C6H12O6(s)+ 6 02(g) 6 CO2(g) + 6 H2O(g) AS ran = +900. J/(mol,n K) Substance Approximate S (J/(mol K)) C6H12O6(s) 209 O2(9) 205 H2O(g) CO2(9) 214 The chemical equation above represents the combustion of glucose, and the table provides the approximate standard absolute entropies, S , for some substances. Based on the information given, which of these equations can be used to calculate an approximation of S for H,O(g) ? S° = [900 + 209 + 205 - 214] J/(mol K) S° = (900 + 209 + (6 x 205) - (6 x 214)] J/(mol K) S° = - 900 - (6 x 214) + 209 + (6 x 205)] J/(mol K)
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
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Question 10
C,H12O6(s) + 6 02(9) → 6 CO2(g) + 6 H2O(g)
AS rzn = +900. J/(mol,n · K)
Substance
Approximate S (J/(mol · K))
C6H12O6(s)
209
O2(9)
205
H2O(g)
CO2(g)
214
The chemical equation above represents the combustion of glucose, and the table provides the approximate standard absolute entropies, S, for some substances. Based on the information given, which of
these equations can be used to calculate an approximation of S° for H,O(g) ?
S° = [900 + 209 + 205 – 214] J/(mol · K)
S° = 900 + 209 + (6 × 205) – (6 x 214)] J/(mol · K)
B
S° = - 900 – (6 × 214) + 209 + (6 x 205)] J/(mol - K)
D
S° = [- 900 – 214 + 209 + 205] J/(mol · K)
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