Question

calculate the enthalpy of vaporization, the energy required to convert one mole of liquid into vapor, for the phase change below.\\(\ce{h2o(l) -> h2o(g)}\\)\

compound	\\(\delta h^\circ_f\\) (kj/mol)	\
----	----	\
\\(\ce{h2o(l)}\\)	-286	\
\\(\ce{h2o(g)}\\)	-242	\

\\(\delta h_{\text{vap}} = ?\\) kj/mol\
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for enthalpy change of a reaction

The enthalpy change of a reaction ($\Delta H$) is given by the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. For the reaction $\ce{H_2O(l) -> H_2O(g)}$, the formula is $\Delta H_{\text{vap}}=\Delta H_f^{\circ}(\ce{H_2O(g)}) - \Delta H_f^{\circ}(\ce{H_2O(l)})$.

Step2: Substitute the given values

We know that $\Delta H_f^{\circ}(\ce{H_2O(g)})=-242\ \text{kJ/mol}$ and $\Delta H_f^{\circ}(\ce{H_2O(l)})=-286\ \text{kJ/mol}$. Substituting these values into the formula:
$\Delta H_{\text{vap}} = (-242) - (-286)$

Step3: Calculate the result

Simplify the expression: $(-242)+286 = 44\ \text{kJ/mol}$.

Answer:

+44