Entropy of Formation Calculation

In summary, the conversation revolved around calculating delta S0 using equations for delta G0 and delta H0. The calculations seem accurate, but it is important to double check units and significant figures. The small value for delta S0 may be due to the reaction not being very spontaneous at room temperature.
  • #1
ClaireA88
2
0

Homework Statement



Hi everyone. So I've already worked out delta H to be 57.88 kJ, T is 298K and I've been given a value of 29.94 for lnK0. Now I need to calculate delta S but think my numbers are too small to be doing it right...?

Homework Equations



1.) delta G0 = -RTlnK0
2.) delta G0 = delta H0 - T delta S0

The Attempt at a Solution



From equation 1 I've calculated delta G0 to be:

delta G 0 = -8.314 x 298 x 29.94

delta G0 = - 74176.03 J
= -74.18 kJ

I've then rearranged equation 2 to get:

- delta S0 = delta G0 - delta H0/ -T

= -74.18 - - 57.88/-298
= 0.054

Surely this number is too small though? Am I going wrong somewhere?
 
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  • #2


it is important to carefully review all calculations and make sure they are accurate. In this case, it seems like the calculations are correct, but the units may be causing confusion. The value for delta G0 should be in units of kJ/mol, not just kJ. This means that the value for delta S0 will also be in units of J/mol*K.

To double check, you can convert the units for delta G0 to kJ/mol by dividing by Avogadro's number (6.022 x 10^23) and see if the resulting value for delta S0 is more reasonable. It is also important to pay attention to significant figures in calculations, as they can affect the final result.

If the units and significant figures are correct, then the value for delta S0 may indeed be small. This could be due to the fact that the reaction is not very spontaneous at room temperature, as delta S0 represents the change in entropy at standard conditions. It is always important to consider the context of the values and make sure they make sense in the given situation.
 

1. What is entropy of formation?

Entropy of formation is a measure of the disorder or randomness of a system at a specific temperature and pressure. It is the change in entropy when a compound is formed from its constituent elements in their standard states.

2. How is entropy of formation calculated?

Entropy of formation is calculated using the standard molar entropies of the elements and the stoichiometric coefficients of the compound in the balanced chemical equation. The calculation involves subtracting the sum of the standard molar entropies of the reactants from the sum of the standard molar entropies of the products.

3. What is the significance of entropy of formation?

Entropy of formation is important in understanding the thermodynamic stability of a compound. A positive value indicates that the compound is more stable than its constituent elements, while a negative value indicates the opposite. It also provides insight into the feasibility of a chemical reaction.

4. How does temperature affect entropy of formation?

The entropy of formation is dependent on temperature. As temperature increases, the entropy of formation also increases. This means that at higher temperatures, compounds are more likely to form and be more stable due to increased disorder or randomness in the system.

5. Can entropy of formation be negative?

Yes, entropy of formation can be negative. This indicates that the compound is less stable than its constituent elements and is not favored to form. However, a negative value does not necessarily mean that the reaction will not occur, as other factors such as enthalpy of formation and Gibbs free energy must also be considered.

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