Question

If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The \(|\Delta H_f^\circ|\) of ethanol at 298.15 K is _______ \(\times 10^2\) kJ \(\cdot mol^{-1}\). (Nearest integer)
Given: Standard enthalpy for combustion of graphite = \(-393.5\) kJ \(\cdot mol^{-1}\)
Standard enthalpy of formation of water (l) = \(-285.8\) kJ \(\cdot mol^{-1}\)
Molar mass in \(g \cdot mol^{-1}\) of C, H, O are 12, 1 and 16 respectively

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
In a bomb calorimeter, the volume is constant, so the heat measured is the internal energy change of combustion (\(\Delta U_c\)). We then relate this to the enthalpy of combustion (\(\Delta H_c\)) and use Hess's Law to find the enthalpy of formation.

Step 2: Key Formula or Approach:
1. \(\Delta U_c (\text{per mole}) = \frac{q \times M}{m}\).
2. \(\Delta H_c = \Delta U_c + \Delta n_g RT\).
3. \(\Delta H_c = [2 \Delta H_f^\circ(CO_2) + 3 \Delta H_f^\circ(H_2O)] - [\Delta H_f^\circ(Ethanol) + 3 \Delta H_f^\circ(O_2)]\).

Step 3: Detailed Explanation:
Molar mass of ethanol (\(C_2H_5OH\)) = 46 g/mol.
Moles of ethanol = \(3.365 / 46 \approx 0.07315\) mol.
\(\Delta U_c = -99.472 / 0.07315 = -1360\) kJ/mol.
Reaction: \(C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)\).
\(\Delta n_g = 2 - 3 = -1\).
\(\Delta H_c = -1360 + (-1)(8.314 \times 10^{-3} \times 298.15) \approx -1362.5\) kJ/mol.
Using Hess's Law:
\(-1362.5 = [2(-393.5) + 3(-285.8)] - [\Delta H_f^\circ(Ethanol)]\)
\(-1362.5 = [-787 - 857.4] - \Delta H_f^\circ(Ethanol)\)
\(-1362.5 = -1644.4 - \Delta H_f^\circ(Ethanol) \implies \Delta H_f^\circ(Ethanol) = -281.9\) kJ/mol.
Value in \(10^2\) units: \(2.819 \times 10^2\). Nearest integer = 3.

Step 4: Final Answer:
The value is 3 \(\times 10^2\) kJ \(\cdot mol^{-1}\).