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In a bomb calorimeter with heat capacity 1400 kJ/K, combustion of 18 g of glucose raises the temperature from 27 degree C to 27.2 degree C. What is the magnitude of the standard enthalpy of combustion of glucose in kJ/mol? [R = 8.314 J/mol-K, M(glucose) = 180 g/mol]

1. 1200
2. 2000
3. 2800
4. 1400

Correct answer: 2800

Solution

The heat released at constant volume: q_v = C * delta_T = 1400 * 0.2 = 280 kJ for 18 g (0.1 mol) glucose. So delta_U_combustion = -280/0.1 = -2800 kJ/mol. For standard enthalpy: delta_H = delta_U + deltaₙ_g * R * T. C6H12O6 + 6O2 -> 6CO2 + 6H2O(l). deltaₙ_g = 6 - 6 = 0. So delta_H = delta_U = -2800 kJ/mol. Magnitude = 2800 kJ/mol.

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