Question

The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hyrdroxide from iron by the following reaction: $$\mathrm{Fe}(s)+2{\mathrm{H}}_2\mathrm{O}(l)⟶\mathrm{Fe}(\mathrm{OH}{)}_2(s)+{\mathrm{H}}_2(g)$$ (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 ${30}^{\circ }\mathrm{C}$.

Short answer

$\mathrm{\Delta }{H}_{reaction}=11.73\mathrm{kJ}/\mathrm{mol}$ The energy required to heat the water is 8328 J. 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.

Step-by-step solution

Part A: Calculate the standard enthalpy change for the reaction

We are given the molar enthalpy of formation of Fe(OH)₂, which is -583.39 kJ/mol. Since the reaction is: $$\mathrm{Fe}(s)+2{\mathrm{H}}_2\mathrm{O}(l)⟶\mathrm{Fe}(\mathrm{OH}{)}_2(s)+{\mathrm{H}}_2(g)$$ We can calculate the standard enthalpy change for the reaction using the enthalpy of formation values for the products and reactants: $$\mathrm{\Delta }{H}_{reaction}=\mathrm{\Delta }{H}_f(\mathrm{Fe}(\mathrm{OH}{)}_2)+\mathrm{\Delta }{H}_f({\mathrm{H}}_2)-[\mathrm{\Delta }{H}_f(\mathrm{Fe})+2\mathrm{\Delta }{H}_f({\mathrm{H}}_2\mathrm{O})]$$ Since the enthalpy of formation for elements in their standard state (Fe, and H₂) is zero, $$\mathrm{\Delta }{H}_{reaction}=\mathrm{\Delta }{H}_f(\mathrm{Fe}(\mathrm{OH}{)}_2)-2\mathrm{\Delta }{H}_f({\mathrm{H}}_2\mathrm{O})$$ 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: $$\mathrm{\Delta }{H}_{reaction}=(-583.39\mathrm{kJ}/\mathrm{mol})-2(-285.83\mathrm{kJ}/\mathrm{mol})$$

Part A: Calculate the value of ΔHₐ

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

Part B: Calculate the energy needed to heat the water

To calculate how much energy is needed to heat 250 mL of water from 22 to 30°C, use the formula: $$q=mc\mathrm{\Delta }T$$ 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) - $\mathrm{\Delta }T$ 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. $$q=(250\mathrm{g})\times (4.18\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C})\times (8^{\circ }\mathrm{C})=8328\mathrm{J}$$

Part B: Calculate the number of moles of Fe

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: $$\text{moles of Fe}=\frac{8328\mathrm{J}}{11.73\cdot {10}^3\mathrm{J}/\mathrm{mol}}=0.709\mathrm{mol}$$

Part B: Calculate the mass of Fe

Finally, we need to calculate the mass of Fe needed. The molar mass of Fe is 55.85 g/mol: $$\text{mass of Fe}=\text{moles of Fe}\cdot \text{molar mass of Fe}=(0.709\mathrm{mol})\times (55.85\mathrm{g}/\mathrm{mol})=39.6\mathrm{g}$$ 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.

Key concepts

Corrosion of Iron

When iron, \(\mathrm{Fe}\), comes into contact with oxygen-free water, it can undergo corrosion, forming iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), and hydrogen gas. This reaction is significant in practices that aim to prevent rusting, as the product \(\mathrm{Fe}(\mathrm{OH})_{2}\) is often undesirable in industrial processes.
Corrosion is essentially the "eating away" of the metal, weakening it over time. This process is accelerated when water and oxygen are present, especially in environments exposed to atmospheric conditions.

• **Iron is susceptible to rust**: Iron reacts with water and oxygen to form rust, which is a form of iron oxide.
• **Formation of iron hydroxide**: In this exercise, the specific product is iron(II) hydroxide, which forms in a water-rich, oxygen-free environment.
• **Role of water**: Water acts as an electron acceptor, facilitating the reaction and the creation of hydroxides.

Understanding the conditions under which iron corrodes is crucial for its protection in various structures and components. This knowledge guides us in implementing preventive treatments like coating, galvanizing, or using corrosion inhibitors.

Enthalpy of Formation

Enthalpy of formation is a key component in calculating the enthalpy change of a chemical reaction. It refers to the energy change when one mole of a compound is formed from its elements in their standard states.
The standard enthalpy change for the reaction, \(\Delta H_{reaction}\), can be determined by using the enthalpies of formation of the reactants and products involved in the reaction:

• **Formula for enthalpy change**: $$\mathrm{\Delta }{H}_{reaction}=\mathrm{\Delta }{H}_f(\mathrm{products})-\mathrm{\Delta }{H}_f(\mathrm{reactants})$$
• **Elemental states**: The enthalpy of formation for an element in its standard state is zero.
• **Reaction specifics**: For iron(II) hydroxide formation, we use the known enthalpy of formation values to compute the reaction's overall enthalpy change.

With this understanding, we can calculate energy transfers in chemical processes, which are important in fields such as thermodynamics and industrial chemistry, ensuring reactions are economically viable and safe.

Specific Heat Capacity

Specific heat capacity is a property of a material that refers to the amount of energy required to raise the temperature of one gram of the material by one degree Celsius.
This concept is essential in various applications, particularly in heat exchange calculations. In the exercise, we use it to determine the energy needed to heat water through the formula:

• **Formula**: $$q=mc\mathrm{\Delta }T$$
• **Components**:
  • $q$: Heat energy in joules.
  • $m$: Mass in grams.
  • $c$: Specific heat capacity of the substance (water here is 4.18 J/g°C).
  • $\mathrm{\Delta }T$: Change in temperature in degrees Celsius.
• **Application**: By determining $q$, we can figure out how much energy must be transferred to attain a specified temperature rise.

This calculation is vital in engineering and environmental science, providing insights into energy management and efficiency in heating systems.