Define these terms: enthalpy, 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
Enthalpy is a property of a system, calculated as the sum of its internal energy and the product of its pressure and volume. Enthalpy of reaction is the change in enthalpy during a chemical reaction, calculated as the difference of enthalpy of the products and reactants. The heat of a reaction equals enthalpy change when the reaction occurs under constant pressure and no non-pressure-volume work is done on or by the system.

Step by step solution

01

Define Enthalpy

Enthalpy (\(H\)) is a thermodynamic property of a system. It is the sum of the internal energy (\(U\)) of the system and the product of its pressure (\(P\)) and volume (\(V\)). This can be represented by the formula: \(H = U + PV\).
02

Define Enthalpy of Reaction

Enthalpy of reaction, or heat of reaction, is the change in enthalpy of a chemical reaction that occurs at a constant pressure. It is represented as \(ΔH\) and calculated as the difference in enthalpy of the products and reactants: \(ΔH = H_{products} - H_{reactants}\).
03

Conditions for Heat of Reaction Equal to Enthalpy Change

The heat of a reaction (\(q\)) is equal to the enthalpy change of the reaction (\(ΔH\)) under conditions of constant pressure and no non-pressure-volume work being done on or by the system. When these conditions are met, we can write: \(q_p = ΔH\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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

I he enthaipy of combustion benzo1c acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .)\)

A gas expands and does \(P-V\) work on the surroundings equal to \(325 \mathrm{~J}\). At the same time, it absorbs \(127 \mathrm{~J}\) of heat from the surroundings. Calculate the change in energy of the gas.

A quantity of 0.020 mole of a gas initially at \(0.050 \mathrm{~L}\) and \(20^{\circ} \mathrm{C}\) undergoes a constant-temperature expansion until its volume is \(0.50 \mathrm{~L}\). Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (c) If the gas in (b) is allowed to expand unchecked until its pressure is equal to the external pressure, what would its final volume be before it stopped expanding, and what would be the work done?

Consider the reaction $$ \begin{aligned} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow & 2 \mathrm{NH}_{3}(g) \\\ \Delta H_{\mathrm{rxn}}^{\circ} &=-92.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ If 2.0 moles of \(\mathrm{N}_{2}\) react with 6.0 moles of \(\mathrm{H}_{2}\) to form \(\mathrm{NH}_{3},\) calculate the work done (in joules) against a pressure of 1.0 atm at \(25^{\circ} \mathrm{C}\). What is \(\Delta E\) for this reaction? Assume the reaction goes to completion.

A 2.10 -mole sample of crystalline acetic acid, initially at \(17.0^{\circ} \mathrm{C}\), is allowed to melt at \(17.0^{\circ} \mathrm{C}\) and is then heated to \(118.1^{\circ} \mathrm{C}\) (its normal boiling point) at 1.00 atm. The sample is allowed to vaporize at \(118.1^{\circ} \mathrm{C}\) and is then rapidly quenched to \(17.0^{\circ} \mathrm{C}\), so that it recrystallizes. Calculate \(\Delta H^{\circ}\) for the total process as described.

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