rule:
- Higher atomic number > Lower atomic number
- For isotopes, higher mass number > Lower mass number
- For compounds, alphabetize and use the first different atom

4) Arrange the substituents in a tetrahedral arrangement around the central carbon.

5) A carbon is asymmetric (chiral) if its four substituents have different priorities and cannot be superimposed on their mirror image by rotation.

6) If two different arrangements of substituents around the carbon are non-superimposable mirror images, that carbon is a stereocenter (asymmetric/chiral).

Additional Notes:
- Double bonds can make a carbon asymmetric if the groups on each side are different
- Check each carbon individually - a molecule can have multiple stereocenters
- Disregard hydrogens unless comparing H to another substituent
- Cis/trans isomers are not chiral, but geometric isomers
- Chiral molecules do not have a plane of symmetry

The key is identifying the carbons bonded to four different substituents, ranking the substituents by the Sequence Rule priorities, and determining if the tetrahedral arrangement is non-superimposable on its mirror image. Practice examining molecules methodically.