Question

Find ΔS for the combustion of methane to carbon dioxide and liquid water.

Is this the sign of ΔS you would get if you only considered the change in mols of gas during the reaction?

Answer 1

-
`\Delta S^o=-242.6(J)/(mol*K)`

- Yes.

Step-by-step explanation:

Hello,

In this case, the combustion of methane is:

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)`

And the standard entropy of combustion is computed as:

`\Delta S^o=s_(CO_2)^o+2*s_(H_2O)^o-s_(CH_4)^o-2*s_(O_2)^o`

Now, looking for those data on the NIST database, we obtain:

`\Delta S^o=213.8(J)/(mol*K) +2*69.92(J)/(mol*K)-186.16(J)/(mol*K)-2*205.04(J)/(mol*K)\\\\\Delta S^o=-242.6(J)/(mol*K)`

Moreover, as the result is negative, it means that the disorder decreased from reactants to products, this is noticeable by realizing that liquid water has a lower entropy than gaseous water, so YES, that also would be the sign if only the change in gaseous moles is considered.

Regards.