Answer: ΔH − TΔS Gibbs free energy: ΔG = ΔH − TΔS at constant T, P.

A spontaneous process at constant T, P has

Answer: ΔG < 0 Negative ΔG means the process can proceed without external input.

What does the marked crossover temperature on the ΔG vs T plot represent?

Answer: The temperature at which ΔG = 0; reaction becomes spontaneous above it ΔG = ΔH - TΔS = 0 when T = ΔH/ΔS; above this T, with ΔH0 and ΔS0, ΔG becomes negative and the reaction is spontaneous.

AP Chemistry · Topic 9.3

Gibbs Free Energy and Thermodynamic Favorability Practice

Q: A spontaneous process at constant T, P has

Answer: ΔG < 0 Negative ΔG means the process can proceed without external input.

Q: What does the marked crossover temperature on the ΔG vs T plot represent?

Answer: The temperature at which ΔG = 0; reaction becomes spontaneous above it ΔG = ΔH - TΔS = 0 when T = ΔH/ΔS; above this T, with ΔH0 and ΔS0, ΔG becomes negative and the reaction is spontaneous.

Part of Applications of Thermodynamics.(ENE-4.C)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$ΔG =$
- A
  $ΔH\cdotTΔS$
- B
  $ΔH + TΔS$
- C
  $TΔS - ΔH$
- D
  $ΔH - TΔS$
  check_circle
Why
$Gibbs free energy: ΔG = ΔH - TΔS at constant T, P.$
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Sample 2difficulty 2/5
$A spontaneous process at constant T, P has$
- A
  $ΔS = 0$
- B
  $ΔG > 0$
- C
  $ΔG < 0$
  check_circle
- D
  $ΔG = 0$
Why
$Negative ΔG means the process can proceed without external input.$
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Sample 3difficulty 3/5
$ΔH > 0 and ΔS > 0 for the reaction.$
$What does the marked crossover temperature on the ΔG vs T plot represent?$
- A
  $The boiling point of the reactants$
- B
  $The temperature at which ΔG = 0; reaction becomes spontaneous above it$
  check_circle
- C
  $The temperature where ΔH = 0$
- D
  $The point of maximum entropy$
Why
$ΔG = ΔH - TΔS = 0 when T = ΔH/ΔS; above this T, with ΔH>0 and ΔS>0, ΔG becomes negative and the reaction is spontaneous.$
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Sample 4difficulty 3/5
$"Exergonic" means$
- A
  $ΔS > 0$
- B
  $ΔH < 0$
- C
  $ΔG < 0$
  check_circle
- D
  $ΔG > 0$
Why
$Exergonic = releases free energy (ΔG < 0); not the same as exothermic (ΔH < 0).$
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Sample 5difficulty 3/5
$The ΔG vs T plot shown is for a reaction with ΔH<0 and ΔS<0.$
$For a reaction with ΔH < 0 and ΔS < 0, the ΔG vs T line slopes upward. What does this imply?$
- A
  $Always spontaneous$
- B
  $Spontaneous at high T only$
- C
  $Spontaneous at low T; non-spontaneous at high T$
  check_circle
- D
  $Always non-spontaneous$
Why
$ΔG = ΔH - TΔS. With ΔH<0 and ΔS<0, the -TΔS term grows positive with T, so ΔG eventually exceeds 0.$
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