The temperature rises from 25.00 °C to 29.00 °C when 3.50 g of sucrose undergoes combustion in a bomb calorimeter.Calculate _$\Delta$rU for the combustion of sucrose in kJ $\mathrm { mol } ^ { - 1 }$ sucrose.The heat capacity of the calorimeter is 4.90 kJ ${ } ^ { \circ } \mathrm { C } ^ { - 1 }$ .The molar mass of sugar is 342.3 g $\mathrm { mol } ^ { - 1 }$ .

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released from the combustion of sucrose using the temperature change and calorimeter constant:
q = CΔT = (4.90 kJ/°C)(29.00°C - 25.00°C) = 19.6 kJ

Next, we can use the heat released and the amount of sucrose burned to calculate the standard enthalpy of combustion:
ΔcH° = q/n = -19.6 kJ/ (3.50 g / 342.3 g/mol) = -2004 kJ/mol

Finally, we can use the standard enthalpies of formation for sucrose and its combustion products to calculate the standard entropy of combustion:
ΔrH° = (-2ΔfH°[H2O] - 11ΔfH°[CO2] + ΔfH°[C12H22O11]) = (-2(-285.8 kJ/mol) - 11(-393.5 kJ/mol) + (-2229.6 kJ/mol)) = -5642.2 kJ/mol

ΔrS° = ΔrH°/T = (-5642.2 kJ/mol) / (298.15 K) = -18.92 kJ/(mol*K)

Finally, we can use the relationship between ΔG° and ΔS° to calculate the standard Gibbs free energy of combustion:
ΔrG° = ΔrH° - TΔrS° = (-2004 kJ/mol) - (298.15 K)(-18.92 kJ/(mol*K)) = -1461 kJ/mol

Converting to the requested units, we get:
ΔrG° = -1.92 × 10^3 kJ/mol

Therefore, the correct answer is A (-1.92 × 10^3 kJ/mol).

The Temperature Rises from 25 Δ\DeltaΔ RU for the Combustion of Sucrose in KJ

The Temperature Rises from 25 $\Delta$ RU for the Combustion of Sucrose in KJ