Why is entropy negative when it should be positive? Very large percent error.
Explain why (delta)G is reversed.
I received this directions from my teacher.
 You are calculating deltaH through the lab. The equation should be written as a ionization of your solid into aqueous ions. Water is not included.
 You need to look up the deltaG of formation for your solid, reverse the sign. You may check my bible if you like for the value.
 Using the Gibbs-Helmholtz equation, your answer to #2 above = calulated deltaH from the lab minus (T)deltaS. Use standard Temperature for your calculation or 25 celcius.
 Calculate deltaS.
 Now you have what the lab is looking for. Many of you have expressed that the answer to this question does not follow or is not what you expect based on what you have learned. I (we) understand that. Therefore, with a little research on your groups part, you will find why and can better explain your results. Farmer and I found online for example exactly what can help explain the results to help you with one search
G of formation is -363 - reverse is 363KJ/mol.
H of formation is 11.88 (accepted is 15).
Final temperature is 292.8K
363 = 11.88 - (292.8)(DS)
Accepted value for DS = 74J/mol.
My friends are getting the same error.
Can anyone explain what's missing or why DS is negative?
1 Answer | Add Yours
I realize I am not following the format for a good Editor, however your question is very specific to a unique condition.
It would help if you included the units for H. Is it in joules of kilojoules? Or is it measured in joules/g? In which case you need to multiply by the mass of your solid.
If it is in kj, then your calculation for Delta S would give
39.3 for your value of H, and 50 is you use the acceted value of 15 for H.
Check your table of values carefully for the correct units of H.
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