/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 The corrosion (rusting) of iron ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 108

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 \(30^{\circ} \mathrm{C}\).

Short Answer

Expert verified
\( \Delta H_{reaction} = 11.73 \,\mathrm{kJ/mol} \) 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掳C.

Step by step solution

01

Part A: Calculate the standard enthalpy change for the reaction

We are given the molar enthalpy of formation of Fe(OH)鈧�, 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 H鈧�) 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 H鈧侽. 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: $$ \Delta H_{reaction} = (-583.39 \,\mathrm{kJ/mol}) - 2(-285.83 \,\mathrm{kJ/mol}) $$
02

Part A: Calculate the value of 螖H鈧�

Now compute the standard enthalpy change for the reaction: $$ \Delta H_{reaction} = (-583.39 \,\mathrm{kJ/mol}) + (2 \cdot 285.83 \,\mathrm{kJ/mol}) = 11.73 \,\mathrm{kJ/mol} $$
03

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掳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掳C) - \(\Delta T\) is the change in temperature (in 掳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掳C. The change in temperature is 30 鈥� 22 = 8掳C. $$ q = (250\,\mathrm{g})\times(4.18\,\mathrm{J/g^{\circ}C})\times(8^{\circ}\mathrm{C}) = 8328\,\mathrm{J} $$
04

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: $$ \text{moles of Fe} = \frac{8328\, \mathrm{J}}{11.73 \cdot 10^3\, \mathrm{J/mol}} = 0.709\, \mathrm{mol} $$
05

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: $$ \text{mass of Fe} = \text{moles of Fe} \cdot \text{molar mass of Fe} = (0.709\, \mathrm{mol})\times(55.85\,\mathrm{g/mol}) = 39.6\, \mathrm{g} $$ 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掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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**: \[\Delta H_{reaction} = \Delta H_{f}(\mathrm{products}) - \Delta H_{f}(\mathrm{reactants})\]
  • **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**: \[q = mc\Delta T\]
  • **Components**:
    • \(q\): Heat energy in joules.
    • \(m\): Mass in grams.
    • \(c\): Specific heat capacity of the substance (water here is 4.18 J/g掳C).
    • \(\Delta T\): 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.

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Most popular questions from this chapter

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix C: (a) \(\mathrm{CH}_{3} \mathrm{OH}(l),\) (b) \(\mathrm{CaSO}_{4}(s),\) (d) \(\mathrm{P}_{4} \mathrm{O}_{6}(s),\) (c) \(\mathrm{NO}(g)\).

Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{~kJ} \\ \mathrm{~B} \longrightarrow \mathrm{C} \quad \Delta H_{I I}=-90 \mathrm{~kJ} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\). (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

Three hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: $$ \begin{array}{llc} \hline \text { Hydrocarbon } & \text { Formula } & \Delta H_{f}^{0}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline \text { Butane } & \mathrm{C}_{4} \mathrm{H}_{10}(g) & -125 \\ \text { 1-Butene } & \mathrm{C}_{4} \mathrm{H}_{8}(g) & -1 \\ \text { 1-Butyne } & \mathrm{C}_{4} \mathrm{H}_{6}(g) & 165 \\ \hline \end{array} $$ (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Calculate the fuel value, in \(\mathrm{kJ} / \mathrm{g}\), for each of these compounds. (c) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: \((\mathbf{a}) q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J}\). (b) A system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system.

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) is \(2.22 \mathrm{~J} / \mathrm{g}\) -K. \((\mathbf{a})\) How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from 10.0 to \(25.0^{\circ} \mathrm{C} ?(\mathbf{b})\) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) by the same amount?
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