Question

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.

Answer 1

Step-by-step explanation:

The given data is as follows.

T = 298 K,
`\Delta H^(o)` = -5645 kJ/mol

`\Delta G^(o)` = -5798 kJ/mol

Relation between
`\Delta H` and
`\Delta G` are as follows.

`\Delta G^(o)` =
`\Delta H^(o) - T \Delta S^(o)`

-5798 kJ/mol = -5645 kJ/mol -
`298 * \Delta S^(o)`

-153 kJ/mol = -
`298 * \Delta S^(o)`

`\Delta S^(o)` = 0.513 kJ/mol K

Now, temperature is
`37^(o)C` = (37 + 273) K = 310 K

Since,
`\Delta G` =
`\Delta H^(o) - T \Delta S^(o)`

=
`-5645 kJ/mol - 310 K * 0.513 kJ/mol K`

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

= -5804.03 kJ/mol

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

`\Delta G = \Delta G_(310 K) - \Delta G_(298 K)`

= -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.