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Apply The molar heat of vaporization of ammonia is 23.3 kJ/mol. What is the molar heat of condensation of ammonia?

Short Answer

Expert verified
The molar heat of condensation of ammonia is \(-23.3\) kJ/mol.

Step by step solution

01

Identify the given information

The given molar heat of vaporization of ammonia is 23.3 kJ/mol.
02

Apply the relationship between molar heat of vaporization and condensation

Molar heat of condensation is equal in magnitude but opposite in sign to the molar heat of vaporization. Therefore, the molar heat of condensation will have a negative sign: \(-23.3\) kJ/mol.
03

State the answer

The molar heat of condensation of ammonia is \(-23.3\) kJ/mol.

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Key Concepts

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

Molar Heat of Condensation
When we talk about the molar heat of condensation, we're referring to the amount of heat released when one mole of a substance transitions from a gas to a liquid. This process is exothermic, meaning it releases energy into the surroundings. Ammonia, for example, releases 23.3 kJ of energy per mole during condensation, the negative sign indicating energy release.

Understanding the molar heat of condensation is crucial for processes like distillation or refrigeration, where controlling phase changes is essential. It shows us how much energy needs to be removed to condense a gas into a liquid without changing the temperature.
  • Molar Heat of Vaporization: Energy absorbed during vaporization
  • Molar Heat of Condensation: Energy released during condensation
Keep in mind that the values for condensation and vaporization are equal in magnitude but opposite in sign.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and energy. It helps us understand how and why energy transfers occur during phase changes, like vaporization and condensation. There are four main laws of thermodynamics, but here we focus on the principles that govern energy transfer.

In the case of ammonia, when it condenses, thermodynamics ensures that energy conservation is maintained. The energy released during condensation is directly related to the energy absorbed during vaporization. These transformations are governed by enthalpy, which is a central concept of thermodynamics.
  • First Law of Thermodynamics: Energy cannot be created or destroyed.
  • Energy transfer takes different forms like heat or work.
  • Phase changes rely heavily on these principles to proceed sustainably.
By understanding these laws, we can predict and control conditions in industrial processes.
Enthalpy Change
Enthalpy change is a measurement of energy change in reactions at constant pressure, crucial for understanding heat changes. When ammonia vaporizes or condenses, there's a change in enthalpy. This change depends on whether the process is absorbing or releasing energy.

For ammonia, the enthalpy change in vaporization is positive, as energy is absorbed to convert liquid to gas. Conversely, condensation has a negative enthalpy change, because energy is released. Knowing the enthalpy change helps in calculating energy requirements or releases for various chemical reactions and processes.
  • Enthalpy (\( \Delta H \)): Total heat content of a system.
  • Positive \( \Delta H \): Endothermic (heat absorbed).
  • Negative \( \Delta H \): Exothermic (heat released).
Understanding enthalpy change is key for designing energy-efficient systems in chemistry and engineering.

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Most popular questions from this chapter

A fruit-and-oatmeal bar contains 142 nutritional Calories. Convert this energy to calories.

How many joules of heat are lost by 3580 kg of granite as it cools from 41.2°C to -12.9°C? The specific heat of granite is 0.803 J/(g·°C).

Use Hess's law and the following thermochemical equations to produce the thermochemical equation for the reaction \(\mathrm{C}(\mathrm{s}, \text { diamond }) \rightarrow \mathrm{C}(\mathrm{s}, \text { graphite).What is }\) \(\Delta H\) for the reaction? a. \(C(s, \text { graphite })+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-394 \mathrm{kJ}\) b. \(C(s, \text { diamond })+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-396 \mathrm{kJ}\)

Design an Experiment Describe a procedure you could follow to determine the specific heat of a 45-g piece of metal.

Challenge Two enthalpy of formation equations, a and b, combine to form the equation for the reaction of nitrogen oxide and oxygen. The product of the reaction is nitrogen dioxide: \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{rxn}}^{\circ}=-58.1 \mathrm{kJ}\) a. \(\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=91.3 \mathrm{kJ}\) b. \(\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=?\) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for Equation \(\mathrm{b}\) ?

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