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Use Hess's law and the following data to calculate the standard enthalpy of formation of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\ell)\) $$\begin{array}{ll} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(g) \rightarrow & \\ 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(\ell) & \Delta H_{\mathrm{ren}}^{\circ}=-1368.2 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) & \Delta H_{f}^{\circ}=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) & \Delta H_{f}^{\circ}=-285.9 \mathrm{kJ} / \mathrm{mol} \end{array}$$

Short Answer

Expert verified
Question: Calculate the standard enthalpy of formation of ethanol using Hess's law and the given information on the complete combustion of ethanol and the standard enthalpies of formation for carbon dioxide and liquid water. Answer: The standard enthalpy of formation of ethanol (CH3CH2OH) is -1368.2 kJ/mol.

Step by step solution

01

Identify the intermediate steps to create ethanol

Write down the reactions that create ethanol and oxygen from their elemental forms: $$ \mathrm{C}(s) + 2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) \\ \mathrm{CH}_{4}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(\ell) \\ $$
02

Determine the enthalpy change for creating CH4

Based on the given reactions, the enthalpy change for creating CH4 (methane) can be determined as a combination of creating two H2O molecules and one CO2 molecule: $$ \Delta H_{f}^{\circ}(\mathrm{CH}_{4}) = 2\Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}) + \Delta H_{f}^{\circ}(\mathrm{CO}_{2}) \\ \Delta H_{f}^{\circ}(\mathrm{CH}_{4}) = 2(-285.9) + (-393.5) \\ \Delta H_{f}^{\circ}(\mathrm{CH}_{4}) = -965.3~\mathrm{kJ/mol} \\ $$
03

Determine the enthalpy change for the complete combustion of CH4

Write down the complete combustion reaction for CH4 and use the given data to compute the enthalpy change: $$ \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(\ell) \\ \Delta H_{ren}^{\circ}(\mathrm{CH}_{4}) = \Delta H_{f}^{\circ}(\mathrm{CO}_{2}) + 2\Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}) - \Delta H_{f}^{\circ}(\mathrm{CH}_{4}) \\ \Delta H_{ren}^{\circ}(\mathrm{CH}_{4}) = -393.5 + 2(-285.9) + 965.3 \\ \Delta H_{ren}^{\circ}(\mathrm{CH}_{4}) = -0.0~\mathrm{kJ/mol} \\ $$
04

Calculate the enthalpy change for creating CH3OH from CH4

Write down the reaction for creating CH3OH from CH4 and use Hess's Law to compute the enthalpy change: $$ \mathrm{CH}_{4}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(\ell) \\ \Delta H_{f}^{\circ}(\mathrm{CH}_{3}\mathrm{OH})= \Delta H_{ren}^{\circ}(\mathrm{CH}_{3}\mathrm{OH}) - \Delta H_{ren}^{\circ}(\mathrm{CH}_{4}) \\ \Delta H_{f}^{\circ}(\mathrm{CH}_{3}\mathrm{OH}) = -1368.2 - (-0.0) \\ \Delta H_{f}^{\circ}(\mathrm{CH}_{3}\mathrm{OH}) = -1368.2~\mathrm{kJ/mol} \\ $$ Thus, the standard enthalpy of formation of ethanol (CH3CH2OH) is -1368.2 kJ/mol.

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Key Concepts

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

Enthalpy of Formation
The enthalpy of formation is a fundamental concept in thermodynamics. It represents the heat change when one mole of a compound is formed from its elements in their standard states. This process occurs under standard conditions of 1 atm pressure and 25°C (298 K).
To understand this concept, consider the formation of water from hydrogen and oxygen. The balanced chemical equation is:
\[\mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2}\mathrm{O}(\ ext{l}) \]The standard enthalpy of formation for water is -285.9 kJ/mol. This indicates that when one mole of water is formed, 285.9 kJ of energy is released.
Key points about enthalpy of formation:
  • It involves elements in their standard states, e.g., \( \mathrm{O}_{2} \) is a gas at room temperature.
  • Negative values indicate that the formation is exothermic (releases energy).
  • It is used to calculate other thermodynamic properties, like reaction enthalpy.
Combustion Reactions
Combustion reactions are a type of chemical reaction where a substance reacts with oxygen, releasing energy. This process is generally highly exothermic, meaning it releases a significant amount of heat.
For example, the combustion of methane can be represented by the equation:
\[\mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(\ell) \]In this reaction, methane burns in the presence of oxygen to produce carbon dioxide and water, releasing energy in the form of heat.
Important aspects of combustion reactions include:
  • They typically produce heat and are considered exothermic.
  • They can involve hydrocarbons, like methane, or other combustible materials.
  • The products are usually \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \).
Understanding combustion reactions is crucial for energy production and environmental considerations.
Enthalpy Change Calculations
Enthalpy change calculations involve determining the heat released or absorbed during a chemical reaction. Hess's Law, which states that the total enthalpy change is the same regardless of the path taken, is a key principle utilized in these calculations.
Applying Hess's Law requires knowing the enthalpy changes of various steps and combining them to find the overall reaction enthalpy. For example, when calculating the enthalpy of formation for ethanol using given reactions, each step's enthalpy change contributes to the final value.
The typical process involves:
  • Writing balanced chemical equations for intermediate steps.
  • Using known enthalpy values, such as formation or reaction enthalpies.
  • Summing the enthalpy changes for all steps to get the total change.
By understanding these principles, you can accurately compute the energy changes in chemical reactions and enhance your grasp of thermochemistry.

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Most popular questions from this chapter

Adding \(2.00 \mathrm{g}\) of \(\mathrm{Mg}\) metal to \(95.0 \mathrm{mL}\) of \(1.00 \mathrm{MHCl}\) in a coffee-cup calorimeter leads to a temperature increase of \(9.2^{\circ} \mathrm{C}\) a. Write a balanced net ionic equation for the reaction. b. If the molar heat capacity of \(1.00 M \mathrm{HCl}\) is the same as that for water \(\left[c_{\mathrm{P}}=75.3 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right],\) what is \(\Delta H_{\mathrm{rxn}} ?\)

A \(100.0 \mathrm{mL}\) sample of \(1.0 \mathrm{M} \mathrm{NaOH}\) is mixed with \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{MH}_{2} \mathrm{SO}_{4}\) in a large Styrofoam coffee cup; the cup is fitted with a lid through which a calibrated thermometer passes. The temperature of each solution before mixing is \(22.3^{\circ} \mathrm{C} .\) After the \(\mathrm{NaOH}\) solution is added to the coffee cup and the mixed solutions are stirred with the thermometer, the maximum temperature measured is \(31.4^{\circ} \mathrm{C} .\) Assume that the density of the mixed solutions is \(1.00 \mathrm{g} / \mathrm{mL},\) the specific heat of the mixed solutions is \(\left.4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\right),\) and no heat is lost to the surroundings. a. Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. b. Is any \(\mathrm{NaOH}\) or \(\mathrm{H}_{2} \mathrm{SO}_{4}\) left in the Styrofoam cup when the reaction is over? c. Calculate the enthalpy change per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the reaction.

Use the following standard heats of formation to calculate the molar enthalpy of vaporization of acetic acid: \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{CH}_{3} \mathrm{COOH}(\ell)\) is \(-484.5 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{CH}_{3} \mathrm{COOH}(g)\) is \(-432.8 \mathrm{~kJ} / \mathrm{mol}\).

The reactor-core cooling systems in some nuclear power plants use liquid sodium as the coolant. Sodium has a thermal conductivity of \(1.42 \mathrm{J} /\) \((\mathrm{cm} \cdot \mathrm{s} \cdot \mathrm{K}),\) which is quite high compared with that of water \(\left.\left[6.1 \times 10^{-3} \mathrm{J} / \mathrm{cm} \cdot \mathrm{s} \cdot \mathrm{K}\right)\right] .\) The respective molar heat capacities are \(28.28 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) and \(75.31 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) .\) What is the advantage of using liquid sodium over water in this application?

What is the \(\Delta H_{\text {rxn }}\) for the precipitation of AgCl if adding \(125 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{AgNO}_{3}\) to \(125 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{NaCl}\) at \(18.6^{\circ} \mathrm{C}\) causes the temperature to increase to \(26.4^{\circ} \mathrm{C} ?\) (Assume \(\left.c_{\mathrm{P}, \text { soln }}=c_{\text {P,water }}\right)\).

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