What is the value for ∆Soreaction for the following reaction, given the standard entropy values?
C6H12O6(s) + 6O2(g) <==> 6CO2(g) + 6H2O(l)
A. +131 J/K
B. -131 J/K
C. +262 J/K
D. -262 J/K
E. +417 J/K
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Without doing any calculations, you can eliminate B and D. This reaction will have a positive increase in entropy since you are (1) creating more moles of products than reactants and (2) going from solid and gas to gas and liquid products.
Now, according to my text, the standard entropy values for each component of the reaction are:
C6H12O6(s) So = 212.1 J/molK
O2(g) So = 205 J/molK
CO2(g) So = 213.6 J/molK
H2O(l) So = 69.9 J/molK
To calculate ∆So for the reaction:
∆So reaction = 6(213.6) + 6(69.9) - [212.1 + 6(205)] = 259 J/K
I'm guessing that your text and mine have slightly different values for one or two of the entropy values, but I think that C will be the correct answer using your values.