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Acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) is made industrially by the reaction of methanol and carbon monoxide. \(\mathrm{CH}_{3} \mathrm{OH}(\ell)+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(\ell)\) $$ \Delta_{\mathrm{r}} H^{\circ}=-135.3 \mathrm{~kJ} / \mathrm{mol} $$ If you produce \(1.00 \mathrm{~L}\) acetic acid \((d=1.044 \mathrm{~g} / \mathrm{mL})\) by this reaction, calculate how much energy is transferred out of the system.

Short Answer

Expert verified
Approximately -2350.71 kJ of energy is released.

Step by step solution

01

Calculating the mass of acetic acid

Use the formula for density to find the mass of acetic acid produced. The density formula is given by \( \text{mass} = \text{density} \times \text{volume} \). Here, the density of acetic acid is 1.044 g/mL, and the volume is 1000 mL (since 1 L = 1000 mL). Therefore, \( \text{mass} = 1.044 \, \text{g/mL} \times 1000 \, \text{mL} = 1044 \, \text{g} \).
02

Converting mass to moles

Calculate the moles of acetic acid by using its molar mass. The molar mass of \( \text{CH}_3 \text{COOH} \) is: Carbon (12.01 g/mol) \(\times 2\) + Hydrogen (1.01 g/mol) \(\times 4\) + Oxygen (16.00 g/mol) \(\times 2\) = 60.05 g/mol. Therefore, moles of acetic acid = \( \frac{1044 \, \text{g}}{60.05 \, \text{g/mol}} \approx 17.38 \, \text{mol} \).
03

Calculating the energy transferred

Use the provided enthalpy change \( \Delta_{\mathrm{r}} H^{\circ} = -135.3 \, \text{kJ/mol} \) to find the total energy by multiplying by the number of moles: Total energy released = \(-135.3 \, \text{kJ/mol} \times 17.38 \, \text{mol} \approx -2350.71 \, \text{kJ} \). The negative sign indicates energy is released from the system.

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Key Concepts

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

Industrial Chemical Reactions
Industrial chemical reactions are the backbone of manufacturing processes that create essential compounds for everyday life. One significant example is the production of acetic acid, a key component used in various industries. This acid, often manufactured via the carbonylation of methanol, is a classic depiction of how industrial processes are designed for efficiency and scalability.
  • The reaction combines methanol (\( \mathrm{CH}_3 \mathrm{OH} \)) and carbon monoxide (\( \mathrm{CO} \)) to yield acetic acid (\( \mathrm{CH}_3 \mathrm{COOH} \)).
  • This process is not only efficient but also showcases how different reactants are methodically transformed into useful products.
  • Critical to this process is the management of conditions such as temperature and pressure to maximize output and safety.
Understanding these reactions involves both chemistry and engineering, balancing cost-efficiency with product demand.
Enthalpy Change Calculations
In chemical reactions, energy changes are fundamental. Enthalpy, symbolized as \( \Delta H \), refers to the heat change at constant pressure. For the acetic acid production reaction, the given enthalpy change is \( \Delta_{\mathrm{r}} H^{\circ}=-135.3 \, \text{kJ/mol} \). Here's what this means:
  • The negative sign indicates the reaction is exothermic, releasing heat to the surroundings.
  • To find the total energy released during acetic acid formation, multiply this enthalpy change by the number of moles produced.
  • For instance, if 17.38 mol of acetic acid are produced, the total energy released would be calculated as \( -135.3 \, \text{kJ/mol} \times 17.38 \, \text{mol} = -2350.71 \, \text{kJ} \).
This calculation is crucial in industries as it affects how heat is managed in an industrial setup, impacting safety and energy efficiency.
Density and Molar Mass Concepts
Density and molar mass are essential concepts that help connect macroscopic and molecular scale insights in chemistry. Density, often expressed in \( \text{g/mL} \) or \( \text{g/cm}^3 \), is a measure of mass per unit volume. Given acetic acid has a density of 1.044 \( \text{g/mL} \), this information is crucial in calculating how much mass is in a specific volume.
  • By using the density equation \( \text{mass} = \text{density} \times \text{volume} \), you determine the mass of acetic acid from 1 L (or 1000 mL) to be 1044 g.
  • Next, understanding molar mass allows us to convert this mass into moles. For acetic acid, its molar mass is calculated by summing atomic masses: 12.01 for Carbon, 1.01 for Hydrogen, and 16.00 for Oxygen resulting in 60.05 \( \text{g/mol} \).
  • With these values, convert mass to moles: \( \frac{1044 \, \text{g}}{60.05 \, \text{g/mol}} \approx 17.38 \, \text{mol} \).
These measurements and conversions are foundational in chemistry, assisting in predicting how substances react and adapt in different scenarios.

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Most popular questions from this chapter

The specific heat capacity of copper is \(0.385 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\), whereas it is \(0.128 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) for gold. Assume you place \(100 . \mathrm{g}\) of each metal, originally at \(25^{\circ} \mathrm{C},\) in a boiling water bath at \(100^{\circ} \mathrm{C}\). If energy is transferred to each metal at the same rate, determine which piece of metal will reach \(100^{\circ} \mathrm{C}\) first.

The freezing point of mercury is \(-38.8{ }^{\circ} \mathrm{C}\). Calculate what quantity of energy, in joules, is released to the surroundings if \(1.00 \mathrm{~mL}\) mercury is cooled from \(23.0^{\circ} \mathrm{C}\) to \(-38.8^{\circ} \mathrm{C}\) and then frozen to a solid. (The density of liquid mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\). Its specific heat capacity is \(0.140 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\) and its heat of fusion is \(11.4 \mathrm{~J} \mathrm{~g}^{-1}\).)

A coffee cup calorimeter can be used to investigate the "cold pack reaction," the process that occurs when solid ammonium nitrate dissolves in water: $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) $$ Suppose \(25.0 \mathrm{~g}\) solid \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) at \(23.0{ }^{\circ} \mathrm{C}\) is added to 250\. \(\mathrm{mL} \mathrm{H}_{2} \mathrm{O}\) at the same temperature. After all of the solid dissolves, the temperature is measured to be \(15.6^{\circ} \mathrm{C}\). Calculate the reaction enthalpy for the cold pack reaction. (Assume that the specific heat capacity of the solution is the same as for water.) Is the reaction endothermic or exothermic?

Explain in your own words why it is useful in thermodynamics to distinguish a system from its surroundings.

What is the value of the standard formation enthalpy for any element under standard conditions?

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