Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Define these terms: enthalpy and enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

Short Answer

Expert verified
The heat of reaction equals the enthalpy change at constant pressure.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Define Enthalpy

Enthalpy, denoted as \( H \), is a thermodynamic quantity that represents the total heat content of a system. It is the sum of the internal energy of the system plus the product of its pressure and volume, i.e., \( H = U + PV \), where \( U \) is the internal energy, \( P \) is pressure, and \( V \) is volume. It is often used to calculate the heat changes in chemical reactions conducted at constant pressure.
02

Define Enthalpy of Reaction

The enthalpy of reaction, \( \Delta H_{reaction} \), is the change in enthalpy observed in a chemical reaction, under standard conditions. It is the difference between the enthalpy of the products and the enthalpy of the reactants. It provides information about whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).
03

Condition for Heat of Reaction = Enthalpy Change

The heat of reaction is equal to the enthalpy change of the same reaction when the reaction occurs at constant pressure and without doing non-PV work. In this case, any heat exchanged with the surroundings is equivalent to the change in enthalpy, as no other forms of work affect the system's energy.

Key Concepts

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

Thermodynamic Quantity
In the world of thermodynamics, a 'thermodynamic quantity' is a property that helps us understand the state and behavior of a system. Enthalpy, denoted by the symbol \( H \), is one such quantity. It provides vital insight into how heat is involved in various processes, particularly chemical reactions. This quantity is a combination of the internal energy, \( U \), plus the product of pressure (\( P \)) and volume (\( V \)). Thus, \( H = U + PV \).

This equation suggests that enthalpy can be thought of as the total heat content of a system under constant pressure. It helps predict whether a process will release or absorb heat, guiding us in analyzing the energetics of reactions. By understanding enthalpy, we can better gauge how energy is exchanged in everyday processes, from heating our homes to running engines.
Enthalpy of Reaction
When we discuss the 'enthalpy of reaction,' we refer specifically to the change in enthalpy during a chemical reaction. This term is written as \( \Delta H_{reaction} \). It represents the difference in heat content between the products and reactants when a reaction is conducted under standard conditions.

The value of \( \Delta H_{reaction} \) helps us determine the thermal nature of the reaction:
  • If \( \Delta H_{reaction} \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings.
  • If \( \Delta H_{reaction} \) is positive, the reaction is endothermic, absorbing heat from the surroundings.
This knowledge is useful for predicting how temperature and heat will affect the reaction environment, influencing everything from industrial processes to biological systems.
Constant Pressure
In thermodynamics, many reactions are analyzed under 'constant pressure,' a condition where the pressure in the system does not change as the reaction proceeds. This scenario is common in open systems exposed to the atmosphere, like laboratory reactions or natural processes occurring outdoors.

Under constant pressure, the heat absorbed or released by a chemical reaction corresponds directly to the change in enthalpy, \( \Delta H \). Thus, the heat of reaction equals the enthalpy change. The reaction can then be simplified to focus solely on changes in heat, without needing to account for work done by the system involving changes in volume.

Operating under constant pressure allows for straightforward thermodynamic calculations as it aligns closely with many real-world scenarios. It simplifies the analysis and is foundational in studying reactions, providing accurate predictions about the energy changes involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

From the following heats of combustion, \(\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$ \mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Br}_{2}(g)\), \(\mathrm{Br}_{2}(l) ;(\mathrm{b}) \mathrm{I}_{2}(g), \mathrm{I}_{2}(s)\).

Explain the cooling effect experienced when ethanol is rubbed on your skin, given that \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \quad \Delta H^{\circ}=42.2 \mathrm{~kJ} / \mathrm{mol}\)

Suggest ways (with appropriate equations) that would allow you to measure the \(\Delta H_{\mathrm{f}}^{\circ}\) values of \(\mathrm{Ag}_{2} \mathrm{O}(s)\) and \(\mathrm{CaCl}_{2}(s)\) from their elements. No calculations are necessary.

For reactions in condensed phases (liquids and solids), the difference between \(\Delta H\) and \(\Delta U\) is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that \(\Delta H\) and \(\Delta U\) can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth's surface. Calculate \(\Delta H-\Delta U\) for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 50,000 atm. The densities of graphite and diamond are \(2.25 \mathrm{~g} / \mathrm{cm}^{3}\) and \(3.52 \mathrm{~g} / \mathrm{cm}^{3},\) respectively.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free