Question

What is the standard entropy of the reaction below at 25 degrees Celsius? H2O(g) + C (graphite, s) --> H2(g) + CO(g)

I see that the equation doesn't need to be balanced, and know that for entropy of reaction it should be products-reactants. I am using this site for values: https://sites.google.com/site/chempendix/thermo. However, I cannot seem to get an acceptable answer on the online homework system I am using for school. I did: (130.7+213.8)-(188.8+5.7)=150 J/(k x mol). That is incorrect. Have I pulled the wrong values or do I have this concept wrong?

Answer 1

Well, as you know, the standard change in molar entropy is for

`"H"_2"O"(g) + "C"("gr") -> "H"_2(g) + "CO"(g)`

Just to mention, yes, this is obviously balanced as-written (so that's not the problem), but on exams, you should check any reaction you see and make sure it is balanced.

And it would be calculated as:

`DeltaS_(rxn)^@ = sum_P n_P S_P^@ - sum_R n_R S_R^@`

where `S^@` are the standard molar entropies in `"J/mol"cdot"K"`, `n` are the stoichiometric coefficients in terms of `"mols"`, and `P//R` stand for products/reactants.

The most common issue is looking up the wrong phase, or sometimes the units. Sometimes even the wrong compound.

I looked them up on your reference site:

`S_(H_2O(g))^@ = "188.8 J/mol"cdot"K"`

`S_("C"(gr))^@ = "5.7 J/mol"cdot"K"`

`S_(H_2(g))^@ = "130.7 J/mol"cdot"K"`

`S_(CO(g))^@ = "197.7 J/mol"cdot"K"`

I am not seeing `"197.7 J/mol"cdot"K"` among your used `S^@` values. You used `S_(CO_2(g))^@`. Once you give it another go, see that you get:

`color(blue)(DeltaS_(rxn)^@) = [(1 cdot 130.7 + 1 cdot 197.7) - (1 cdot 188.8 + 1 cdot 5.7)] " J/K"`

`=` `color(blue)("133.9 J/K")`

It is possible that it may have wanted it in `"J/mol"cdot"K"`, so check that as well... sometimes `DeltaS_(rxn)^@` is defined so that the coefficients are unitless.