Question

Determine $∆E$for the process $H_{2}O\left(l\right)\to H_{2}O\left(g\right)$ at 25^oC and 1 atm.

Short answer

The $∆E$for the process is 44.0 kJ.

Step-by-step solution

Enthalpy of formation of substance

The chemical reaction is,

$H_{2}O\left(l\right)\to H_{2}O\left(g\right)$

The enthalpy of formation of the substance present in the reaction is,

$$\begin{gathered} \Delta H_f^0 of H_{2}O\left(l\right)=-286.0kJ/mol \\ \Delta H_f^0 of H_{2}O\left(g\right)=-242.0kJ/mol \end{gathered}$$

Determination ΔE  for the reaction

The $∆E$ is equal to $∆H$for reactions that does not involves gases.

$$\begin{gathered} \Delta H^o=\Delta H_f^o\left(product\right)-\Delta H_f^o\left(reactant\right) \\ \Delta H^o=\left(242.0kJ\right)-\left(-286kJ\right) \\ \Delta H^o=44.0kJ \end{gathered}$$

Thus, the$∆E$for the reaction is 44.0 kJ.