Question

Help me please

Substance	S° (J/mol×K)
Al (s)	28.33
Al2O3 (s)	50.92
Cl2 (g)	223.07
H2 (g)	130.68
N2 (g)	191.61
O2 (g)	205.14
Si (s)	18.83
SiCl4 (g)	330.73
SiHCl3 (g)	330.73
SiH2Cl2 (g)	285.71
SiH3Cl (g)	313.87
SiH4 (g)	204.62
Si3N4 (s)	101.3
SiO2 (s)	41.84

1. Predict the sign of $\Delta$ S°rxn for the following processes. Then calculate $\Delta$ S°rxn using the standard values just given.

	sign of $\Delta$ S°rxn	$\Delta$ S°rxn
a. SiO2 (s) $\rightarrow$ Si (s) + O2 (g)
b. Al2O3 (s) $\rightarrow$ 2 Al (s) + 3/2 O2 (g)
c. Si (s) + 2 Cl2 (g) $\rightarrow$ SiCl4 (g)
d.3 Si (s) + 2 N2 (g) $\rightarrow$ Si3N4 (s)

Thank you!

Answer 1

	sign of $\Delta$S°rxn	$\Delta$S°rxn
a. SiO2 (s) $\rightarrow$ Si (s) + O2 (g)	+ ve (positive)
b. Al2O3 (s) $\rightarrow$ 2 Al (s) + 3/2 O2 (g)	+ ve (positive)
c. Si (s) + 2 Cl2 (g) $\rightarrow$ SiCl4 (g)	- ve (negative)
d.3 Si (s) + 2 N2 (g) $\rightarrow$ Si3N4 (s)	- ve (negative)

Explanation :

a) SiO2 (s) $\rightarrow$ Si (s) + O2 (g)

$\Delta$S°rxn = ( 18.83 + 205.14 ) - 41.84

$\Delta$S°rxn   = 182.13 J/mol K

b) b. Al2O3 (s) $\rightarrow$ 2 Al (s) + 3/2 O2 (g)

$\Delta$S°rxn = (2 x 28.33 + 3/2 x 205.14) - 50.92

             = 313.45 J/mol . K

c)

$\Delta$S°rxn = 330.73 - (18.83 + 2 x 223.07)

              = - 134.24 J/mol K

d)

$\Delta$S° rxn = 101.3 - 18.83 +2 x 191.61

               = - 300.75 J/mol K