Thermodynamic and Electrochemistry

All out of Energy

100

How does the molar entropy of a substance change with increasing temperature?

The molar entropy of a substance increases with increasing temperature.

200

Calculate ∆G° for the reaction                             SO₂(g) +2H₂(g) → S (s)+ 2H₂O (g)

∆G°(SO₂(g)) =-300 kJ/mol  , ∆G°(H₂O (g)) = -230 kJ/mol

-160 kJ/mol

300

For the reaction at 298K 

2NO₂(g) ⇌ N₂O₄ (g) , ∆H°=-58.03 kJ, ∆S°=-176.6 J/K

What is the value of ∆G° at 298K?

At what temperature will the reach have equilibrium?

∆G°=-5.43 kJ

T=328.6K

400

Calculate ∆G° for H₂O(g) +0.5O₂(g) ⇌H₂O₂(g) at 600K

use 

H₂(g) +O₂(g) ⇌H₂O₂(g) K=2.3 x 10⁶

2H₂(g) +O₂(g) ⇌ 2H₂O(g) K = 1.8 x 10³⁷

∆G° = 140.89 kJ/mol

500

Calculate ∆G° for the following reaction:

6C(s) + 3H₂(g)  →C₆H₆(l)

given that 

2C₆H₆(l) +15O₂→ 12CO₂(g) +6H₂O(l) ∆G°=-6399 kJ

C(s) + O₂(g) → CO₂(g)                  ∆G° = -394 kJ

H₂(g) + 0.5O₂(g) → H₂O (l)             ∆G° = -237 kJ

 ∆G° = 124 kJ

Process

100

The value of ∆G° for A →B is greater than zero. What can be concluded for  B → A ?

It will be spontaneous, since ∆G° will be negative.

200

Will the following be spontaneous at constant T and P?

A) ∆H = +25 kJ, ∆S = 5.0 J/K , T=300k 

B) ∆H=25 kJ , ∆S=100 J/K , T=300K

A) No

B) Yes

300

At which temperature will the following process be spontaneous?

A) ∆H= -18 kJ and ∆S = -60 J/K

B) ∆H = 18 kJ and ∆S = 60 J/K

C) ∆H= 18 kJ and ∆S = -60 J/k

D) ∆H= -18 kJ and ∆S= 60 J/K

A) Spontaneous less than 300K

B) Spontaneous above 300K

C) Non spontaneous at all temperatures, -300K does not exist.

D) Spontaneous at all temperatures

400

Which process(es) are driven by an increase in the entropy of the surroundings?

A) condensation of water

B)the sublimation of dry ice

C) the freezing of water

A) condensation of water,

C) the Freezing of water

500

Which process is spontaneous at 298K.

∆G°_rxn=8.5

A) H₂O (l) → H₂O(g, 1 atm)

B) H₂O (l) → H₂O(g, 0.10 atm)

C) H₂O (l) → H₂O(g, 0.010 atm)

C) H₂O (l) → H₂O(g, 0.010 atm)

Electric Shock

100

What is the substance being oxidized and reduced? 

2Ag(s)+H₂S⟶Ag₂S(g)+H₂(g)

Oxidized: Ag

Reduced: H

200

Cr³⁺ +Cl₂ → Cr₂O₇^-2 +Cl^-

What is the E° for this reaction?

Use 

Cr₂O₇^-2+ 14H⁺ +6e^- → 2Cr³⁺ +7H₂O E°= 1.33

Cl₂ +2e^-  → 2Cl^- E° = 1.36

E°_cell=0.03

300

Is this reaction spontaneous?

MnO₄^- + I^- → I₂ + Mn²⁺

Use

MnO₄^- +8H⁺ +5e^- → Mn²⁺ +4H₂O E=1.51 V

2I^-→ I₂ +2e^- E=-0.54

E_cell = 0.97 V, yes this is spontaneous

400

Give the complete reaction and determine E_cell

use

Mn²⁺ +2e^- → Mn E=-1.18 V

Fe³⁺ + 3e^- → Fe  E= -0.036

3Mn + 2Fe³⁺ → 2Fe +3Mn²⁺

E_cell=3.468

500

A galvanic cell is base on these half reactions at 25 Celsius. Predict E given the concentrations, [Ag⁺] =1.00 M, 

[H₂O₂] = 2.0 M, [H⁺] = 2.0 M

Ag⁺ + e^-→Ag E=0.80 v

H₂O₂ + 2H⁺ + 2e^- → 2H₂O E=1.78 V

E = 1.01 V approximately

My friend K

100

What can be concluded about the equilibrium constant (K) for a reaction if ∆G° is negative?

K is greater than 1.

200

Ni²⁺(aq) + Zn(s) ⇌ Zn²⁺ (aq) +Ni(s)

Initial concetrations of [Ni²⁺] = 1.50 M, [Zn²⁺ ] = 0.100M. What is initial cell potential?

0.56 V

300

Zn²⁺ +2e^- → Zn E = -0.76

Fe²⁺ + 2e^- → Fe E= -0.44

Calculate E°_cell, ∆G°, and K, at 25 Celsius

E°_cell = 0.32 v

∆G° = -61750.4 J = -61.75 kJ

K = 6.7 x 10¹⁰

400

2SO₂(g) +O₂(g)  → 2SO₃(g) 

At equilibrium partial pressures are pO₂ =0.50 atm, pSO₃= 2.0. Determine partial pressure of SO₂.

∆G°(SO₂(g)) =-300 kJ, ∆G°(SO₃(g))= -371 kJ

pSO₂= 10^-12 atm

500

Hgb + O₂ →  HgbO₂ ∆G° =-70 kJ

Hgb + CO → HgbCO  ∆G°= -80kJ

Estimate the equilibrium constant at 25 Celsius for: 

HgbO₂ + CO  ⇌ HgbCO + O₂ 

 K= 56.60

Entropically Enthused

100

For which will entropy decrease the most?

Zn(s) +Cu²⁺ (aq) ⇌ Zn²⁺(aq) + Cu(s)

H₂CO₃(aq) ⇌ H₂O (l) + CO₂(g) 

Fe³⁺ (aq) +H₂O (l) ⇌ Fe(OH)²⁺(aq) + H⁺(aq)

AgCl(s) +Br^-(aq) ⇌ AgBr(s) +Cl^-(aq)

SO₃(g) +H₂O(l) ⇌ H⁺(aq) +HSO₄^-(aq)

SO₃(g) +H₂O(l) ⇌ H⁺(aq) +HSO₄^-(aq)

200

Which reaction will have entropy increase the most?

Solid ⇌ Liquid

Gas ⇌ Liquid

2H₂ (g) + O₂(g) ⇌ 2H₂O (g)

HCl(g) +NH₃(g) ⇌ NH₄Cl(s)

2SO₂(g) + O₂(g) ⇌2SO₃(g)

Solid ⇌ Liquid

300

C₂H₄ (g)+ H₂O (g)→CH₃CH₂OH(l)

C₂H₆(g)+ H₂O (g)→CH₃CH₂OH(l) +H₂(g)

Which would be more thermodynamically feasible ?

C₂H₄ (g)+ H₂O (g)→CH₃CH₂OH(l)

400

Predict sign of entropy

A) Si₆(aq) +H₂ (g) →  2HF (g) +SiF₄(g)

B) 4Al(s) +3O₂(g) → 2Al₂O₃(s)

A) +

B) -

500

Determine S° for Fe(CO)₅(g) given:

Fe(s)+ 5CO(g) →  Fe(CO)₅(g) ∆S°= ?

 Fe(CO)₅(l)  → Fe(CO)₅(g)     ∆S°=107 J/K

Fe(s) + 5CO (g) → Fe(CO)₅(l)  ∆S°=-677 J/K

 S°(CO(g)) =198 J/K,   S(Fe(s))= 27 J/K

 S°(Fe(CO)₅(g))=447 J/K