Question

For the complete combustion of ethanol, \(\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\), the amount of heat produced as measured in bomb calorimeter is \(1364.47 \text{ kJ mol}^{-1}\) at \(25^\circ\text{C}\). Assuming ideality the enthalpy of combustion, \(\Delta H_C\), for the reaction will be \((\text{R} = 8.314 \text{ JK}^{-1}\text{mol}^{-1})\)

• A. \( -1366.95 \text{ kJ mol}^{-1} \)
• B. \( -1361.95 \text{ kJ mol}^{-1} \)
• C. \( -1460.50 \text{ kJ mol}^{-1} \)
• D. \( -1350.50 \text{ kJ mol}^{-1} \)

Hint:

Always remember the basic definition: Heat at constant volume (\(q_v\)) = \(\Delta U\) (measured via Bomb Calorimeter). Heat at constant pressure (\(q_p\)) = \(\Delta H\) (measured via Coffee-cup Calorimeter).

Answer 1

The Correct Option is A

Concept: A bomb calorimeter operates at a constant volume. Therefore, the heat produced or exchanged in a bomb calorimeter is equal to the internal energy change (\(\Delta U\)) of the reaction. Because heat is *produced* (exothermic combustion), \(\Delta U\) is negative. The relation between the enthalpy of combustion (\(\Delta H_C\)) and the internal energy change (\(\Delta U\)) is given by: \[ \Delta H_C = \Delta U + \Delta n_g RT \] Where:
• \(\Delta n_g\) = (Sum of moles of gaseous products) \(-\) (Sum of moles of gaseous reactants)
• \(R\) = Universal gas constant = \(8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}\)
• \(T\) = Absolute temperature in Kelvin Step 1: Determining \(\Delta U\), \(\Delta n_g\), and converting temperature to Kelvin.
Given that the heat produced at constant volume is \(1364.47 \text{ kJ mol}^{-1}\): \[ \Delta U = -1364.47 \text{ kJ mol}^{-1} \] From the balanced chemical equation: \[ \text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \] The gaseous components are \(2\text{ moles of } \text{CO}_2(g)\) and \(3\text{ moles of } \text{O}_2(g)\). Liquids are omitted. \[ \Delta n_g = 2 - 3 = -1 \] Convert the given temperature to Kelvin: \[ T = 25^\circ\text{C} + 273.15 = 298.15\text{ K} \]

Step 2: Calculating the enthalpy of combustion (\(\Delta H_C\)).
Substitute the parameters into the relation: \[ \Delta H_C = -1364.47 + \left[(-1) \times \left(8.314 \times 10^{-3}\right) \times 298.15\right] \] \[ \Delta H_C = -1364.47 - \left[\frac{8.314 \times 298.15}{1000}\right] \] \[ \Delta H_C = -1364.47 - 2.4788 \approx -1366.95 \text{ kJ mol}^{-1} \]