Entropy P2 help

[Scienuk]

For part (c) I got 335 J K^- mol^- which is correct according to the mark scheme, however when I tried to see if I get the same answer when using the standard entropy values (products minus reactants) provided in the table I get 70.8 J K^-1 mol^- which is a completely different answer, why is that so?

Edited August 6, 2016 by Scienuk

[kw0573]

In IB, delta G = delta H - T * delta S is used most of the time. G = H - T*S is seldomly used. delta values (G and S) are specified in relation to the values of the reaction that forms carbon monoxide/methanol from elements. But the given entropy values are in reference to 0K. When each variable is specified by different references, the equation delta G = delta H - T*S cannot hold true. So if you try to plug in given values, the equation will not hold for either species. To find the standard entropy change, which is a difference of the standard entropies of formation of carbon monoxide and methanol, we need delta G_f, and delta H_f.

[Msj Chem]

When using the equation products - reactants, are you adding the value for the H2? I found a value online of 131 JK-1mol-1. This gives you (126.8) - (131x2 + 197) = - 333 JK-1mol-1 which is pretty close to the value you calculated. 

[Scienuk]

7 minutes ago, Msj Chem said:

When using the equation products - reactants, are you adding the value for the H2? I found a value online of 131 JK-1mol-1. This gives you (126.8) - (131x2 + 197) = - 333 JK-1mol-1 which is pretty close to the value you calculated. 

Forgot to do that, thanks.

[kw0573]

1 hour ago, Msj Chem said:

When using the equation products - reactants, are you adding the value for the H2? I found a value online of 131 JK-1mol-1. This gives you (126.8) - (131x2 + 197) = - 333 JK-1mol-1 which is pretty close to the value you calculated. 

Right yeah. So if we are using absolute entropy to calculate entropy change, as what the question gives, then we need to consider H2. But if we use standard entropies of formation, then that of H2 is 0

 

[Msj Chem]

6 hours ago, kw0573 said:

Right yeah. So if we are using absolute entropy to calculate entropy change, as what the question gives, then we need to consider H2. But if we use standard entropies of formation, then that of H2 is 0

 

Don't you mean standard enthalpy of formation? Standard enthalpy of formation values for elements are zero. 

[kw0573]

13 hours ago, Msj Chem said:

Don't you mean standard enthalpy of formation? Standard enthalpy of formation values for elements are zero. 

Although that's a weird sounding name, I can justify my reasoning, even if standard entropy is very rarely used. (just like absolute Gibb's free energy and absolute enthalpy are also seldom used)

The standard enthalpy/entropy/Gibb's free energy, are some of the state function changes needed to synthesize one mole of a compound (minus those of the other products) in 100 kPa (I'm quite sure IB also uses 100 kPa but it's that or 101.3 kPa). Because elements need not be synthesize in such a pressure, then all three of enthalpy/entropy/Gibb's are 0. That is, standard (state function) values provides the difference between the compound and its constituent elements, so by definition, each element is not different in standard states from itself. Similarly, the standard entropy of formation can be used in other syntheses reactions of compounds from pure elements. 

So I was saying that If we use the provided entropy values to calculate the standard entropy change of CO + 2H₂ ---> CH₃OH, then there are two problems 1) the absolute entropy value for H₂ is not given, so we can't use (products) - (reactants) for all 3 species 2) the standard entropy of formation for neither CO or CH₃OH is given, so we can't use (products) - (reactants) for just the 2 compounds whose standard entropies of formation are not 0. Although two different reference frames, both approaches when given enough information, will lead to same answer, because the change is the same no matter where we set it to 0. 

It just means whichever equation we use, each variable must be measured in the same conditions for it to work. Either it's all in reference to 0K or standard states, but don't mix them up! It's the reason why ΔG is not equal to ΔH - T*S for each row of CO or CH₃OH. We can't mix absolute and relative values. The 131 J•K^-1•mol^-1 mentioned is the absolute entropy of H₂, which allow us to do (products) - (reactants) based on absolute entropy. 

Edited August 7, 2016 by kw0573