A)The expansion work is -1234 kJ/mol and the total work is 1570 kJ/mol. B)The expansion work is -1234 kJ/mol and the total work is 1570 kJ/mol.
Ingesting glucose is more energy-effective than sucrose because glucose is a simpler sugar and can be more easily broken down by the body for energy production.
To calculate the non-expansion work, expansion work, and total work that can be obtained from the combustion of 1.0 kg of sucrose under standard conditions at 25 C when the product includes (a) water vapor, (b) liquid water, we need to use the following thermodynamic equations:
ΔG = ΔH - TΔS
w = -ΔG
q = ΔH
where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, w is the work, and q is the heat.
We are given that:
ΔG = -1570 kJ/mol
Sucrose + O2 + CO2 + H2O → CO2 + H2O
The balanced equation shows that 1 mole of sucrose produces 9 moles of water. Therefore, the combustion of 1.0 kg of sucrose produces 9 kg of water.
(a) When the product includes water vapor:
The non-expansion work is zero because the reaction is not expanding against a constant external pressure. The expansion work can be calculated as follows:
ΔG = ΔH - TΔS
-1570 kJ/mol = ΔH - (298 K)(ΔS)
ΔH = -1570 kJ/mol + (298 K)(ΔS)
The enthalpy change can be calculated as follows:
ΔH = q + PΔV
q = ΔH - PΔV
q = -228.57 kJ/mol - (9 kg/mol)(18.01528 g/mol)(298 K)(1.01325 × 10^5 Pa/m^2)
where PΔV is the work done by the system against the external pressure of the atmosphere, which is equal to the product of the pressure (1 atm = 1.01325 × 10^5 Pa) and the change in volume of the system.
The total work can be calculated as follows:
w = -ΔG
w = -(-1570 kJ/mol)
w = 1570 kJ/mol
Therefore, the expansion work is -1234 kJ/mol and the total work is 1570 kJ/mol.
(b) When the product includes liquid water:
The non-expansion work is zero because the reaction is not expanding against a constant external pressure. The expansion work can be calculated as follows:
ΔG = ΔH - TΔS
-1570 kJ/mol = ΔH - (298 K)(ΔS)
ΔH = -1570 kJ/mol + (298 K)(ΔS)
The enthalpy change can be calculated as follows:
ΔH = q
q = ΔH
q = -228.57 kJ/mol
The total work can be calculated as follows:
w = -ΔG
w = -(-1570 kJ/mol)
w = 1570 kJ/mol
Therefore, the expansion work is -1234 kJ/mol and the total work is 1570 kJ/mol.
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(a) When water vapor is produced as a product, the total work obtained from the combustion of 1.0 kg of sucrose is 228.57 kJ/mol.
(b) When liquid water is produced as a product, the total work obtained is 455.7 kJ/mol.
To determine whether it is more energy effective to ingest sucrose or glucose, we can compare the non-expansion work, expansion work, and total work that can be obtained from the combustion of 1.0 kg of sucrose.
First, we need to convert the mass of sucrose to moles. The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol, so 1.0 kg of sucrose is equal to 1000 g / 342.3 g/mol ≈ 2.92 mol.
Given the standard Gibbs free energy change (ΔG°) for the combustion of sucrose, we can calculate the non-expansion work (W_ne), expansion work (W_e), and total work (W_total) under the specified conditions.
a) Water vapor as a product:
The balanced chemical equation for the combustion of sucrose when water vapor is produced is:
C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g)
Non-expansion work (W_ne):
W_ne = -ΔG° = -(-228.57 kJ/mol) = 228.57 kJ/mol
Expansion work (W_e):
No expansion work is performed when water vapor is the product because it does not undergo a volume change.
Total work (W_total):
W_total = W_ne + W_e = 228.57 kJ/mol
b) Liquid water as a product:
The balanced chemical equation for the combustion of sucrose when liquid water is produced is:
C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
Non-expansion work (W_ne):
W_ne = -ΔG° = -(-1570 kJ/mol) = 1570 kJ/mol
Expansion work (W_e):
To calculate the expansion work, we need to consider the change in volume of water from vapor to liquid state. The change in volume (ΔV) can be estimated using the ideal gas law and the density of water.
The density of water at 25°C is approximately 997 kg/m^3, which is equivalent to 997 g/L. Since 11 moles of water are produced per mole of sucrose, the volume change is given by:
ΔV = 11 mol * 997 g/L / 997 g/mol ≈ 11 L
The expansion work is calculated as:
W_e = -PΔV
W_e = -(101.3 kPa * 11 L)
W_e = -1114.3 kJ/mol
Total work (W_total):
W_total = W_ne + W_e
W_total = 1570 kJ/mol + (-1114.3 kJ/mol)
W_total = 455.7 kJ/mol
Therefore, in terms of energy effectiveness, it is more favorable to ingest sucrose when the product includes liquid water rather than water vapor.
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