Chapter 6: Problem 122

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 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 $T$.)

Short Answer

Expert verified

The change in internal energy for the given process at 298K and 1 atm is approximately -284.6 kJ/mol.

Step by step solution

From thermodynamics, we know that the change in enthalpy, $\Delta H$, is related to the change in internal energy, $\Delta E$, and the work done in a process, $W$, as follows: $$ \Delta H = \Delta E + W $$ We want to find $\Delta E$. So we need to calculate $W$ and rearrange the equation: $$ \Delta E = \Delta H - W $$ #Step 2: Derive the work done using the ideal gas equation#

The ideal gas equation is given by: $$ PV=nRT $$ Where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature. Now, let's consider the work done, $W$, during the expansion of the gas. It can be expressed as: $$ W = -P\Delta V $$ Using the ideal gas equation, we can derive an expression for $\Delta V$ in terms of $n$, $R$, and $T$: $$ \Delta V = \frac{\Delta (nRT)}{P} $$ Substitute this expression back into the work equation: $$ W = -P \left(\frac{\Delta (nRT)}{P}\right) $$ Cancel out the pressure terms: $$ W = -\Delta(nRT) $$ #Step 3: Calculate the work done in the given process#

In the given process, we have the following stoichiometric changes in number of moles: $$ \Delta n = 1 + \frac{1}{2} - 1 = \frac{1}{2} $$ Now, we can substitute the values of $\Delta n$, $R$, and $T$ into the work equation: $$ W = - \left(\frac{1}{2}\cdot 8.314 \ \mathrm{J/(mol\cdot K)} \cdot 298 \ \mathrm{K} \right) $$ Calculate the work done: $$ W \approx -1243 \ \mathrm{J/mol} $$ #Step 4: Calculate the change in internal energy#

Now that we have the work done, we can calculate the change in internal energy using the equation from Step 1: $$ \Delta E = \Delta H - W $$ Substitute the given values for $\Delta H$ and the calculated value for $W$: $$ \Delta E = -285.8 \ \mathrm{kJ/mol} - (-1243 \ \mathrm{J/mol}) $$ Convert the second term to kJ/mol: $$ \Delta E = -285.8 \ \mathrm{kJ/mol} + 1.243 \ \mathrm{kJ/mol} $$ Calculate the change in internal energy: $$ \Delta E \approx -284.6 \ \mathrm{kJ/mol} $$ Thus, the change in internal energy for the given process at 298K and 1 atm is approximately -284.6 kJ/mol.