Question

The standard enthalpy of combustion of C (graphite), H\(_2\) (g) and CH\(_3\)OH (l) respectively are \(-393\), \(-286\) and \(-726\) kJ mol\(^{-1}\). What is the standard enthalpy of formation of methanol?

• A. \(-726\) kJ mol\(^{-1}\)
• B. \(-239\) kJ mol\(^{-1}\)
• C. \(-96\) kJ mol\(^{-1}\)
• D. \(+96\) kJ mol\(^{-1}\)
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Hint:

Use Hess’s law to find formation enthalpy: \[ \Delta H_f = \sum \Delta H_{\text{combustion of products}} - \sum \Delta H_{\text{combustion of reactants}}. \]

Answer 1

The Correct Option is B

Step 1: Writing the Combustion Reactions The combustion reactions are: \[ C (graphite) + O_2 \rightarrow CO_2 \quad \Delta H = -393 \text{ kJ mol}^{-1} \] \[ H_2 + \frac{1}{2}O_2 \rightarrow H_2O \quad \Delta H = -286 \text{ kJ mol}^{-1} \] \[ CH_3OH + \frac{3}{2} O_2 \rightarrow CO_2 + 2H_2O \quad \Delta H = -726 \text{ kJ mol}^{-1} \] Step 2: Applying Hess’s Law \[ \Delta H_f \text{(CH}_3\text{OH)} = \Delta H_{\text{combustion}}(\text{CH}_3\text{OH}) - \left[ \Delta H_{\text{combustion}}(C) + 2 \times \Delta H_{\text{combustion}}(H_2) \right] \] \[ = -726 - \left[ -393 + 2(-286) \right] \] \[ = -726 + 393 + 572 \] \[ = -239 \text{ kJ mol}^{-1} \] \bigskip