Question

For the process,${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$at 298 K and 1.0 atm. $\mathbf{∆}\mathbf{H}$ is more positive than $\mathbf{∆}\mathit{E}$by$\mathbf{2}\mathbf{.}\mathbf{5}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$. What does the$\mathbf{2}\mathbf{.}\mathbf{5}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$quantity represent?

Short answer

The$2.5kJ/mol$quantity is the work done.

Step-by-step solution

Equation of enthalpy

Enthalpy is a thermodynamic property which is sum of system’s internal energy and product of its pressure and volume.

$$\begin{gathered} ∆H=∆E+W \\ ∆H=∆E+P∆V \end{gathered}$$

Where,$∆H$= Change in Enthalpy

$∆E$= Change in Internal energy

W = Work done by the system

Meaning of in question

From, the sentence ‘$∆H$is more positive than$∆E$by$2.5kJ/mol$. And equation of Enthalpy, we conclude that the$2.5kJ/mol$quantity is the work done.