Calculating Free Energy: Standard Enthalpy and Entropy at 32C

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The discussion focuses on calculating Gibbs free energy for a reaction with a standard enthalpy of -196 kJ/mol and standard entropy of 151 J/K mol at 32°C. The formula used is G = H - TS, leading to a calculation of G = -196 kJ/mol - [-305.15 K * 0.15] = -242 kJ. Participants clarify the conversion of entropy from joules to kilojoules, emphasizing the importance of consistent units in calculations. Misunderstandings about unit conversion were addressed, confirming that the conversion was correctly applied. The conversation highlights the significance of accurate unit handling in thermodynamic calculations.
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Calculate the gibbs free energy for a reaction which has a standard enthalpy = -196kJ/mol and standard entropy = 151 J/K mol at 32C.

G = -196 KJ/MOL -[-305.15 K*0.15] = -242KJ

Thank you.
 
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Isn't that what I did?
 
You're adding kiloJoules to Joules...
 
dav2008 said:
You're adding kiloJoules to Joules...

No, I think he did not. Because he converted 151 J to 0.15 kJ:

Soaring Crane said:
G = -196 KJ/MOL -[-305.15 K*0.15] = -242KJ
 
PPonte said:
No, I think he did not. Because he converted 151 J to 0.15 kJ:
Oops didn't notice that.
 
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