Thermodynamics II Flashcards

1
Q

Gibbs free energy

A

G = H – TS therefore ∆G = ∆H - T ∆S

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2
Q

What does each letter mean?

A

G = Gibbs Free Energy (kJ mol-1)
H = Enthalpy (kJ mol-1)
S = Entropy (J K-1 mol-1)
T = Temperature (K)

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3
Q

Gibbs Free Energy can inform us if the reaction is_________ or not (Will it occur of it’s own accord)

A

spontaneous

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4
Q

∆G negative:

A

Spontaneous in Forward direction only (irreversible)

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5
Q

∆G zero:

A

Reaction at equilibrium, can proceed in either direction (reversible)

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6
Q

∆G positive:

A

Non-spontaneous, will not proceed in the forward direction unless coupled with an energetically favourable reaction.

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7
Q

Coupled reactions:

A

Other reaction must have a larger, negative ∆G value. Sometimes called tandem reactions.

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8
Q

T is what units?

A

Kelvin

C > K = +273(.15)

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9
Q

Endothermic +∆H / -∆S, spon?

A

+∆G always
Spontaneous = Never

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10
Q

Endothermic +∆H / +∆S, spon?

A

∆G + low T, - high T
Spontaneous = on heating

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11
Q

Exothemic -∆H / +∆S, spon?

A

-∆G always
Spon = Always

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12
Q

Exothermic -∆H / -∆S, spon?

A

+∆G (high T), - (low T)
Spon = on cooling

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13
Q

Free Energy of Formation (∆Gf⁰)

A

The free energy change which results from 1 mol of substance prepared from its elements at standard pressure (1 atm) and a given temperature (usually 298 K). The units are kJ/mol

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14
Q

The ∆Gf⁰ of chemical elements is always ____

A

0 = zero

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15
Q

What does the little 0 with a line through mean?

A

In standard conditions

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16
Q

Oxidative Metabolism (cellular respiration)

A

Process by which nutrients from our diet are converted into energy

17
Q

There are several intermediate steps

A
  • Glycolysis + Kreb’s cycle - but overall reaction for cellular oxidation of glucose is:

∆G = ∆H - T ∆S

18
Q

Glucose + 6O2 —> 6CO2 + 6H2O
Is this process endo/ excothermic?

A

Exothermic

19
Q

Glucose + 6O2 —> 6CO2 + 6H2O
Increase or decrease in entropy?

A

Increase

Goes from 7 molecules (gas and solid) to 12 (liquid and gas molecules)

20
Q

Glucose + 6O2 —> 6CO2 + 6H2O
Spontaneous reaction?

A

∆G = ((6 x -394) + (6 x 2370) – (-917.2) + 0)
∆G = -2871.4 kJ/mol

-∆H +∆S
∆G will always be -ive