Question 16.3: CALCULATING THE STANDARD ENTROPY OF REACTION Calculate the s...

CALCULATING THE STANDARD ENTROPY OF REACTION

Calculate the standard entropy of reaction at 25 °C for the Haber synthesis of ammonia:

N2(g) + 3 H2(g)  2 NH3(g)N_{2}(g)  +  3  H_{2}(g)  →  2  NH_{3}(g)

STRATEGY
To calculate ΔS° for the reaction, subtract the standard molar entropies of all the reactants from the standard molar entropies of all the products. Look up the S° values in Table 16.1 or Appendix B, and remember to multiply the value for each substance by its coefficient in the balanced chemical equation.

TABLE 16.1 Standard Molar Entropies for Some Common Substances at 25 °C

Substance               Formula     S°[J/K · mol)] Substance                 Formula     S°[J/K · mol)]
Gases

Acetylene                        C2H2C_{2}H_{2}                 200.8

Ammonia                      NH3NH_{3}                  192.3

Carbon dioxide            CO2CO_{2}                      213.6

Carbon monoxide         CO                     197.6

Ethylene                      C2H4C_{2}H_{4}                   219.5

Hydrogen                    H2H_{2}                        130.6

Methane                      CH4CH_{4}                      186.2

Nitrogen                      N2N_{2}                         191.5

Nitrogen dioxide        NO2NO_{2}                     240.0

Dinitrogen tetroxide    N2O4N_{2}O_{4}                 304.3

Oxygen                        O2O_{2}                            205.0

Liquids

Acetic acid              CH3CO2HCH_{3}CO_{2}H               160

Ethanol                  CH3CH2OHCH_{3}CH_{2}OH             161

Methanol                CH3OHCH_{3}OH                     127

Water                      H2OH_{2}O                          69.9

Solids

Calcium carbonate    CaCO3CaCO_{3}                  91.7

Calcium oxide             CaO                       38.1

Diamond                      C                             2.4

Graphite                       C                              5.7

Iron                              Fe                            27.3

Iron(III) oxide            Fe2O3Fe_{2}O_{3}                    87.4

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ΔS° = 2 S°(NH3) – [S°(N2) + 3 S°(H2)]S°(NH_{3})  –  [S°(N_{2})  +  3  S°(H_{2})]

= (2 mol)(192.3 JK  mol) – [(1 mol)(191.5JK   mol)+ (3 mol)(130.6JK  mol)](2  mol)\left(192.3  \frac{J}{K  ·  mol}\right)  –  \left[(1  mol)\left(191.5 \frac{J}{K  ·   mol}\right) +  (3  mol)\left(130.6 \frac{J}{K  ·  mol}\right)\right]

= -198.7 J/K

BALLPARK CHECK
As predicted in Worked Example 16.1c, ΔS° should be negative because the reaction decreases the number of gaseous molecules from 4 mol to 2 mol.

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