The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hyrdroxide from iron by the following reaction: $$ \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ (a) Calculate the standard enthalpy change for this reaction (the molar enthalpy of formation of \(\mathrm{Fe}(\mathrm{OH})_{2}\) is $-583.39 \mathrm{~kJ} / \mathrm{mol})$ (b) Calculate the number of grams of Fe needed to release enough energy to increase the temperature of \(250 \mathrm{~mL}\) of water from 22 to $30^{\circ} \mathrm{C}$.

We are given the molar enthalpy of formation of Fe(OH)₂, which is -583.39 kJ/mol. Since the reaction is: $$ \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ We can calculate the standard enthalpy change for the reaction using the enthalpy of formation values for the products and reactants: $$ \Delta H_{reaction} = \Delta H_{f}(\mathrm{Fe}(\mathrm{OH})_{2}) + \Delta H_{f}(\mathrm{H}_{2}) - [\Delta H_{f}(\mathrm{Fe}) + 2\Delta H_{f}(\mathrm{H}_{2} \mathrm{O})] $$ Since the enthalpy of formation for elements in their standard state (Fe, and H₂) is zero, $$ \Delta H_{reaction} = \Delta H_{f}(\mathrm{Fe}(\mathrm{OH})_{2}) - 2\Delta H_{f}(\mathrm{H}_{2} \mathrm{O}) $$ Now we need the enthalpy of formation of H₂O. The molar enthalpy of formation of gaseous water is -241.8 kJ/mol so for liquid water, it is about -285.83 kJ/mol. Plugging these values into the above equation, we get: $$ \Delta H_{reaction} = (-583.39 \,\mathrm{kJ/mol}) - 2(-285.83 \,\mathrm{kJ/mol}) $$

Now compute the standard enthalpy change for the reaction: $$ \Delta H_{reaction} = (-583.39 \,\mathrm{kJ/mol}) + (2 \cdot 285.83 \,\mathrm{kJ/mol}) = 11.73 \,\mathrm{kJ/mol} $$

To calculate how much energy is needed to heat 250 mL of water from 22 to 30°C, use the formula: $$ q = mc\Delta T $$ Where: - \(q\) is the energy required (in J); - \(m\) is the mass of the water (in g) - \(c\) is the specific heat capacity of water (in J/g°C) - \(\Delta T\) is the change in temperature (in °C) The mass of the water is equal to its volume (250 mL) multiplied by its density (1 g/mL), which is 250 g. The specific heat capacity of water is 4.18 J/g°C. The change in temperature is 30 – 22 = 8°C. $$ q = (250\,\mathrm{g})\times(4.18\,\mathrm{J/g^{\circ}C})\times(8^{\circ}\mathrm{C}) = 8328\,\mathrm{J} $$

Next, calculate the number of moles of Fe needed to release 8328 J of energy. We know that release of 1 mol of Fe causes 11.73 kJ of energy to be released: $$ \text{moles of Fe} = \frac{8328\, \mathrm{J}}{11.73 \cdot 10^3\, \mathrm{J/mol}} = 0.709\, \mathrm{mol} $$

Finally, we need to calculate the mass of Fe needed. The molar mass of Fe is 55.85 g/mol: $$ \text{mass of Fe} = \text{moles of Fe} \cdot \text{molar mass of Fe} = (0.709\, \mathrm{mol})\times(55.85\,\mathrm{g/mol}) = 39.6\, \mathrm{g} $$ Approximately 39.6 grams of Fe are needed to release enough energy to increase the temperature of 250 mL of water from 22 to 30°C.