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For the reaction A + B ⇌ C + D, if ΔH = -150 kJ and ΔS = -200 J/K, what is the t

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Question: For the reaction A + B ⇌ C + D, if ΔH = -150 kJ and ΔS = -200 J/K, what is the temperature at which the reaction becomes non-spontaneous?

Options:

  1. 750 K
  2. 750 °C
  3. 300 K
  4. 150 K

Correct Answer: 750 K

Solution: 

At equilibrium, ΔG = 0; ΔG = ΔH - TΔS = 0; T = ΔH/ΔS = 150,000 J / 200 J/K = 750 K.

For the reaction A + B ⇌ C + D, if ΔH = -150 kJ and ΔS = -200 J/K, what is the t

Practice Questions

Q1
For the reaction A + B ⇌ C + D, if ΔH = -150 kJ and ΔS = -200 J/K, what is the temperature at which the reaction becomes non-spontaneous?
  1. 750 K
  2. 750 °C
  3. 300 K
  4. 150 K

Questions & Step-by-Step Solutions

For the reaction A + B ⇌ C + D, if ΔH = -150 kJ and ΔS = -200 J/K, what is the temperature at which the reaction becomes non-spontaneous?
  • Step 1: Understand the reaction and the given values. We have a reaction A + B ⇌ C + D with ΔH = -150 kJ and ΔS = -200 J/K.
  • Step 2: Convert ΔH from kJ to J. Since 1 kJ = 1000 J, we have ΔH = -150 kJ = -150,000 J.
  • Step 3: Write the equation for Gibbs free energy (ΔG): ΔG = ΔH - TΔS.
  • Step 4: Set ΔG to 0 because we want to find the temperature at which the reaction becomes non-spontaneous (equilibrium). So, we have 0 = ΔH - TΔS.
  • Step 5: Rearrange the equation to solve for T: T = ΔH / ΔS.
  • Step 6: Substitute the values of ΔH and ΔS into the equation: T = -150,000 J / -200 J/K.
  • Step 7: Calculate T: T = 150,000 J / 200 J/K = 750 K.
  • Step 8: Conclude that the temperature at which the reaction becomes non-spontaneous is 750 K.
  • Gibbs Free Energy – Understanding the relationship between enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG) to determine spontaneity of a reaction.
  • Equilibrium Conditions – Recognizing that a reaction is non-spontaneous when ΔG is greater than zero, which occurs at equilibrium.
  • Temperature Dependence – Calculating the temperature at which the reaction shifts from spontaneous to non-spontaneous based on ΔH and ΔS.
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