Question textQ4 V32.568 g of yellow sulfur (S8) is burned in a bomb calorimeter with excess oxygen. The temperature of the calorimeter and its contents increases for this burning from 21.00 oC to 31.00 oC.The calibration factor of the calorimeter was previously found to be 725 J K-1What is the enthalpy change (ΔH) for the following the reaction? (Mr(S) = 32.1 g mol-1, Mr(O) = 16.0 g mol-1)S8(s) + 8O2(g) [math: ⟶]\ce{ - } 8SO2(g) Answer 1 Question 4[input] kJ mol-1.

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Question textQ4 V32.568 g of yellow sulfur (S8) is burned in a bomb calorimeter with excess oxygen. The temperature of the calorimeter and its contents increases for this burning from 21.00 oC to 31.00 oC.The calibration factor of the calorimeter was previously found to be 725 J K-1What is the enthalpy change (ΔH) for the following the reaction? (Mr(S) = 32.1 g mol-1, Mr(O) = 16.0 g mol-1)S8(s) +  8O2(g) [math: ⟶]\ce{ -> }  8SO2(g) Answer 1 Question 4[input] kJ mol-1.

BlackTom AI · Accepted Answer

The question provides a bomb calorimeter experiment and asks for the enthalpy change (ΔH) for the combustion reaction of S8 to SO2 under the given conditions.

First, determine the heat exchanged with the calorimeter: the calorimeter constant C = 725 J K-1 and the temperature rise ΔT = 31.00 °C − 21.00 °C = 10.00 K. The heat absorbed by the calorimeter is q_cal = C × ΔT = 725 J/K × 10.00 K = 7250 J.

In a bomb calorimeter, the heat measured corresponds to the heat released by the reaction at constant volume, so q_rxn = −q_cal = −7250 J for the amount of substance burned.

Next, compute the amount of S8 that reacted. The sample mass is 2.568 g of S8. The molar mass of S8 is M(S8) = 8 × M(S) = 8 × 32.1 g/mol = 256.8 g/mol. Therefore, n(S8) = 2.568 g / 256.8 g/mol = 0.0100 mol (i.e., 1.0 × 10−2 mol).

For the reaction S8(s) + 8 O2(g) → 8 SO2(g), the gas moles change is Δn_gas = 8 (SO2) − 8 (O2) = 0, since S8 is solid. Because Δn_gas = 0, the relationship ΔH = ΔU + Δn_gas RT reduces to ΔH ≈ ΔU under these conditions. Thus, the heat released at constant volume is essentially the same as the enthalpy change per mole of reaction for this specific stoichiometry.

Convert the total reaction heat to a per-mole basis: ΔH per mole of S8 = (−7250 J) / (0.0100 mol) = −725,000 J/mol = −725 kJ/mol.

As a result, the enthalpy change for the reaction per mole of S8 combusted is −725 kJ per mole.

MUF0042 Chemistry Unit 2 - Semester 2, 2025

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