Question

The enthalpy of formation of \( \text{CO}_2(\text{g}) \) and \( \text{H}_2\text{O}(l) \) are \(-393.5\) and \(-286 \text{ kJ mol}^{-1}\) respectively. If the heat of combustion of \( \text{CH}_3\text{OH}(l) \) is \(-749 \text{ kJ mol}^{-1}\), the enthalpy of formation of \( \text{CH}_3\text{OH}(l) \) \(in \text{kJ mol}^{-1}\) is

• A. +216.5
• B. -216.5
• C. -355.5
• D. +355.5

Hint:

Always ensure the chemical equation is balanced before applying the enthalpy of reaction formula. Remember that the standard enthalpy of formation for elements in their most stable state (e.g., $\text{O}_2(\text{g})$, $\text{N}_2(\text{g})$, $\text{C(graphite)}$) is zero. Pay close attention to signs and unit conversions (if any are needed).

Answer 1

The Correct Option is B

Step 1: Write the Balanced Chemical Equation for the Combustion of \( \text{CH}_3\text{OH}(l) \)
The combustion of methanol ($\text{CH}_3\text{OH}$) involves its reaction with oxygen ($\text{O}_2$) to produce carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$). \[ \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(l) \] Step 2: State the Formula for Enthalpy of Reaction from Enthalpies of Formation
The enthalpy of a reaction ($\Delta H_{\text{reaction}}$) can be calculated from the standard enthalpies of formation ($\Delta H_f^\circ$) of the products and reactants using the following formula: \[ \Delta H_{\text{reaction}} = \sum (\text{stoichiometric coefficients} \times \Delta H_f^\circ \text{ products}) - \sum (\text{stoichiometric coefficients} \times \Delta H_f^\circ \text{ reactants}) \] Step 3: List the Given Enthalpies of Formation and Reaction
Given: \begin{itemize} \item Enthalpy of formation of $\text{CO}_2(\text{g})$, $\Delta H_f^\circ (\text{CO}_2(\text{g})) = -393.5 \text{ kJ mol}^{-1}$ \item Enthalpy of formation of $\text{H}_2\text{O}(l)$, $\Delta H_f^\circ (\text{H}_2\text{O}(l)) = -286 \text{ kJ mol}^{-1}$ \item Heat of combustion of $\text{CH}_3\text{OH}(l)$, $\Delta H_{\text{combustion}} = -749 \text{ kJ mol}^{-1}$ \item Enthalpy of formation of $\text{O}_2(\text{g})$ is 0 (as it's an element in its standard state). \end{itemize} Let the enthalpy of formation of $\text{CH}_3\text{OH}(l)$ be $\Delta H_f^\circ (\text{CH}_3\text{OH}(l))$. Step 4: Substitute the Values into the Enthalpy of Reaction Formula
Using the balanced equation from Step 1 and the formula from Step 2: \[ \Delta H_{\text{combustion}} = [1 \times \Delta H_f^\circ (\text{CO}_2(\text{g})) + 2 \times \Delta H_f^\circ (\text{H}_2\text{O}(l))] - [1 \times \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) + \frac{3}{2} \times \Delta H_f^\circ (\text{O}_2(\text{g}))] \] Substitute the known values: \[ -749 = [1 \times (-393.5) + 2 \times (-286)] - [\Delta H_f^\circ (\text{CH}_3\text{OH}(l)) + \frac{3}{2} \times 0] \] \[ -749 = [-393.5 - 572] - \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) \] \[ -749 = -965.5 - \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) \] Step 5: Solve for \( \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) \)
Rearrange the equation to solve for $\Delta H_f^\circ (\text{CH}_3\text{OH}(l))$: \[ \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) = -965.5 + 749 \] \[ \Delta H_f^\circ (\text{CH}_3\text{OH}(l)) = -216.5 \text{ kJ mol}^{-1} \] Step 6: Analyze Options
\begin{itemize} \item Option (1): +216.5. Incorrect. \item Option (2): -216.5. Correct, as it matches our calculated enthalpy of formation. \item Option (3): -355.5. Incorrect. Option (4): +355.5. Incorrect. \end{itemize}