Question

eNaICUrate Equilibrium Constants TABLE 26.1 Standard molar Gibbs energies of formation, A,G", for various substances at 298.15 K and one bar. 1057 Substance Formula 4,G"/kJ-mol acetylene сН.0 NH,(g) ammonia 209.20 benzene -16.367 с,н.Ф 124.35 bromine Br (g) 3.126 butane C,Hog) -17.15 carbon(diamond) C(s) 2.900 carbon(graphite) C(s) carbon dioxide cO(g) -394.389 carbon monoxide COg) -137.163 ethane СН, С Н, ОнФ сН. Ф C,H,0,(s) HBr(g) НСK -32.82 ethanol -174.78 ethene 68.421 glucose -910.52 hydrogen bromide hydrogen chloride hydrogen fluoride hydrogen iodide -53.513 -95.300 HF(g) -274.646 1.560 HI(g) -105.445 Н.0,0 hydrogen peroxide 19.325 1(g) CH,(g) iodine -50.768 methane CH, ОНФ) CH, ОН -166.27 methanol -161.96 86.600 nitrogen oxide nitrogen dioxide dinitrogen tetraoxide NO(g) 51.258 NO,(g) N,O,(g) N.O,0) С,Н, Ф 97.787 97.52 -23.47 propane 1544.65 sucrose -300.125 s0,(g) s0,(g) CCL,@) CCL(g) Н.о0 Н, O) sulfur dioxide -371.016 sulfur trioxide -65.21 tetrachloromethane 53.617 -237.141 -228.582 water

Answer 1

Answer - Given,

Temperature = 25^$\degree$C or 298 K [0°C + 273 = 273K]

Pressure = 1 bar

1 kJ = 1000 J

$\Delta$S°_rxn = ?

Some useful information,

Species	$\Delta$Hf°(kJ/mol)	$\Delta$Gf°(kJ/mol)
C (s)	0	0
O2 (g)	0	0
CO2 (g)	-110.525	-137.168
CH4 (g)	-74.81	-50.72
C2H2 (g)	226.73	209.2
H2 (g)	0	0
H2O (l)	-285.83	-237.129
C2H4 (g)	52.26	68.15

A)

C (s) + O₂ (g)  $\rightarrow$ CO₂ (g)

We know that,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 1 * $\Delta$ G^$\degree$_f(CO₂ (g)) - 1 * $\Delta$ G^$\degree$_f(O₂ (g)) - 1 *$\Delta$G^$\degree$_f(C (s))

$\Delta$G^$\degree$_rxn = [1 * (-137.168 kJ/mol)] - [1 * (0 kJ/mol)] - [1 * (0 kJ/mol)]

$\Delta$G^$\degree$_rxn = -137.168 kJ/mol

Also,

$\Delta$H^$\degree$_rxn = $\sum$ n * $\Delta$ H^$\degree$_f(products) - $\sum$ n * $\Delta$ H^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$H^$\degree$_rxn = 1 * $\Delta$ H^$\degree$_f(CO₂ (g)) - 1 * $\Delta$ H^$\degree$_f(O₂ (g)) - 1 *$\Delta$H^$\degree$_f(C (s))

$\Delta$H^$\degree$_rxn = [1 * (-110.525 kJ/mol)] - [1 * (0 kJ/mol)] - [1 * (0 kJ/mol)]

$\Delta$H^$\degree$_rxn = -110.525 kJ/mol

Now,

We know that,

$\Delta$G^$\degree$_rxn = $\Delta$ H^$\degree$_rxn - T $\Delta$ S^$\degree$_rxn

Put the values,

-137.168 kJ/mol = -110.525 kJ/mol - 298 K * $\Delta$ S^$\degree$_rxn

So,$\Delta$S^$\degree$_rxn = (137.168 kJ/mol - 110.525 kJ/mol )/ 298 K

$\Delta$S^$\degree$_rxn = 0.0894 kJ/mol.K or 89.4 J/mol.K [Answer]

B)

CH₄ (g) + 2 O₂ (g)  $\rightarrow$ CO₂ (g) + 2 H₂O (l)

We know that,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 2 * $\Delta$ G^$\degree$_f(H₂O (l)) + 1 * $\Delta$ G^$\degree$_f(CO₂ (g)) - 1 *$\Delta$G^$\degree$_f(CH₄ (g)) - 2 *$\Delta$G^$\degree$_f(O₂ (g))

$\Delta$G^$\degree$_rxn = [2 * (-237.129 kJ/mol)] + [1 * (-137.168 kJ/mol)] - [1 * (-50.72 kJ/mol)] - [2 * (0 kJ/mol)]

$\Delta$G^$\degree$_rxn = -560.706 kJ/mol

Also,

$\Delta$H^$\degree$_rxn = $\sum$ n * $\Delta$ H^$\degree$_f(products) - $\sum$ n * $\Delta$ H^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$H^$\degree$_rxn = 2 * $\Delta$ H^$\degree$_f(H₂O (l)) + 1 * $\Delta$ H^$\degree$_f(CO₂ (g)) - 1 *$\Delta$H^$\degree$_f(CH₄ (g)) - 2 *$\Delta$H^$\degree$_f(O₂ (g))

$\Delta$H^$\degree$_rxn = [2 * (-285.83 kJ/mol)] + [1 * (-110.525 kJ/mol)] - [1 * (-74.81 kJ/mol)] - [2 * (0 kJ/mol)]

$\Delta$H^$\degree$_rxn = -607.375​​​​​​​ kJ/mol

Now,

We know that,

$\Delta$G^$\degree$_rxn = $\Delta$ H^$\degree$_rxn - T $\Delta$ S^$\degree$_rxn

Put the values,

-560.706​​​​​​​ kJ/mol  = -607.375​​​​​​​ kJ/mol - 298 K * $\Delta$ S^$\degree$_rxn

So,$\Delta$S^$\degree$_rxn = (560.706​​​​​​​ kJ/mol - 607.375​​​​​​​ kJ/mol)/ 298 K

$\Delta$S^$\degree$_rxn = -0.15661 kJ/mol.K or -156.61 J/mol.K [Answer]

C)

C₂H₂ (g) + H₂ (g)  $\rightarrow$ C₂H₄ (g)

We know that,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 1 * $\Delta$ G^$\degree$_f(C₂H₄ (g)) - 1 * $\Delta$ G^$\degree$_f(C₂H₂ (g)) - 1 *$\Delta$G^$\degree$_f(H₂ (g))

$\Delta$G^$\degree$_rxn = [1 * (68.15 kJ/mol)] - [1 * (209.2 kJ/mol)] - [1 * (0 kJ/mol)]

$\Delta$G^$\degree$_rxn = -141.05 kJ/mol

Also,

$\Delta$H^$\degree$_rxn = $\sum$ n * $\Delta$ H^$\degree$_f(products) - $\sum$ n * $\Delta$ H^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$H^$\degree$_rxn = 1 * $\Delta$ H^$\degree$_f(C₂H₄ (g)) - 1 * $\Delta$ H^$\degree$_f(C₂H₂ (g)) - 1 *$\Delta$H^$\degree$_f(H₂ (g))

$\Delta$H^$\degree$_rxn = [1 * (52.26 kJ/mol)] - [1 * (226.73 kJ/mol)] - [1 * (0 kJ/mol)]

$\Delta$H^$\degree$_rxn = -174.47 kJ/mol

Now,

We know that,

$\Delta$G^$\degree$_rxn = $\Delta$ H^$\degree$_rxn - T $\Delta$ S^$\degree$_rxn

Put the values,

-141.05 kJ/mol = -174.47 kJ/mol - 298 K * $\Delta$ S^$\degree$_rxn

So,$\Delta$S^$\degree$_rxn = (141.05 kJ/mol - 174.47 kJ/mol)/ 298 K

$\Delta$S^$\degree$_rxn = -0.11215 kJ/mol.K or -112.14 J/mol.K [Answer]