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At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol. Assume ∆H does not change to estimate the additional non-expansion work that may be obtained by raising the temperature to blood temperature, 37o C. Enter your answer in kJ/mol to two significant figures and do not enter the units.

Respuesta :

Explanation:

The given data is as follows.

             T = 298 K,          [tex]\Delta H^{o}[/tex] = -5645 kJ/mol

          [tex]\Delta G^{o}[/tex] = -5798 kJ/mol

Relation between [tex]\Delta H[/tex] and [tex]\Delta G[/tex] are as follows.

          [tex]\Delta G^{o}[/tex] = [tex]\Delta H^{o} - T \Delta S^{o}[/tex]    

             -5798 kJ/mol = -5645 kJ/mol - [tex]298 \times \Delta S^{o}[/tex]

                       -153 kJ/mol = -[tex]298 \times \Delta S^{o}[/tex]

                    [tex]\Delta S^{o}[/tex] = 0.513 kJ/mol K

Now, temperature is [tex]37^{o}C[/tex] = (37 + 273) K = 310 K

Since,        [tex]\Delta G[/tex] = [tex]\Delta H^{o} - T \Delta S^{o}[/tex]

                            = [tex]-5645 kJ/mol - 310 K \times 0.513 kJ/mol K[/tex]

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            [tex]\Delta G = \Delta G_{310 K} - \Delta G_{298 K}[/tex]

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

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