Question

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

C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)
+131 j/k
-131 j/k
+262j/k
-262 j/k
+417j/k

Answer 1

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.

Answer 2

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