/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 What is the difference between \... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 16

What is the difference between \(\Delta H\) and \(\Delta E ?\)

Short Answer

Expert verified
The difference between \(\Delta H\) (change in enthalpy) and \(\Delta E\) (change in internal energy) depends on the change in the product of pressure and volume during a process or reaction. While \(\Delta H\) represents the total energy of a system, including both its internal energy and the energy related to its pressure and volume, \(\Delta E\) refers only to the total energy stored in a system. The relationship between \(\Delta H\) and \(\Delta E\) is given by: \(\Delta H = \Delta E + \Delta (PV)\). The difference between the two depends on the specific conditions of the process or reaction.

Step by step solution

01

Define Enthalpy (H) and Internal Energy (E)

Enthalpy (H) is a thermodynamic property that represents the total energy of a system, including both its internal energy (E) and the energy related to its pressure and volume. It is defined as: \(H = E + PV\) Where, \(H\) is enthalpy, \(E\) is internal energy, \(P\) is pressure, and \(V\) is volume. Internal Energy (E) is the total energy stored in a system, including kinetic energy, potential energy and other forms of energy (including molecular bonds, electron configuration, etc.).
02

Define the Changes in Enthalpy (\(\Delta H\)) and Internal Energy (\(\Delta E\))

The change in enthalpy (\(\Delta H\)) represents the difference in the enthalpy of a system before and after a process or reaction has taken place. Similarly, the change in internal energy (\(\Delta E\)) represents the difference in the internal energy of a system before and after a process or reaction has occurred. Mathematically, these changes are represented as: \(\Delta H = H_{final} - H_{initial}\) \(\Delta E = E_{final} - E_{initial}\)
03

Explain the Relationship between \(\Delta H\) and \(\Delta E\)

From the definition of enthalpy, we know that: \(H = E + PV\) Taking the change in enthalpy and change in internal energy, we can write: \(\Delta H = \Delta E + \Delta (PV)\) This equation shows that the difference between the changes in enthalpy (\(\Delta H\)) and internal energy (\(\Delta E\)) depends on the change in the product of pressure and volume during a process or reaction. In some cases, \(\Delta H\) and \(\Delta E\) may be equal or approximately equal, such as when the process or reaction occurs under constant volume or constant pressure conditions, and the system does not perform any work. In general, the difference between the two depends on the specific conditions of the process or reaction.

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Most popular questions from this chapter

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion \(=-802 \mathrm{kJ} /\) \(\mathrm{mol} \mathrm{CH}_{4}\) in the bomb. The temperature changed by \(10.8^{\circ} \mathrm{C} .\) a. What is the heat capacity of the bomb? b. A 12.6 -g sample of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) produced a temperature increase of \(16.9^{\circ} \mathrm{C}\) in the same calorimeter. What is the energy of combustion of acetylene (in \(\mathrm{kJ} / \mathrm{mol} )\) ?

Given: $$ \begin{array}{ll}{2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s)} & {\Delta H^{\circ}=-288 \mathrm{kJ}} \\\ {\mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuO}(s)+\mathrm{Cu}(s)} & {\Delta H^{\circ}=11 \mathrm{kJ}}\end{array} $$ Calculate the standard enthalpy of formation \(\left(\Delta H_{f}^{\circ}\right)\) for \(\mathrm{CuO}(s) .\)

A balloon filled with 39.1 moles of helium has a volume of 876 \(\mathrm{L}\) at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of 998 \(\mathrm{L}\) , the pressure remaining constant. Calculate \(q, w,\) and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{mol.} )\)

Using the following data, calculate the standard heat of formation of ICl \((g)\) in \(\mathrm{kJ} / \mathrm{mol} :\) $$\begin{array}{ll}{\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} & {\Delta H^{\circ}=242.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{I}(g)} & {\Delta H^{\circ}=151.0 \mathrm{kJ}} \\ {\mathrm{ICl}(g) \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g)} & {\Delta H^{\circ}=211.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)} & {\Delta H^{\circ}=62.8 \mathrm{kJ}}\end{array}$$
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