Exam 3 Practice Questions

Chapter 16: Thermodynamics

1. ΔS = Change in entropy

2. ΔH = Change in Enthalpy

3. ΔG = Change in Free Energy

Free energy in action

Question 1

At what temperature (K) will a reaction become nonspontaneous when ΔH = +50.20 kJ mol^-1 and ΔS = +20.50 J K^-1 mol^-1?

• > 1200K

• < 1200 K

• > 2448 K

• < 2448 K

Answer

Solve for when ΔG is 0

ΔG = 0 = ΔH – TΔS

So T = (ΔH/ΔS)

T = 50.20 kJ mol^-1)/(0.02050 kJ K^-1 mol^-1)

T = 2448 K when ΔG is 0 and when T < 2448 K reaction becomes non-spontaneous (ΔG becomes positive)

• < 2448 K

Question 2

For the reaction 2NO(g) + O­2(g) —> 2NO₂(g),

ΔH° = -113.1 kJ/mol and ΔS° = -145.3 J/K mol.

Which of these statements is true?

• The reaction is spontaneous at all temperatures.

• The reaction is only spontaneous at low temperatures.

• The reaction is only spontaneous at high temperatures.

• The reaction is at equilibrium at 25°C under standard conditions.

Answer

• The reaction is only spontaneous at low temperatures.

Calculation of Entropy Changes

Example 3

Calculate ΔS° for the reduction of aluminum oxide by hydrogen gas

Al₂O₃(s) + 3H₂(g) → 2Al(s) + 3H₂O(g)

Answer

ΔS° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 J/K )

ΔS° = 179.9 J/K

Practice Question

What is the entropy change for the following reaction?

Ag⁺(aq) + Cl^-(aq) → AgCl(s)

S° 72.68 56.5 96.2 JK^-1 mol^-1

• +32.88 JK^-1 mol^-1

• –32.88 JK^-1 mol^-1

• –32.88 Jmol^-1

• +112.38 JK^-1 mol^-1

Answer

ΔS ° = [96.2 – (72.68 + 56.5)] JK^-1 mol^-1 = –32.9 JK^-1 mol^-1

• –32.88 JK^-1 mol^-1

Gibbs Free Energy under Standard Conditions

Standard conditions = 1 atm and 25°C (298K)

Example 5

Calculate ΔG° for the reaction below at 1 atm and 25 °C, and determine if it is spontaneous under standard conditions, given ΔH° = –246.1 kJ/mol, ΔS° = 377.1 J/(mol K).
H₂C₂O₄(s) + ½O₂(g) → 2CO₂(g) + H₂O(l)

Answer

ΔG°= ΔH – TΔS

ΔG°= (–246.1 – 112.4) kJ/mol

ΔG°= –358.5 kJ/mol, yes spontaneous

Practice 1

Calculate ΔG° for the following reaction, H2O2(l) → H2O(l) + O2(g)

ΔH°= –196.8 kJ mol^-1 and ΔS °= +125.72 JK^-1 mol^-1.

• –234.3 kJ mol^-1

• +234.3 kJ mol^-1

• 199.9 kJ mol^-1

• 3.7 × 105 kJ mol^-1

Answer

ΔG ° = –196.8 kJ mol^-1 – 298K (0.12572 kJ K^-1 mol^-1)

ΔG ° = –234.3 kJ mol

• –234.3 kJ mol^-1

Standard Free Energy Changes

ΔG⁰_rxn = standard free-energy of reaction

Practice Questions

Calculate the standard free-energy changes for the following reactions at 25°C.

1. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
   ΔG°rxn = [Sum of ΔG°f of Product] - [Sum of ΔG°f of Reactant] = [ΔG°f (CO2) + 2ΔG°f (H2O)] - [ΔG°f (CH4) + 2ΔG°f (O2)]

   We insert the appropriate values from Appendix 3:

   ΔG°rxn =[(-394.4 kJ/mol) + (2)(-237.2 kJ/mol)] - [(-50.8 kJ/mol) + (2) (0 kJ/mol)]

   = -818.0 kJ/mol

2. 2MgO(s) → 2Mg(s) + O2(g)

   ΔG°rxn = [Sum of ΔG°f of Product] - [Sum of ΔG°f of Reactant] = [2ΔG°f (Mg) + ΔG°f (O2)] - [2ΔG°f (MgO)]

   From data in Appendix 3 we write:

   ΔG°rxn = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-569.6 kJ/mol)]

   = 1139 kJ/mol

Gibbs Free Energy at Non-Standard vs. Equilibrium Conditions

• non-standard (away from equilibrium): ΔG = ΔG⁰ + RT * lnQ

• At equilibrium: ΔG⁰ = - RT lnK —> K = e^(ΔG°/RT)

  
  R is the gas constant (8.314 J/K•mol)

  T is the absolute temperature (K)

  Q is the reaction quotient

Example 7

Calculate ΔG at 298 K for the Haber process

N2(g) + 3H2(g) <-> 2NH3(g)

ΔG° = –33.3 kJ

For a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2 and 0.5 atm NH3

Answer

Step 1: Calculate Q

Step 2: Calculate ΔG = ΔG° + RT*lnQ

Iclicker Question

Calculate the equilibrium constant for the decomposition of water at 298K given ΔG° = 474.4 kJ mol.

A. 1.47 × 10^-83

B. 1.21 x 10^0

C. 7 × 10^-84

D. 1.17 × 10^41

Answer

C. 7 × 10^-84

ATP as energy currency of cell

Iclicker Question

Answer

A. Cells often convert some of the energy from ATP hydrolysis into other usable forms.

Chapter 17: Electrochemistry

Question 1

The cathode in a galvanic cell has which type of polarity?

• A. Negative

• B. Positive

• C. Neutral

Answer

• B. Positive

Question 2

Which of the following species would be appropriate for a salt bridge solution?

• A. AgCl

• B. C12H22O11 sucrose (sugar)

• C. NaCl

• D. C6H6

Answer

The solution needs to be an electrolyte. AgCl is not soluble and the organic compounds are not ionic.

• C. NaCl