Determine the enthalpy of vaporization, in [tex]$kJ / mol$[/tex], of [tex]$CH_3OH$[/tex] if 14.5 kJ of heat is needed to vaporize 0.500 moles of [tex][tex]$CH_3OH$[/tex][/tex].
To determine the enthalpy of vaporization of [tex]\( \text{CH}_3\text{OH} \)[/tex] (methanol), follow these steps:
1. Identify and list the given data: - The amount of heat needed to vaporize methanol, [tex]\( Q \)[/tex], is 14.5 kJ. - The number of moles of methanol, [tex]\( n \)[/tex], is 0.500 moles.
2. Recall the formula for the enthalpy of vaporization ([tex]\( \Delta H_{\text{vap}} \)[/tex]): [tex]\[
\Delta H_{\text{vap}} = \frac{Q}{n}
\][/tex] where: - [tex]\( \Delta H_{\text{vap}} \)[/tex] is the enthalpy of vaporization in kJ/mol. - [tex]\( Q \)[/tex] is the heat required in kJ. - [tex]\( n \)[/tex] is the number of moles of the substance.
3. Substitute the given values into the formula: [tex]\[
\Delta H_{\text{vap}} = \frac{14.5 \text{ kJ}}{0.500 \text{ moles}}
\][/tex]
4. Perform the division to find the enthalpy of vaporization: [tex]\[
\Delta H_{\text{vap}} = \frac{14.5}{0.500} = 29.0 \text{ kJ/mol}
\][/tex]
Therefore, the enthalpy of vaporization of [tex]\( \text{CH}_3\text{OH} \)[/tex] is [tex]\( 29.0 \text{ kJ/mol} \)[/tex].
