Is My Delta S Calculation Too Small for Entropy of Formation?

[ClaireA88]

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?

 

[Jhero]

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.