Class 5: Le Chatelier's Principle and the Connection Between Free Energy and Equilibrium

Question 1

Predict how changes in concentration, temperature, and the presence of a catalyst affect equilibrium.

Answer 1

Changes in Concentration:
* Increasing concentration of a reactant shifts equilibrium towards products (Le Chatelier’s Principle)
* Increasing concentration of a product shifts equilibrium towards reactants
* Decreasing concentration does the opposite of increasing it
* The equilibrium constant (K) is not affected by changes in concentration

Changes in Temperature:
* If the reaction is endothermic (ΔH > 0), raising temperature shifts equilibrium towards products
* If the reaction is exothermic (ΔH < 0), raising temperature shifts equilibrium towards reactants
* Changing temperature does change the value of K

Presence of a Catalyst:
* A catalyst does not affect the equilibrium concentrations
* It does not shift the position of the equilibrium
* A catalyst increases the rates of both the forward and reverse reactions equally
* It allows a system to reach equilibrium faster, but does not change the equilibrium state itself
* A catalyst has no effect on the value of the equilibrium constant K

In summary:
- Changing concentrations shifts position of equilibrium
- Changing temperature shifts position and changes K value
- A catalyst does not shift the equilibrium position or change K

Question 2

Calculate new equilibrium concentrations from any of the above changes.

Answer 2

Changes in Concentration:
* Use reaction quotient Q = [products]/[reactants]
* Substitute initial concentrations into Q
* If Q ≠ K, equilibrium is disturbed
* Calculate new concentrations by setting Q = K and solving

Changes in Temperature:
* Calculate new K value using van’t Hoff equation: ln(K2/K1) = (-ΔH°/R)[(1/T1) - (1/T2)]
* Substitute old K, new T, ΔH° into equation to get new K
* Use new K and initial concentrations to calculate new equilibrium concentrations

Addition of Catalyst:
* A catalyst does not affect equilibrium concentrations
* It only speeds up the rate at which equilibrium is reached
* No calculation for new concentrations needed
* Equilibrium concentrations remain the same as before catalyst

General Steps:
* Identify the change (concentration, temperature, catalyst)
* For concentration changes, use Q = K
* For temperature changes, calculate new K first
* Substitute appropriate values into equilibrium constant expression
* Solve for new equilibrium concentration(s)

The key is using reaction quotient Q, equilibrium constant K, and the balanced chemical equation to relate concentrations before and after a stress.

Question 3

Describe how certain changes cause the equilibrium to shift in one direction or the other and propose an explanation for why this happens (rates of reactions and collisions).

Answer 3

Concentration Changes:
* Increasing reactant concentrations increases forward reaction rate more than reverse rate
* More reactant collisions per unit time drives equilibrium towards products
* Increasing product concentrations increases reverse reaction rate more than forward
* More product collisions per unit time drives equilibrium back towards reactants

Temperature Changes:
* Raising temperature increases rates of both forward and reverse reactions
* But it increases the endothermic reaction rate more due to more molecular collisions having activation energy
* So if reaction is endothermic, raising temperature favors products
* If reaction is exothermic, raising temperature favors reactants

Catalyst Addition:
* Catalysts provide alternative pathway with lower activation energy
* This increases rates of both forward and reverse reactions equally
* But equilibrium position is not affected, as rates increase by same factor

Explanations:
* Reactions occur from molecular collisions with sufficient energy (activation energy)
* Increasing concentrations increases collisions per unit time, affecting one direction more
* Temperature increases molecular kinetic energies, affecting endothermic/exothermic differently
* Catalysts lower the energy barrier, increasing collision frequencies in both directions equally

So in essence, these changes alter the rates of forward/reverse reactions differentially by changing the frequencies and energies of molecular collisions.

Question 4

Graphically represent how changes in concentration or temperature affect concentrations.

Answer 4

Concentration Changes:
* Use a graph with concentration on the y-axis and time on the x-axis
* Initially plot reactant and product concentrations at equilibrium
* If a reactant concentration increases:
- Shift reactant line up and product line down initially
- Then show product line increasing and reactant line decreasing toward new equilibrium
* If a product concentration increases:
- Shift product line up and reactant line down initially
- Then show reactant line increasing and product line decreasing toward new equilibrium

Temperature Changes:
* Use a graph with ln(Q) or ln(K) on the y-axis and 1/T on the x-axis
* Initially plot ln(K) value at the given temperature
* If temperature increases for an endothermic reaction:
- Positive slope, ln(K) increases with 1/T
* If temperature increases for an exothermic reaction:
- Negative slope, ln(K) decreases with 1/T
* Can calculate and plot multiple ln(K) points at different temperatures

Key Points:
* For concentrations, show initial shift then return to new equilibrium
* For temperature, plot ln(K) vs 1/T to show trend with increasing T
* Slopes indicate endo/exothermic based on van’t Hoff equation

So in summary, use appropriate x/y axes to qualitatively or quantitatively represent how equilibrium concentrations or constants change.

Question 5

Use the Van’t Hoff equation to relate equilibrium constant to temperature.

Answer 5

Here are concise bulleted notes on using the van’t Hoff equation to relate the equilibrium constant (K) to temperature:

• The van’t Hoff equation is: ln(K2/K1) = (-ΔH°/R)((1/T1) - (1/T2))
• It shows the logarithmic relationship between the equilibrium constant ratio and inverse absolute temperatures
• ΔH° is the standard enthalpy change (heat of reaction)
• R is the universal gas constant (8.314 J/mol*K)
• T1 and T2 are the initial and final absolute temperatures in Kelvins
• ln(K2/K1) can be plotted vs (1/T) to get a straight line with slope = -ΔH°/R

Key Points:
- At constant ΔH°, ln(K) varies linearly with 1/T
- Positive ΔH° (endothermic) gives positive slope
- Negative ΔH° (exothermic) gives negative slope
- Slope allows calculating ΔH° from ln(K) vs 1/T plot

Using the Equation:
- Measure K at two different temperatures
- Substitute K1, K2, T1, T2 into van’t Hoff equation
- Can calculate ΔH° from ln(K2/K1) and temperatures
- Or predict K2 at new T2 if ΔH° is known

So in essence, the van’t Hoff equation quantifies how K changes with temperature based on the sign and magnitude of ΔH°.

Question 6

Describe what is meant by the standard state. Give examples of elements and compounds in their standard states.

Answer 6

• The standard state refers to a set of reference conditions used to determine thermodynamic properties like enthalpy, entropy, and Gibbs free energy.
• For gases, the standard state is an ideal gas at 1 bar pressure.
  
  • Examples: O2, N2, CO2, CH4 gases at 1 bar
• For pure liquids and solids, the standard state is the pure substance at 1 bar.
  
  • Examples: H2O(l), C(s graphite), Fe(s)
• For solutes, the standard state is a 1 M ideal solution.
  
  • Examples: 1 M NaCl(aq), 1 M C6H12O6(aq)
• Elements in their most stable forms at 25°C and 1 bar are in their standard states.
  
  • Examples: O2(g), H2(g), C(s graphite), S8(s)
• Compounds are in their standard states if all elements are in their standard states.
  
  • Examples: CO2(g), H2O(l), NaCl(s)
• The standard state allows comparing different substances on a common basis.

So in summary, the standard state provides reference conditions (usually 25°C, 1 bar) for reporting thermodynamic data, with specific definitions for gases, liquids, solids, and solutions.

Question 7

Write and apply the mathematical relationship between ΔG° and K. Explain the difference between ∆G° and ∆G.

Answer 7

Mathematical Relationship:
* ΔG° = -RT ln K
* R is the universal gas constant (8.314 J/mol*K)
* T is the absolute temperature in Kelvins
* ln K is the natural log of the equilibrium constant

Applying the Relationship:
* If ΔG° and T are known, can calculate K
* If K and T are known, can calculate ΔG°
* Negative ΔG° means K > 1, reaction favors products
* Positive ΔG° means K < 1, reaction favors reactants

Difference between ΔG° and ΔG:
* ΔG° is the standard Gibbs free energy change
- Calculated from ΔH° and ΔS° under standard state conditions
* ΔG is the actual Gibbs free energy change
- Depends on reaction quotient Q and deviates from ΔG° at non-standard concentrations
* ΔG = ΔG° + RT ln Q
* At equilibrium, when Q = K, then ΔG = 0

So in summary:
- Use ΔG° = -RT ln K to interconvert between ΔG° and K
- ΔG° applies under standard state conditions
- ΔG accounts for non-standard concentrations via Q
- ΔG = 0 at equilibrium when Q = K

Question 8

Predict how changes in concentration, temperature, pressure and the presence of a catalyst affect equilibrium.

Answer 8

If you increase the concentration of a reactant or decrease the concentration of a product, then it will increase the forward reaction
If you increase the concentration of a product or decrease the amount of reactant, then it will increase the reverse reaction
AMT OF EACH DIFFERENT< BUT THE RATIO IS THE SAME
If the rxn is exothermic: increasing heat is like adding product, so the reverse reaction is favored. Removing heat would be like removing product, so the forward reaction is favored
If the rxn is endothermic: increasing heat is like adding to the reactant side, so the forward reaction is favored. Decreasing heat is like removing reactant, so the reverse reaction is favored
If there is a reduction of pressure, the side with more molecules will increase. If there is an increase in pressure, the side with less molecules with be favored.

Question 9

Describe how certain changes cause the equilibrium to shift in one direction or the other and propose an explanation for why this happens (rate of reaction and collisions)

Answer 9

With an increase in concentration in reactants or products, its more likely that they will bump collide with others to create products or reactants

Question 10

Use the Van’t Hoff equation to relate equilibrium constant to temperature

Answer 10

ln(Kt2/Kt1)=-(∆H/R)(1/T2 - 1/T1)
R=8.314 J mol^-1 K^-1

Question 11

Describe what is meant by the standard state. Give examples of elements and compounds in their standard states

Answer 11

The standard states of elements are the forms they adopt at room temp and a pressure of 1 ATM
The heat of formation of an element in its standard state is zero

Question 12

Write and apply the mathematical relationship between delta G and K

Answer 12

If K > 1, ∆G is negative — runs forward
If K=1, ∆G = 0 — at equilibrium
If K < 1, ∆G is positive — runs backward
K and ∆G HAVE AN INVERSE RELATIONSHIP
If K=negative, runs backward
If K=positive, runs forward

Question 13

Explain the difference between ∆G° and ∆G.

Answer 13

∆G° refers to to the ∆G per mole of compound in standard state conditions (Q=1)
It is ∆G when the reactants and products are in their standard states
T=25 degrees C (NOT FOR OUR CLASS)
p= 1 atm
Q=1
PROOF THAT ∆G°=0 at equilibrium:
∆G= RT ln (Q/K) = RT ln (1/1) = RT ln(1)=0

Question 14

REVIEW TEST 2022 & 2023

Answer 14

Exam questions about ln(k2/k1)=-∆H/R(1/T2-1/T1):
∆H should be IN JOULES
Requiring a spark DOES NOT MEAN THAT IT IS NONSPONTANEOUS
Important equations to remember
∆G°=∆H°-T∆S°
ln(k2/k1)=-∆H/R(1/T2-1/T1) —> NEED TO CONVERT ∆H to JOULES—R is in JOULES
∆G°=-RT ln K → GIVES ∆G IN JOULES
∆G=RT ln(Q/K) → GIVES ∆G IN JOULES
∆H=Ea of forward-Ea of reactant (absolute value)
∆G=0 at equilibrium
THE EQUILIBRIUM CONSTANT DOES NOT CHANGE WITH CHANGES IN CONCENTRATION
∆G standard DOES NOT DEPEND ON CONCENTRATION
IF T/VOLUME CHANGES—CHANGES K and ∆G
UNITS
R=J/molK
∆G=kj/molK
∆H=kj/mol
∆S=j/molK
T=K
K/Q=none
Kreverse=1/Kfoward

Question 15

The reaction below is conducted in a 1.00 L flask. The equilibrium concentrations were determined
to be 0.050 M H2, 0.040 M I2, and 0.16 M CH3OH

H2(g) + I2(g) ⇄ 2 HI(g)

K = 12.8 @ 448 °C

Answer 15

a. If we add 0.10 mol of H2 to the system, will the system still be at equilibrium? Explain.
No! You have added more of one of the reactants but none of the products (or the other
reactant) so the EQ ratio can no longer be equal to 12.8.

b. Based on your knowledge of equilibrium, predict the direction of the shift that the system must
take to relieve the stress of the additional reactant.
Too much reactant so must shift to the right (reactants → products)

Question 16

Provide a scientific explanation for the shift in equilibrium position seen with a change in
concentration.

Answer 16

Reasoning: The reaction would shift to the right because the number of collisions between
H2 and I2 would increase and therefore the reaction rate of the forward reaction would
increase. From a mathematical point of view, the denominator of the K expression would
make the ratio lower than K and shifting to the right would bring the ratio back to 12.8.

Question 17

Adding heat shifts the EQ to in the endothermic direction.

Answer 17

EQ shifts left (Ice forms); b. removing heat shifts in the exothermic direction

Question 18

K is temperature dependent.

Question 19

enthalpy positive - endothermic

Question 20

Delta H in Joules for vant hoff

Question 21

Given the same starting concentrations of CO and H2O, will more or less H2 be obtained at
equilibrium if the reaction is at higher temperatures? Less

Answer 21

When the reaction involves the formation of hydrogen gas (H2) from carbon monoxide (CO) and water (H2O), increasing the temperature will actually lead to less hydrogen being obtained at equilibrium. This phenomenon is governed by Le Chatelier’s Principle, which states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust to counteract that change.In this case, the reaction CO+H2O⇌CO2+H2CO + H_2O \rightleftharpoons CO_2 + H_2CO+H2​O⇌CO2​+H2​ favors the formation of hydrogen gas. However, this reaction is also endothermic, meaning it absorbs heat. When the temperature is increased, according to Le Chatelier’s Principle, the system will shift in the direction that absorbs heat.Since the forward reaction (formation of hydrogen) is endothermic, increasing the temperature will favor the reverse reaction (formation of carbon dioxide and water) in order to absorb the additional heat. As a result, less hydrogen will be obtained at equilibrium when the reaction is at higher temperatures.

Question 22

9. The standard free energy change can be calculated several ways, but two of them are listed below.

Answer 22

i. Gibbs Free Energy Equation: ∆G = ∆H - T∆S
ii. ∆G = -RT ln K

Question 23

a. According to the second equation, what will the sign of ∆G be if:
K>1? K<1? K = 1?
- + 0

Question 23

c) What are the advantages to running this reaction at 450°C vs. 25°C?
- The advantage of running this reaction at 450oC is that at this temperature, there are
going to be more collisions (and those collisions will be more energetic). This will
establish equilibrium more quickly than if we ran the reaction at 25oC.
- HOWEVER, this would not be very helpful, because raising the temperature this high
severely affects the equilibrium position and reduces our total reaction yield (less
products will be formed at the equilibrium position because, for exothermic reactions, K
decreases at the temperature increases).

Answer 23

b) What are the advantages to running this reaction at 25°C vs. 450°C?
- The equilibrium constant is 20 orders of magnitude higher at 25oC than at 450oC. Using
LeChatelier’s Principle, we see that raising the temperature of the reaction gives a lower
equilibrium constant (K). Therefore, heat must be a product (adding a product gives a
leftwards shift to re-establish equilibrium). Since heat is a product, this reaction is
exothermic.
- The advantage of running the reaction at 25oC is that at this temperature, the products
are strongly favored. In this case, SO3 is our desired compound, so favoring the products
increases the yield of the reaction.

Question 24

At “all reactants,” Q = 0 (because there are no products)
- At “all products,” Q = Indeterminant (goes to infinity as the concentration of the
products goes to zero)

Answer 24

G is a state function, meaning that this is the energy that will be released if one mole
of PCl3 reacts with one mole of Cl2 to form one mole of PCl5, completely. We know this is
an equilibrium reaction (can go forwards and backwards). The actual equilibrium
position is not “all products” (100% conversion).

Equilibrium is defined as where G = 0. The slope of the curve is 0 at this point. This
means you should look for where on the curve G is the lowest and pick that point. The
equilibrium position of any system will be the lowest in free energy.
Recall, G is the energy given off per mole at standard conditions assuming the
reaction goes to completion. G is the change in Gibb’s free energy for the reaction
(whatever its specific conditions are).

Question 25

The equilibrium position is where G is at its lowest (IE., G = 0). If Q ≠ K, then the
reaction position at that moment has a higher G than the equilibrium position.
Spontaneous means that G <0. Since all positions besides equilibrium are higher in
Gibb’s free energy than the equilibrium position, all shifts to the equilibrium will have
G <0 (must be spontaneous).
- One helpful analogy is to imagine the curve in this graph as a ramp. Place an imaginary
ball anywhere on the ramp. Where is the ball going to end up? No matter where you
place it, it will spontaneously roll to the bottom position and stay there. The ball likes to
minimize its energy (gravitational potential energy in this analogy of a ball on a track);
reactions also want to minimize energy (Gibb’s free energy) and spontaneously move to
the lowest energy position