2.In human bodies, energy is generated when glucose is oxidized in the metabolic reactionC6H₁2O6(s) +60₂(g) = 6H₂O()+6CO₂(g).At T = 298 K, the standard enthalpy change for the reaction is AH° = -2801 kJ/mol. The absolute entropySo and molecular masses for the substances are given bya.b.C.d.C6H12O6(S)O₂(g)CO₂(g)H₂O()Entropy So in Unit of (J-mol¹K¹)212.13205213.669.9Molar Mass (g/mol)180324418Calculate the entropy change AS for the glucose oxidization reaction.Calculate the free energy change AG, at T = 298 K.Calculate the equilibrium constant of the reaction Keg.Calculate the energy generated by oxidizing a 10 g piece of glucose.173

Answered: 2. In human bodies, energy is generated…

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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2.
In human bodies, energy is generated when glucose is oxidized in the metabolic reaction
C6H₁2O6(s) +60₂(g) = 6H₂O()+6CO₂(g).
At T = 298 K, the standard enthalpy change for the reaction is AH° = -2801 kJ/mol. The absolute entropy
So and molecular masses for the substances are given by
a.
b.
C.
d.
C6H12O6(S)
O₂(g)
CO₂(g)
H₂O()
Entropy So in Unit of (J-mol¹K¹)
212.13
205
213.6
69.9
Molar Mass (g/mol)
180
32
44
18
Calculate the entropy change AS for the glucose oxidization reaction.
Calculate the free energy change AG, at T = 298 K.
Calculate the equilibrium constant of the reaction Keg.
Calculate the energy generated by oxidizing a 10 g piece of glucose.
173

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A chemical equation for the oxidation of glucose is given as follows:

$C subscript 6 H subscript 12 O subscript 6 space left parenthesis s right parenthesis space plus space 6 O subscript 2 space left parenthesis g right parenthesis space left right double arrow space 6 H subscript 2 O space left parenthesis l right parenthesis space plus space 6 C O subscript 2 space left parenthesis g right parenthesis$

The change in standard enthalpy of reaction $increment H to the power of o$ is given as $negative 2801 space k J divided by m o l$ .

The relationship between kJ and J is given by an expression:

$1 space k J space equals space 1000 space J$

Converting $increment H to the power of o$ in J/mol.

$increment H to the power of o space end exponent left parenthesis i n space J divided by m o l right parenthesis space equals space open parentheses fraction numerator 1000 space J over denominator 1 space k J end fraction close parentheses space cross times space left parenthesis negative 2801 space k J divided by m o l right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space minus 2801 space cross times space 10 cubed space J divided by m o l$

The temperature of the reaction is given as $298 space K$ .

The values of standard entropy, S^o for C₆H₁₂O₆, O₂, CO₂, and H₂O are given as follows:

	S^o(J.mol^-1.K^-1)
C₆H₁₂O₆(s)	212.13
O₂(g)	205
CO₂ (g)	213.6
H₂O (l)	69.9

a. We have to calculate $increment S to the power of o$ for the oxidation reaction of glucose.

b. We have to calculate $increment G to the power of o$ at 298 K.

c. We have to calculate K_eq.

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