CHEM 122, UNIT 2 - SCALAR/PODER/PRATICE EXAM Questions

ΔG° = ΔH° - TΔS° At the boiling point of water, the reaction: H2O(l) H2O(g+44 kJ/mol) is at equilibrium, with P(H2O) = 1 atm. Calculate the boiling point of water given the following data: S°(H2O)l = 69.9 J mol-1K-1 S°(H2O)g = 188.8 J mol-1K-1 ΔH°= +44 kJ/mol

At boiling point ΔG° = 0 0 = ΔH° - TΔS° TΔS° = ΔH° T = ΔH° / ΔS° T = (+44 kJ/mol) / ((188.8 mol^(-1)K^(-1)) - (69.9 J mol^(-1)K^(-1))) T ≈ 370.20 K

Lecture 1

steady-state equilibrium, Dynamicequilibrium via Legos, Law of mass action;Keq, Kc, and Kp,

H2 (g) + I2 (g) = 2HI(g) T(K) 500 550 650 700 720 Kp 6.25×10-3 | 8.81×10-3 | 1.49×10-2 | 1.84×10-2 | 1.98×10-2 - As T increases, are more products or reactants present at equilibrium? - As T increases, how do kf and kr change? - If Keq increases as T increases, compare Δkf to Δkr - Is this reaction endothermic or exothermic?

- As T increases, are more products or reactants present at equilibrium? More products because this is an endothermic reaction - As T increases, how do kf and kr change? The forward and reverse rates increase - If Keq increases as T increases, compare Δkf to Δkr The equilibrium constant is becoming larger meaning the forward rate is favored over the reverse rate - Is this reaction endothermic or exothermic? As T increases, if Keq increases, it suggests that the reaction is endothermic (absorbs heat). Conversely, if Keq decreases as T increases, the reaction is exothermic (releases heat)

Using your kinetic molecular understanding of dynamic equilibrium, briefly explain what is happening at the molecular level as the temperature is raised (at constant pressure) from point A to point B to point C in the following phase diagram for CO2 .

- As the temperature increases from point A to B, the rate of evaporation of the liquid increases. - However, at point A the rate of condensation remains constant and still exceeds the rate of evaporation. - At point B, the rates of evaporation and condensation become equal, and the liquid starts to boil (gas-liquid equilibrium). - As the temperature continues to rise to point C, the rate of evaporation further increases, while the rate of condensation remains the same and the liquid turns entirely into gas

Using arguments from the Kinetic Molecular Theory and the concept of dynamic equilibrium, explain why, at a given applied pressure, there can be one and only one temperature, the boiling point, at which a specific liquid and its vapor can be in equilibrium.

- Dynamic equilibrium occurs when the rates of evaporation and condensation are equal (both liquid and gas present) - At a specific vapor pressure, the rate of condensation depends on gas density. - Evaporation rate is determined by the fraction of liquid molecules with enough energy to escape into the gas phase (related to T) - At equilibrium, the rates match, leading to one temperature - the boiling point - for a given pressure.

Even more molecules - How many microstates (W) exist if there are 4 molecules? - What is the probability that all 4 will be on one side? - What is the most probable macrostate? Why? - How many microstates (W) exist if there are N molecules? - How does the probability of an ordered state change as N increases?

- How many microstates (W) exist if there are 4 molecules? 16 = 2^4 x^n x = number of sides/arrangements n = number of molecules - What is the probability that all 4 will be on one side? 1/8 = 1/16 all on L + 1/16 all on R - What is the most probable macrostate? Why? 14/16 = 7/8 disordered - How many microstates (W) exist if there are N molecules? W = 2^n - How does the probability of an ordered state change as N increases? decreases exponentially

Using dynamic equilibrium arguments, explain why the vapor pressure of a liquid is independent of the volume available for the vapor above the liquid.

- The rate of condensation of molecules depends on Vapor pressure which affects the collision frequency of vapor molecules hitting the liquid surface. - At a fixed temperature, vapor pressure is proportional to gas density (n/V) and not Volume alone - Thus, regardless of volume equilibrium will occur at the same particle density, same vapor pressure

Explain why the vapor pressure curve of water is both a graph of the boiling point of liquid water as a function of applied pressure and a graph of the vapor pressure of liquid water as a function of temperature.

- The vapor pressure curve of water can also serve as a boiling point curve - By varying temperature at constant pressure, the curve can show where the vapor pressure = the applied pressure (boiling point) - This is where the liquid and gas phases are in equilibrium - Conversely, when keeping the temperature constant and varying the pressure, the curve shows shows the vapor pressure at which the liquid and gas phases are in equilibrium.

Consider the following reaction equations: (1) N2 (g) + 3H2(g) ⇌ 2NH3 (g) (2) 2N2 (g) + 6H2 (g) ⇌ 4NH3 (g)If the Kp of Reaction (1) is 6.2×10 5, What is the value of the equilibrium constant for Reaction (2)? Of course, the pressures at equilibrium do not depend on whether the reaction is balanced as in Reaction (1) or as in Reaction (2). Explain why this is true, even though the equilibrium constant can be written differently and have a different value

- Write out equilibrium expressions for both reactions - Kp(2) is the square of Kp(1). - Kp(2) = [Kp(1)]^2 - The individual pressures don't change at equilibrium, we have simply squared the value of Kp and its expression in terms of partial pressures

CaCO3 (s) = CaO (s) + CO2 (g)

- solids and liquids are not to be considered in equilibrium expressions = 1 - Thus, Kp/Kc = P(CO2)/[CO2] (respectively)

N2(g) + 3H2(g) = 2NH3(g), T = 298 K Using the data below, calculate: 1) Kp using equilibrium partial pressures 2) [N2], [H2], [NH3] at equilibrium 3) Kc using the [eq] values from Step 2) 4) Kc using Kp = Kc(RT)^n from Step 1)

1) Kp using equilibrium partial pressures - Kp expression = (PNH3)^2/[(PN2)(PH2)^3] - Plug in values: (4.82)^2/[(0.0342)(0.1027)^3] = 6.27 x 10^5 2) [N2], [H2], [NH3] at equilibrium - PV = nRT, n/V (concentration) = P/RT - [N2] = 0.0342/(0.08206 x 298) = 1.40 x 10^-3 M - [H2] = 4.20 x 10^-3 M - [NH3] = 1.97 x 10^7 M 3) Kc using the [eq] values from Step 2) - Kp expression = [NH3]^2/[N2][H2]^3 - Plug in values: [1.97 x 10^7 M]^2/[1.40 x 10^-3 M][4.20 x 10^-3 M]^3 = 3.75 x 10^8 4) Kc using Kp = Kc(RT)^n from Step 1) Kp/(RT^delta(n)) = 6.27 x 10^5/ (0.08206 x 298)^(-2) = 3.75 x 10^8

Describe the transitions between the listed points on the phase diagram of water below (in terms of forward and reverse rates and dynamic equilibrium): (Lines on the graph represent phase equilibria) 1. A -> B 2. B -> F 3. F -> D - What's the initial state - What changes from the beginning to the end of the reaction - How does the change affect the rates - What's the final state

1. A -> B (melting) vs freezing (change in T) - at point A, rate f > rate m -> solid phase is favored (vice versa) - As T ↑, KE ↑ to disrupt IMFs (rate of melting will increase as rate of freezing remains unchanged) 2. B -> F (evaporation) vs condensation (change in P) - At point B, rate c > rate e -> liquid phase is favored (vice versa) - As P ↓, rate of condensation ↓ b/c P(vap) ↓, while rate of evaporation remains unchanged - As rate of condensation decreases, with decreasing moles of gas 3. F -> D (deposition) vs sublimation (change in T) - At point F, rate d < rate s -> gas phase is favored (vice versa) - As T ↓, rate of sublimation ↓ while rate of d remains unchanged - less gas is made

Predict which substance would have the higher IMFs in either scenario. 1. Neon or Argon 2. alcohol or ether

1. Argon - bigger molecule more LDFs, more IMFs 2. alcohol - Hydrogen bonding

Vapor Pressure of Water at Fixed T Using your understanding of dynamic equilibrium, explain why P(vap) remained constant even when V increased.

1. Identify rates: rate(e) vs. rate(c) 2. Identify disturbance: Initially volume of gas increases, so the pressure of the gas decreases 3. Predict response: - As the pressure of the gas decreases, fewer particles are colliding with the liquid surface (rate of condensation decreases) - Rate of evaporation remains unchanged since T is constant - As rate(e) > rate(c), more liquid evaporates to produce more water vapor, thus increasing the vapor pressure, which simultaneously increases the rate of condensation until P(vap) is restored to its original state (rate of evaporation = rate of condensation) - Kp = P(vap) unchanged b/c T is constant

Interpreting a Vapor Pressure Curve Using your kinetic molecular understanding, predict and explain what will occur and if each of the following changes were made starting from the red dot (i.e. how does each change affect the rates of condensation and evaporation?): 1. T increases 2. T decreases 3. P increases 4. P decreases

1. T increases - As T ↑, KE ↑ (rate of evaporation increases) - gas exists at new T, P 2. T decreases - As T ↓, KE ↓ (rate of evaporation decreases) - liquid exists at new T, P 3. P increases - As P ↑, gas density ↑ (rate of condensation increases) - liquid exists at new T, P 4. P decreases - As P ↓, gas density ↓ (rate of condensation decreases) - gas exists at new T, P

H2 (g) + CO2 (g) = CO(g) + H2O(g) ΔHf°(CO) = -110.5 kJ/mol ΔHf°(CO2) = -393.5 kJ/mol ΔHf°(H2O) = -241.8 kJ/mol - Calculate ΔH°rxn - Given that Kp = 0.236 at 800 K, estimate Kp at 1200 K.

1. ΔH°rxn = (-241800 - 110500) - (393500) 2. van't Hoff equation Kp = 0.0163

Consider a liquid-vapor equilibrium. If the volume containing the vapor is increased, what will happen once equilibrium is restored? A) The vapor pressure will remain the same, since it depends only on the temperature. B) The vapor pressure will drop, since pressure is inversely proportional to volume. C) The vapor pressure will increase to fill the increased volume. D) The vapor will condense into the liquid, since the pressure is below the vapor pressure. E) The temperature will drop, since temperature is proportional to pressure.

A) The vapor pressure will remain the same since it depends only on the temperature. Why? - When you increase the volume, this and more molecules to evaporate from the liquid to a gaseous state - When the rate of evaporation begins exceeding the rate of condensation, gas particles will return to their liquid states (back and forth) to maintain equilibrium and the same vapor pressure

What are ways to increase product yield?

A) add reactant B) remove the product C) decrease volume D) increase temperature

If 0 = ΔG° + RT ln K, when would K be large? A. ΔG° < 0 B. ΔG° > 0 C. ΔG° = 0 D. High T E. Low T

A. ΔG° < 0 As one side get larger the other has to ger smaller to subside it "richer/poorer analogy"

DEMO- Concentration Effects Co(H2O)6^(2+) + 4Cl- = CoCl4^(2-) + 6H2O - Adding HCl - Adding H2O

Adding HCl - increasing reactants, increases products - Q < K (shift towards products) - blue color observed - Adding H2O - concentrations diluted - increasing products, increases reactant - Q > K (shifts towards reactants) - pink color observed numerator to denominator - when the numerator is greater than the denominator Q > K - when the numerator is less than the denominator Q < K - Think of Q like a numerator (products) - Think of K like the denominator (reactants)

RICE TABLES 2CH4(g) = C2H2(g) + 3H2(g) A vessel was filled with 3.2 atm of CH4 and then heated at a constant temperature to decompose to acetylene and hydrogen gas (above). After the mixture reaches equilibrium, it was determined that the partial pressure of CH4 is 1.2 atm. Determine the value of KP. A) 9.0 B) 19 C) 38 D) 17 E) 6.0

B) 19

COCl2(g) -> Cl2(g) + CO(g) Kp = 3.3 Assume you start with 1.0 atm partial pressures of all three gases. What will happen as we approach equilibrium? A) More COCl2 is produced B) More Cl2 is produced C) Cannot predict without more data What if we started with 2.0 atm of all three gases? What if we started with 4.0 atm of all three gases?

B) More Cl2 is produced Q = ((1)(1))/(1) = 1 Kp = 3.3 Q < K This favors the formation of products (like Cl2) As the reaction proceeds the reaction will proceed to the right 2<3.3 -> Q<K 4>3.3 -> Q>K More Cl2(g)+CO(g) are made in Scenario I; but more COCl2 is made in Scenario II

Heating Curve of water

B. Boiling B. Liquid

Temperature effects If the temperature of the water is increased, what will happen once equilibrium is restored? A. The vapor pressure will increase, since pressure is proportional to temperature. B. The rate of condensation will decrease because the gas molecules have higher kinetic energy. C. The vapor pressure will increase so that the rate of condensation matches the increased rate of evaporation . D. The vapor pressure will not change, because the gas and liquid return to equilibrium.

C. The vapor pressure will increase so that the rate of condensation matches the increased rate of evaporation. (Think about Le Chatelier's Principle) - The rate of evaporation increases initially - However, in response, the rate of condensation will increase until it matches the increased rate of evaporation

Vapor Pressure of Water at Fixed T Would changing n(vap) affect Pvap?

Changing n has no effect on Pvap at equilibrium 1. rate(e) vs rate(c) 2. - As n increases for liquid, there is no change to either rate - As n increases for gas, P(gas) increases as well as the rate of condensation, while the rate of evaporation stays the same (rate (e) < rate (c)), - As more gas condenses (turns to liquid), vapor pressure decreases, which will decrease the rate of condensation until rate(e) = rate (c)

Lecture 5

Free Energy and Thermodynamic Equilibrium, Reaction Equilibrium in the Gas Phase

The larger the ΔH associated with a phase transition is for a substance... A. The stronger the intermolecular forces are between the molecules in the substance. B. The more intermolecular forces there are between the molecules per unit volume of a substance. C. The longer the time it takes at a constant heating rate for the substance to change phases. D. All of the above.

D. All of the above

Consider Liquids A and B at a fixed temperature. If Pvap(A) > Pvap(B), the boiling point of A is _____ than that of B, which suggests that the intermolecular forces in A are _____ than those in B. A. Higher; stronger B. Higher; weaker C. Lower; stronger D. Lower; weaker

D. Lower; weaker - The vapor pressure of a liquid is determined by the tendency of its molecules to escape from the liquid phase and enter the gas (vapor) phase. -A higher vapor pressure indicates that more molecules have enough energy to overcome the IMFs and transition to the gas phase., weaker IMFs

Using kinetic molecular theory arguments, explain why: • Evaporation is an endothermic process. • Condensation is an exothermic process.

Evaporation is an endothermic process. - delta H>0 - in the liquid phase particle density is high -> high IMFs - In order to evaporate, energy is required to provide particles with enough KE (need to move further away from each other to no longer be attracted to one another) to overcome IMFs Condensation is an exothermic process. - delta H<0 - in the gas phase particle density of low -> low IMFs - In order to condense, energy is released when gas particle density increases, thus inter-particle distances decrease - This allows allowing for IMFs to overcome significantly

SCALAR QUESTIONS

In-class practice problems

Identify the types of IMFs experienced by: MgSO4, NH3, and O2, and discuss what factors would increase each type of IMF.

MgSO4: ionic NH3: dipole-dipole O2: LDFs

DEMO- Concentration Effects Co(H2O)6^(2+) + 4Cl- = CoCl4^(2-) + 6H2O Concentration and Rate Graphs

N graphs rxn progress - [Co(H2O)6^(2+)] concentration starts off at a higher point (straight line) - [CoCl4^(2-)] concentration starts off at zero (straight line) - [Co(H2O)6^(2+)] concentration begins to go down - [CoCl4^(2-)] concentration begins to go up - [Co(H2O)6^(2+)] concentration begins to go up now - [CoCl4^(2-)] concentration begins to down now rate graphs rxn progress - rates of both Co(H2O)6^(2+) , and CoCl4^(2-) start off at zero - Co(H2O)6^(2+) rate starts off at a higher point and begins going down - CoCl4^(2-) rate start off at zero and begins going up - they intersect at an overall lower position than the previous graph... - CoCl4^(2-) rate starts of at higher point and begins going down - Co(H2O)6^(2+) rate starts off at a lower point and begins going up - they intersect

If ∆G = RT ln (Q/K), What happens to ∆G when: Q < K Q > K Q = K

Q < K: ∆G = gets smaller Q > K: ∆G = gets larger Q = K: ∆G = 0 Q < K leads to ∆G < 0, favoring the forward reaction. Q > K leads to ∆G > 0, favoring the reverse reaction. Q = K leads to ∆G = 0, indicating equilibrium.

Lecture 2

RICE calculations; Reaction Quotient (Q),Q vs. K; Le Chatelie

Lecture 3

Review Le Chat; intermolecular forces, enthalpy of phase transitions; Vapor-liquid equilibrium; Clasius Clayperon

A CHEM 122 student was asked to discuss intermolecular forces and the enthalpy of transition from solid to liquid and responded with the following: "If a substance has polar bonds, it will have a large dipole moment. This means that the individual molecules will repel each other, and the melting point will be lower than a substance without dipoles. When the substances melt, their temperature does not change, so the change in enthalpy must be zero. Thus, intermolecular forces have no effect on the enthalpy of melting for a substance." Assess the accuracy of the student's response. For each sentence, briefly explain whether the reasoning presented is logical, noting what information is correct or incorrect and providing correct logical reasoning where needed.

S1: Incomplete; A substance with polar bonds MAY have a large dipole moment. For example, If the polar bonds are symmetrically distributed, their dipole can cancel out and have a net dipole of zero S2: Incorrect; If a substance has a large dipole, then the molecules will be attracted to one another, NOT repelled. Thus, the substance has large IMFs and will have a higher melting point than a nonpolar substance. S3: When a substance is melting, the T doesn't change. However, the process is endothermic, since heat must be absorbed during melting. Energy is required to disrupt the IMFs in the solid to make it melt. Enthalpy change > 0 S4: Incorrect; IMFs have a significant effect on the enthalpy of melting. Larger IMFs require more energy to break up, and so cause a higher enthalpy of melting.

This question refers to the hypothetical reaction shown below, for which the equilibrium constant is K=0.10. A(g) + 2B(g) ↔ D(g) A student was asked to predict the effect on the equilibrium partial pressures of all gases if the volume of the reaction container was halved. The student responded... "Halving the volume doubles all of the partial pressures of the gases. Therefore, at equilibrium, the reaction should be written as 2 A(g) + 4 B(g) ↔ 2D(g). The equilibrium constant for this reaction is K=0.01. Since K is smaller, the equilibrium must favor more reactants. Therefore, the equilibrium shifts to produce more reactants and less product." Assess the accuracy and logic of the student's response and graph:

S1: The first sentence is true. S2: Not chronologically logical, The balanced equation represents the ratios of the reactant and product molecules but does not represent the actual partial pressures. So changing of the partial pressures doesn't require rewriting of the reaction S3: True; value of K and its expression is dependent on how the reaction is written S4: Not a logical conclusion from the previous sentence Since both K and its expression are doubled at equilibrium the individual values are not changed it is just rather the square of K1. S5: Incorrect; When the partial pressures are increased, Q becomes less than K since there are more moles of reactants, so the reaction shifts to products

DEMO- Concentration Effects Co(H2O)6^(2+) + 4Cl- = CoCl4^(2-) + 6H2O When the solution is blue, Keq is ____ its value when the solution is pink. A. Larger than B. Less than C. The same as

The equilibrium constant (Keq) is determined solely by the temperature and the balanced chemical equation. It is independent of the concentration or color of the solution. The value of Keq remains constant regardless of the color of the solution. Therefore, when the solution is blue or pink, the equilibrium constant Keq for the reaction remains the same. C. The same as

Lecture 4

Vapor pressure curves; boiling points, phase diagrams, Entropy and the Second Law of Thermodynamics

2NO(g) = N2 (g) + O2 (g) T (K) 2000 2100 2200 2300 2400 Kp 2451 1458 909 592 398 Calculate ΔH°rxn

We are using the van't Hoff Eqaution 1. Plug in Values from chart into the equation ΔH°rxn = -Rln(K2/K1)/((1/T2)-(1/T1)) = -53319 J/mol

Vapor Pressure of Water at Fixed T What affects the rate of evaporation? What affects the rate of condensation?

What affects the rate of evaporation? - Temperature (which is proportional to KE) affects the rate of evaporation - need KE to overcome IMFs What affects the rate of condensation? - Particle density (which is proportional to pressure) affects the rate of condensation (collision frequency hitting liquid surface)

N2O4(g) = 2NO2(g) Kp(1) = 0.148 at 298 K What is Kp(3) for 2NO2(g) = N2O4(g) at 298 K? A) Kp(3) is unrelated to Kp(1). B) Kp(3) = Kp(1) C) Kp(3) = 1/Kp(1) D) Kp(3) = -Kp(1)

Write out the Kp expressions Kp(1) = [NO2]^2/[N2O4] Kp(3) = [N2O4]/ [NO2]^2 1. How do I obtain Kp(1) from Kp(3) 2. You take the inverse 3. Thus, Kp(3) = 1/Kp(1) C) Kp(3) = 1/Kp(1)

N2O4(g) = 2NO2(g) Kp(1) = 0.148 at 298 K What is Kp(2) for (1⁄2)N2O4 = NO2 at 298 K? A) Kp(2) is unrelated to Kp(1) B) Kp(2) = Kp(1) C) Kp(2) = 1⁄2Kp(1) D) Kp(2) = (Kp(1))^(1/2)

Write out the Kp expressions Kp(1) = [NO2]^2/[N2O4] Kp(2) = [NO2]/[N2O4]^(1/2) 1. How do I obtain Kp(2) from Kp(1) 2. square Kp(2) it to get Kp(1) 3. Thus, Kp(2) = (Kp(1))^(1/2) D) Kp(2) = (Kp(1))^(1/2)

* IMPORTANT FOR EXAM * PCl5(g) = PCl3(g) + Cl2(g) Kp = 1.86; T = 250 °C; V = 1.00 L; n(PCl5)i = 0.0129 mol - Write the expression of Kp - Calculate the partial pressures for all three gases at equilibrium

Write the expression of Kp 1. Kp = (PPCl3)(PCl2)e/(PPCl5) Kp = 1.48 Calculate the partial pressures for all three gases at equilibrium 1. Draw Rice Table - Using initial moles of PCl5, calculate for Initial Pressure of PCl5 (P = nRT/V) - (0.0129)(0.08206)(523.15) = 0.554 atm - To calculate C in the Rice table see the change in moles in terms of each individual reactant/product - For instance, moles of PCl3 = moles Cl2, therefore PCl5 change is -x and +x for PCl3 and Cl2 2. Calculate partial pressures using Kp expression at equilibrium x^2/(0.554 -x) = 1.48 Solve for x using the quadratic formula x = (-b +/- sqrt (b^2-4ac))/2a x= 0.447 atm 3. Plug into Equilibrium expressions for each of the components x = PPCl3 = PCl2 = 0.447 atm PPCl5 = 0.554-0.447 = 0.107 atm

2BrCl(soln) = Br2(soln) + Cl2(soln) Kc = 0.145 The above equilibrium takes place in CCl4 solution at 298 K. What are the equilibrium concentrations of the products and reactants, given the following initial concentrations? [BrCl] = 0.050 M [Br2] = 0.035 M [Cl2] = 0.035 M

[BrCl] = 0.0682 M [Br2] = 0.0259 M [Cl2] = 0.0259 M

RICE TABLES C2H4(g) + H2(g) = C2H6(g) Kc = 0.98 If 1.5 atm of C2H6 was trapped in a 2.0 L container at 298 K, what would the gas composition be (in M) after the reaction is allowed to reach dynamic equilibrium?

[C2H4]=[H2] = x = 0.0582 M [C2H6] = (0.0615 - x) = 0.0033 M

The following reaction is performed, and at equilibrium the solution is pink. Subsequently, concentrated hydrochloric acid is added to the solution, and the solution turns blue at equilibrium. When additional water is added, the solution turns pink again at equilibrium. Co(H2O)6^(2+) + 4Cl- = CoCl4^(2-) + 6H2O ΔH > 0 a) Assess the accuracy of the following line of reasoning explaining the changes in color: "When the solution is blue, [CoCl 42-] is higher than [Co(H2 O)62+]. Since Kc is equal to the concentration ratio of products to reactants, the KC of the blue solution is higher than the KC of the pink solution." b) If the solution starts out purple, what color would you expect to see if the following occurred: Add NaCl: Add KNO3 : Place flask in an ice bath: Increase pressure of the system:

a) S1: True S2: - Kc is equal to the equilibrium concentration ratios of products to reactants raised by their stoichiometric coefficients - Write out Kc expression - At constant T, The Kc of the solution will be the same regardless of the color of the solution - Difference in color is a result of the different concentrations of the Co complexes, however, the concentration of the chlorine in the expression serves to keep the Kc constant b) Add NaCl: blue Add KNO3: purple Place flask in an ice bath: endothermic reaction, decreasing temperature will slow down forward rate -> pink Increase pressure of the system: Purple -> no effect solution has no gases

For the reaction: A2 (g) + B2 (g) = 2AB(g), ∆G°= 2.6kJ/mol at 25°C. a) Calculate ∆G when the partial pressures are: 1.0 atm of AB, 3.0 atm of A2 , 3.0 atm of B2 b) Predict what will happen in the reaction.

a) Calculate ∆G when the partial pressures are: 1.0 atm of AB, 3.0 atm of A2 , 3.0 atm of B2 1. Solve for Q (1)^2 / (3)(3) = 1/9 = 0.1111 2. Plug into this equation -> ∆G = ∆G° + RT ln(Q) 3. Convert to kJ to J and solve 2,600 J/mol + (8.314 J/mol·K)(298.15 K) ln(0.1111) = -246.770 J/mol b) Predict what will happen in the reaction. delta G is negative meaning this is a spontaneous process (high to low in the forward direction)

Consider the reaction: N2(g) + 3H2(g) = 2NH3 (g), T = 298 K a) Predict the sign of S° for this reaction. b) For a favorable reaction, what must the sign of H° be? c) Given the following table of ΔS°, calculate ΔS°rxn: d) If ΔH°= -92.2 kJ/mol, calculate ΔG°rxn. Is this reaction favorable? e) Is the reaction spontaneous?

a) Entropy is is decreasing delta S = (-) because moles are decreasing b) delta H = (-) ΔG° = ΔH° - TΔS° (delta S° is less than zero so delta H° is less than zero) c)2ΔS°(NH3) - [ΔS°(N2) + 3ΔS°(H2)] 2(192.4) - 191.5 - 3(130.6) ΔS° = -198.5 J/mol K d) (-92200) - (298 x -198.5) = -33.0 kJ/mol ΔG° < 0 (favorable) e) Not able to determine without more data

4NH3(g) + 7O2(g) = 4NO2(g) + 6H2O(g) ΔH° = +73 kJ In which direction will the equilibrium shift when: a) NH3 is removed? b) the temperature increases? c) the volume increases? d) NO2 is added? e) A catalyst is added? f) Gaseous Ar is added? - For each of the above, explain your reasoning using your understanding of reaction rates. - List ways to increase the product yield for the reaction.

a) NH3 is removed: According to Le Chatelier's principle, if NH3 is removed from the system, Q > K (reaction shifts towards the reactants) b) the temperature increases? Q < K (reaction shifts towards the products) c) the volume increases Q > K (reaction shifts towards the reactants) - reduces both rates, but reduces forward rate since there are more moles to react d) Q > K (reaction shifts towards reactants) e) catalysts do not appear in equilibrium expressions, because forward and reverse rates are equally affected f) has no effect on reaction rate

Using Le Châtelier's principle, predict whether the specified "stress" will produce an increase or a decrease in the amount of product observed at equilibrium for the following reaction: 2H2 (g) + CO(g) ⇌ CH3 OH(g) ΔH° = -91 kJ/mol a) Volume of container is increased. b) Helium is added to container. c) Temperature of container is raised. d) Hydrogen is added to container. e) CH3 OH is extracted from container as it is formed.

a) Volume of container is increased. shifts to reactants to increase total pressure because it has more moles b) Helium is added to container. no effect c) Temperature of container is raised. shifts to reactants d) Hydrogen is added to container. increase the amount of products Q<K e) CH3 OH is extracted from container as it is formed. increase the amount of products Q<K

Draw a graph of G versus Q where the reaction has K>1. a) For which range of Q is ΔG<0? How does this compare with K? What happens in this range? b) For which range of Q is ΔG>0? How does this compare with K? What happens in this range? c) When is ΔG=0? What happens at this point? d) What is the sign of ΔG°, e.g. the value of ΔG when Q=1? e) What does the sign of ΔG° tell you? Draw a graph of G versus Q where the reaction has K<1. Repeat the above questions!

a) ΔG < 0 is when Q < K (favors the formation of products) b) ΔG > 0 is when Q > K (favors the formation of reactants) c) When is ΔG=0? What happens at this point? Q = K - equilibrium d) What is the sign of ΔG°, e.g. the value of ΔG when Q=1? = 0 e) The sign tells you whether or not the reaction is favorable

Match each substance to its vapor pressure curve (EXPLAIN)

i -> C (LDS, H bonding) ii -> B (H bonding, Low LDFs) iii -> A Boiling point increases with increasing IMFs

Consider the following reaction: N2O4 (g) ⇌ 2NO2 (g) For the reaction, the number of moles of NO2 at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms of dynamic equilibrium, and give a reason why the rates of the forward and reverse reactions might be affected differently by changes in the volume.

rate of reverse reaction depends on volume because ↑V, ↓collisions of NO2, ↓ rate of reaction rate of forward reaction is not affected since it is only dependent on number of N2O4 molecules Thus, r(forward) > r(reverse) The reaction will shift to ↑ moles of NO2 until the reaction reaches equilibrium (it will shift to products because it has more moles of gas and will thus increase total pressure)

Can H2O(l) = H2O(g) achieve dynamic equilibrium below the boiling point? ΔH°= +44 kJ/mol ΔS= 118.9 J mol-1K-1 At 298K, ΔG°= ? What does this tell us? In this "reaction,"Q = P(H20) What if P(H2O) = 7.6 torr = 0.010 atm What if P(H2O) = 76 torr = 0.10 atm At what P(H2O) would there be an equilibrium? ΔG= 0

ΔG° ≈ 8,618.8 J delta G is positive meaning that the conversion from liquid to gas isn't favorable at equilibrium P at equilibrium ΔG = ΔG° + RTln(Q) At equilibrium ΔG = 0 ΔG° + RTln(P(H2O)) = 0 P(H2O) = 0.038 atm

What is ΔG° of N2O4 (g) = 2NO2 (g) at 25 °C? ΔS°(N2O4 ) = 304.4 J mol-1K-1 ΔS°(NO2 ) = 240.1 J mol-1K -1 ΔHf°(N2O4 ) = 11.1 kJ mol-1 ΔHf°(NO2 ) = 33.2 kJ mol-1 ΔH° rxn ΔS° rxn ΔS° surrounding ΔS° universe ΔG° rxn What prediction can you make from this calculation?

ΔH° rxn = (2(33.2)) - (11.1) = 55,300 J/ mol ΔS° rxn = (2(240.1)) - (304.4) = 175.8 J/mol*K ΔS° surrounding = 55,300 J/ mol/ 298.15 K = -185.22 J/mol*K ΔS° universe = 175.8 - 185.22 = -9.42 J/mol*K ΔG° rxn = ΔH° rxn - TΔS° rxn = 2,573 J mol^(-1) -> not spontaneous, this reaction isn't favorable so there is no question of favorability ΔG° > 0 The equilibrium lies more towards the side of the reactant, not favorable in the forward direction