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:
A bomb calorimeter measures the heat of reaction at constant volume, which corresponds to the change in internal energy ($\Delta U_c^\circ$). We must convert this experimental data into molar internal energy, then use the relation $\Delta H = \Delta U + \Delta n_g RT$ to find the standard enthalpy of combustion ($\Delta H_c^\circ$). Finally, Hess's Law relates combustion enthalpy to the enthalpies of formation.
Step 2: Key Formula or Approach:
1. Moles $n = \frac{\text{Mass}}{\text{Molar Mass}}$
2. $\Delta U_c^\circ = \frac{\text{Heat}}{\text{moles}}$
3. $\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT$
4. $\Delta H_c^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$
Step 3: Detailed Explanation:
Molar mass of ethanol ($C_2H_5OH$) = $(2 \times 12) + (6 \times 1) + 16 = 46 \text{ g/mol}$.
Moles of ethanol burnt $n = \frac{3.365 \text{ g}}{46 \text{ g/mol}} = 0.073152 \text{ mol}$.
Heat produced at constant volume ($\Delta U$) for these moles is $-99.472 \text{ kJ}$ (negative because heat is released).
Molar internal energy of combustion:
$\Delta U_c^\circ = \frac{-99.472}{0.073152} = -1359.8 \text{ kJ/mol} \approx -1360 \text{ kJ/mol}$.
Write the balanced combustion equation for liquid ethanol:
$C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)$
Change in gaseous moles ($\Delta n_g$) = Moles of gaseous products - Moles of gaseous reactants
$\Delta n_g = 2 (CO_2) - 3 (O_2) = -1$.
Convert $\Delta U_c^\circ$ to $\Delta H_c^\circ$:
$\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT$
$\Delta H_c^\circ = -1360 \text{ kJ/mol} + (-1) \times (8.314 \times 10^{-3} \text{ kJ/K}\cdot\text{mol}) \times (298.15 \text{ K})$
$\Delta H_c^\circ = -1360 - 2.478 = -1362.478 \text{ kJ/mol}$.
Use Hess's Law to find the enthalpy of formation of ethanol:
$\Delta H_c^\circ = [2 \Delta H_f^\circ (CO_2) + 3 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (C_2H_5OH) + 3 \Delta H_f^\circ (O_2)]$
We know $\Delta H_f^\circ (O_2) = 0$.
The enthalpy of formation of $CO_2$ is identical to the combustion of graphite = $-393.5 \text{ kJ/mol}$.
$-1362.478 = [2(-393.5) + 3(-285.8)] - \Delta H_f^\circ (C_2H_5OH)$
$-1362.478 = [-787.0 - 857.4] - \Delta H_f^\circ (C_2H_5OH)$
$-1362.478 = -1644.4 - \Delta H_f^\circ (C_2H_5OH)$
$\Delta H_f^\circ (C_2H_5OH) = -1644.4 + 1362.478 = -281.922 \text{ kJ/mol}$.
Rounding to the nearest integer, we get $-282 \text{ kJ/mol}$.
Step 4: Final Answer:
The value is -282.