Calculate the standard entropy change for the following reactions at 25°c.

(a) c2h5oh(l) + 3 o2(g) 2 co2(g) + 3 h2o(g)

(b) cs2(l) + 3 o2(g) co2(g) + 2 so2(g)

(c) 2 c6h6(l) + 15 o2(g) 12 co2(g) + 6 h2o(g)

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.