Question

Predict the sign and calculate ASⓇ for a reaction. Consider the reaction 2H S(g) + 302(g) +2H2O(g) + 25029) Based upon the stoichiometry of the reaction the sign of Srce should be Using standard thermodynamic data (in the Chemistry References), calculate AS rx at 25°C. AS J/Kºmol

Answer 1

The given reaction is

2H25(g) + 302(g) + 2H2O(g) + 2.5O2(g)

Note that there are 2+2 = 4 moles of gaseous products formed from 2+3=5 moles of gaseous reactants.

Hence, there are more number of gases in the reactant side. So, the system is undergoing decrease in degrees of freedom as less number of product molecules are being formed compared to the reactant molecules. Hence, the entropy change will be negative.

Hence, the sign of
AS! стта
should be negative.

Now, for a generic reaction

aA + bBC+ dD

We can calculate the
AS! стта
from the individual standard entropies of the reactants and products as follows

all products = cx Sº(C) + d x S°(D) - ax Sº (A) – 6 x S°(B) ASP

Now, for our reaction,

2H25(g) + 302(g) + 2H2O(g) + 2.5O2(g)

The standard entropies are

S°(H2O(g)) = 188.8 J/ K.mol

S°(H2S)) = 205.8 J/ Kimol

Sº (O2(g)) = 205.2 J/ K·mol

Sº (SO2(g)) = 248.2 J/ K·mol

Hence, we can calculate the standard entropy change of the reaction as follows:

ASn = nS - mS° all products all reactants Asim = 2 x Sº(H2O(g)) + 2 x Sº (SO2(g)) - 2 x Sº(H2S)) - 3x Sº(O2(g)) → Asim = 2 x 188.8 J/K mol + 2 x 248.2 J/K mol – 2 x 205.8 J/K - mol – 3 x 205.2 J/K mol → AS = 377.6 J/K.mol +496.4 J/K.mol-411.6 J/K.mol-615.6 J/K- mol → AS →AS = 874.0 J/K mol – 1027.2 J/K mol = -153.2 J/K mol

Hence, the standard entropy change for the reaction is
- 153.2 J/K mol
.

Note: the values of the standard entropies might differ slightly from source to source. Hence, do check your source if it matches. Otherwise the process will be same.