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

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?

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.