The temperature rises from 25.00 °C to 29.00 °C when 3.50 g of sucrose undergoes combustion in a bomb calorimeter. Calculate Δ_rU for the combustion of sucrose in kJ mol^-1 sucrose. The heat capacity of the calorimeter is 4.90 kJ °C^-1. The molar mass of sugar is 342.3 g mol^-1.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released by the combustion of 3.50 g of sucrose:
q = CΔT = 4.90 kJ/°C × (29.00 °C - 25.00 °C) = 19.6 kJ
Next, we can calculate the moles of sucrose burned:
n = m/M = 3.50 g / 342.3 g/mol = 0.0102 mol
The molar enthalpy of combustion of sucrose is given by:
ΔU° = -q/n = -19.6 kJ / 0.0102 mol = -1922.0 kJ/mol
Finally, we can calculate the molar entropy change of combustion using:
ΔS° = ΔU°/T = -1922.0 kJ/mol / 302.15 K = -6.36 kJ/(mol·K)
Converting the units to J/(mol·K):
ΔS° = -6360 J/(mol·K)
Therefore, the answer is A: -1.92 × 10^3 J/(mol·K) = -1.92 × 10^-3 kJ/(mol·K).

The Temperature Rises from 25