Question

What is the value for ΔS°reaction for the following reaction, given the standard entropy values

Answer 1

Standard entropy of the compounds of interest are as follows, 
$S^{0} (C6H12O6)$ = 212.1 J/K.mol
$S^{0} (O2)$ = 205.0 J/K.mol
$S^{0} (CO2)$ = 213.6 J/K.mol
$S^{0} (H2O)$ = 69.9 J/K.mol

Now for the reaction: 
C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)

ΔS°reaction = ∑$S^{0} products$ - ∑$S^{0} reactants$
∴ ΔS°reaction =( 6 $S^{0} (CO2)$ + 6 $S^{0} (H2O)$ ) - ($S^{0} (C6H12O6)$ + 6 $S^{0} (O2)$ = 205.0 J/K.mol)
∴ ΔS°reaction = [6 (213.6) + 6 (69.9 )] - [(212.1) + 6(205.0)]
∴ ΔS°reaction = 258.9 ≈ 262 J/ K mol.

Thus, correct answer is option C

Answer 2

The answer is C. 262 J/ K mol.

Molar mass:
C6H12O6212.1 J/K.mol
O2= 205.0 J/K.mol
CO2 = 213.6 J/K.mol
H2O= 69.9 J/K.mol

THE BALANCE IS:
C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)

= [6 (213.6) + 6 (69.9 )] - [(212.1) + 6(205.0)]
= 258.9 
=262 J/ K mol.