The standard enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 298 \(\mathrm{K}\) is \(-285.8 \mathrm{kJ} / \mathrm{mol}\) . Calculate the change in internal energy for the following process at 298 \(\mathrm{K}\) and \(1 \mathrm{atm} :\) $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta E^{\circ}=? $$ (Hint: Using the ideal gas equation, derive an expression for work in terms of \(n, R,\) and \(T\) )

During the given reaction, the change in moles of gas produced is equal to the moles of hydrogen and oxygen gas produced minus the moles of liquid water reacted. We can write this as: $$ \Delta n = \text{moles of }H_2 + \frac{1}{2}\text{ moles of }O_2 - \text{moles of }H_2O $$ Since 1 mole of liquid water produces 1 mole of hydrogen gas and \(\frac{1}{2}\) moles of oxygen gas, the change in moles is: $$ \Delta n = 1 + \frac{1}{2} - 1 = \frac{1}{2} $$

For an ideal gas, the work done on the system during a reaction can be expressed as: $$ w = -P_{ext}\Delta V $$ Since we are given that the reaction occurs at constant temperature and pressure, we can apply the ideal gas equation which states that: $$ PV = nRT $$ We can rearrange this equation to solve for \(\Delta V\): $$ \Delta V =\frac{nRT}{P} $$ Next, we can express \(\Delta V\) in terms of the change in moles of gas, i.e. \(\Delta n\): $$ \Delta V = \Delta n\frac{RT}{P} $$ Now we can substitute this expression for \(\Delta V\) back into the work equation: $$ w = -P_{ext}\left(\Delta n\frac{RT}{P}\right) $$ Since the pressure is constant, we can cancel out the \(P_{ext}\) terms: $$ w = -\Delta nRT $$

Substitute the values of \(\Delta n, R, T\) into the expression derived in Step 2: $$ w = -\left(\frac{1}{2}\right)(8.314\,\mathrm{J/mol\,K})(298\,\mathrm{K}) $$ Solving for the work done, we get: $$ w \approx -1236.9\,\mathrm{J/mol} $$

We are given the standard enthalpy change of formation for water as -285.8 kJ/mol. To maintain consistency in units, we need to convert this value to J/mol: $$ \Delta H = -285.8\,\mathrm{kJ/mol} \times 1000\,\mathrm{J/kJ} = -285800\,\mathrm{J/mol} $$

We can now use the enthalpy change, \(\Delta H\), and the work done, \(w\), to find the change in internal energy, \(\Delta E\), using the following formula: $$ \Delta E = \Delta H - w $$ Substitute the values of \(\Delta H\) and \(w\) into this equation: $$ \Delta E = -285800\,\mathrm{J/mol} - {-1236.9\,\mathrm{J/mol}} $$ Solving for the change in internal energy, we get: $$ \Delta E \approx -284563.1\,\mathrm{J/mol} $$ The change in internal energy for the given process at 298 K and 1 atm is approximately -284563.1 J/mol.