How Can The Enthalpy Change Be Determined For A Reaction In An Aqueous Solution
5.iv: Enthalpy of Reaction
- Folio ID
- 21722
Learning Objectives
- To empathise how enthalpy pertains to chemic reactions
We have stated that the alter in energy (\(ΔU\)) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is chosen pressure-volume work, (or just \(PV\) piece of work ). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in full-bodied nitric acrid. The chemical equation for this reaction is as follows:
\[ \ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(fifty) + 2NO2(k)} \nonumber\]
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will ascent as nitrogen dioxide gas is formed (Figure \(\PageIndex{1}\)). The system is performing work by lifting the piston against the downward strength exerted by the atmosphere (i.e., atmospheric pressure). We observe the corporeality of \(PV\) work done by multiplying the external pressure \(P\) by the change in book acquired past movement of the piston (\(ΔV\)). At a constant external pressure level (here, atmospheric pressure),
\[westward = −PΔV \characterization{five.iv.ii}\]
The negative sign associated with \(PV\) work washed indicates that the arrangement loses energy when the volume increases. If the volume increases at abiding pressure (\(ΔV > 0\)), the work washed by the organisation is negative, indicating that a organization has lost free energy by performing piece of work on its surround. Conversely, if the volume decreases (\(ΔV < 0\)), the work washed by the organisation is positive, which means that the environment have performed piece of work on the system, thereby increasing its energy.
The internal free energy \(U\) of a system is the sum of the kinetic energy and potential energy of all its components. Information technology is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually utilize a related thermodynamic quantity called enthalpy (\(H\)) (from the Greek enthalpein, meaning "to warm"). The enthalpy of a system is defined as the sum of its internal energy \(U\) plus the product of its pressure \(P\) and volume \(Five\):
\[H =U + PV \label{5.four.3}\]
Considering internal energy, pressure, and volume are all state functions, enthalpy is likewise a state function. So nosotros can define a change in enthalpy (\(\Delta H\)) accordingly
\[ΔH = H_{final} − H_{initial} \nonumber\]
If a chemical alter occurs at constant pressure (i.e., for a given \(P\), \(ΔP = 0\)), the change in enthalpy (\(ΔH\)) is
\[ \begin{align} ΔH &= Δ(U + PV) \\[4pt] &= ΔU + ΔPV \\[4pt] &= ΔU + PΔV \label{5.4.4} \stop{align} \]
Substituting \(q + due west\) for \(ΔU\) (Offset Law of Thermodynamics) and \(−due west\) for \(PΔV\) (Equation \(\ref{5.4.2}\)) into Equation \(\ref{5.4.4}\), we obtain
\[ \begin{align} ΔH &= ΔU + PΔV \\[4pt] &= q_p + \abolish{w} −\cancel{westward} \\[4pt] &= q_p \characterization{5.iv.5} \end{align} \]
The subscript \(p\) is used here to emphasize that this equation is truthful only for a process that occurs at constant pressure. From Equation \(\ref{five.4.5}\) we see that at constant pressure the modify in enthalpy, \(ΔH\) of the organisation, is equal to the rut gained or lost.
\[ \brainstorm{marshal} ΔH &= H_{final} − H_{initial} \\[4pt] &= q_p \label{5.iv.half dozen} \finish{marshal} \]
Merely as with \(ΔU\), considering enthalpy is a land function, the magnitude of \(ΔH\) depends on but the initial and last states of the arrangement, not on the path taken. Most important, the enthalpy change is the aforementioned even if the procedure does not occur at abiding pressure.
To observe \(ΔH\) for a reaction, measure \(q_p\).
When nosotros study free energy changes in chemical reactions, the virtually important quantity is usually the enthalpy of reaction (\(ΔH_{rxn}\)), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acrid). If estrus flows from a system to its surroundings, the enthalpy of the system decreases, and so \(ΔH_{rxn}\) is negative. Conversely, if estrus flows from the surround to a system, the enthalpy of the system increases, so \(ΔH_{rxn}\) is positive. Thus:
- \(ΔH_{rxn} < 0\) for an exothermic reaction, and
- \(ΔH_{rxn} > 0\) for an endothermic reaction.
In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic procedure. The sign conventions for heat flow and enthalpy changes are summarized in the following table:
| Reaction Type | q | ΔHrxn |
|---|---|---|
| exothermic | < 0 | < 0 (heat flows from a system to its surroundings) |
| endothermic | > 0 | > 0 (estrus flows from the surroundings to a organisation) |
If ΔH rxn is negative, so the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Effigy \(\PageIndex{2}a\)). Conversely, if ΔH rxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure \(\PageIndex{2b}\)). Two important characteristics of enthalpy and changes in enthalpy are summarized in the post-obit discussion.
Bail breaking E'er requires an input of energy; bond making ALWAYS releases energy.y.
- Reversing a reaction or a procedure changes the sign of Δ H . Ice absorbs heat when it melts (electrostatic interactions are broken), and then liquid water must release heat when it freezes (electrostatic interactions are formed):
\( \begin{matrix}
heat+ H_{ii}O(due south) \rightarrow H_{2}O(fifty) & \Delta H > 0
\end{matrix} \label{5.4.seven} \)\( \begin{matrix}
H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0
\end{matrix} \characterization{v.four.8} \)In both cases, the magnitude of the enthalpy modify is the same; only the sign is dissimilar.
- Enthalpy is an extensive property (like mass). The magnitude of \(ΔH\) for a reaction is proportional to the amounts of the substances that react. For instance, a large burn produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy alter for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation \(\ref{5.iv.9}\), which describes the reaction of aluminum with iron(3) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, ii mol of Iron, one mol of Al2O3, and 851.5 kJ of rut are produced for every 2 mol of Al and 1 mol of Atomic number 262Othree consumed:
\[ \ce{2Al(s) + Fe2O3(due south) -> 2Fe (southward) + Al2O3 (southward) } + 815.5 \; kJ \characterization{five.iv.nine} \]
Thus ΔH = −851.5 kJ/mol of Fe2Othree. We tin can also draw ΔH for the reaction equally −425.viii kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we split up −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written later on the reaction, as in Equation \(\ref{5.iv.x}\), it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
\[ \ce{ 2Al(southward) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.ten} \]
If 4 mol of Al and 2 mol of \(\ce{Fe2O3}\) react, the change in enthalpy is 2 × (−851.five kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
\[ - \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.four.6a} \]
The relationship between the magnitude of the enthalpy alter and the mass of reactants is illustrated in Example \(\PageIndex{1}\).
Example \(\PageIndex{1}\): Melting Icebergs
Certain parts of the world, such as southern California and Kingdom of saudi arabia, are short of freshwater for drinking. 1 possible solution to the trouble is to tow icebergs from Antarctica and so melt them as needed. If \(ΔH\) is 6.01 kJ/mol for the reaction at 0°C and abiding pressure:
\[\ce{H2O(s) → H_2O(l)} \nonumber\]
How much energy would be required to cook a moderately large iceberg with a mass of 1.00 1000000 metric tons (1.00 × x6 metric tons)? (A metric ton is 1000 kg.)
Given: free energy per mole of ice and mass of iceberg
Asked for: energy required to melt iceberg
Strategy:
- Calculate the number of moles of water ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
- Calculate the energy needed to cook the ice by multiplying the number of moles of ice in the iceberg past the corporeality of energy required to melt ane mol of ice.
Solution:
A Considering enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—and so we need to calculate the number of moles of ice in the iceberg and multiply that number past ΔH (+6.01 kJ/mol):
\[ \begin{marshal*} moles \; H_{2}O & = 1.00\times ten^{half-dozen} \; \cancel{\text{metric ton }} \ce{H2o} \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{\text{metric ton}}} \correct ) \left ( \dfrac{1000 \; \abolish{g}}{1 \; \abolish{kg}} \right ) \left ( \dfrac{ane \; mol \; H_{ii}O}{18.015 \; \cancel{g \; H_{ii}O}} \right ) \\[4pt] & = 5.55\times 10^{10} \; mol \,\ce{H2O} \end{align*} \]
B The energy needed to melt the iceberg is thus
\[ \left ( \dfrac{six.01 \; kJ}{\cancel{mol \; H_{two}O}} \right )\left ( 5.55 \times 10^{x} \; \cancel{mol \; H_{2}O} \right )= three.34 \times x^{11} \; kJ \nonumber \]
Because so much energy is needed to melt the iceberg, this programme would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of unlike free energy sources equivalent to the amount of energy needed to melt the iceberg are shown below.
Possible sources of the approximately \(3.34 \times ten^{xi}\, kJ\) needed to melt a \(ane.00 \times 10^6\) metric ton iceberg
- Combustion of 3.eight × 10iii ftthree of natural gas
- Combustion of 68,000 barrels of oil
- Combustion of 15,000 tons of coal
- \(ane.1 \times ten^8\) kilowatt-hours of electricity
Alternatively, we can rely on ambient temperatures to slowly melt the iceberg. The principal issue with this thought is the cost of dragging the iceberg to the desired identify.
Practice \(\PageIndex{1}\): Thermite Reaction
If 17.3 m of powdered aluminum are immune to react with excess \(\ce{Fe2O3}\), how much estrus is produced?
- Reply
-
273 kJ
Enthalpies of Reaction
Ane manner to written report the estrus captivated or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of try. Fortunately, since enthalpy is a country office, all we have to know is the initial and final states of the reaction. This allows united states of america to summate the enthalpy change for almost whatsoever conceivable chemical reaction using a relatively small-scale set of tabulated data, such as the following:
- Enthalpy of combustion (ΔHcomb) The change in enthalpy that occurs during a combustion reaction. Enthalpy changes take been measured for the combustion of virtually any substance that will fire in oxygen; these values are ordinarily reported every bit the enthalpy of combustion per mole of substance.
- Enthalpy of fusion (ΔHfus) The enthalpy modify that acompanies the melting (fusion) of ane mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for about all the elements and for virtually simple compounds.
- Enthalpy of vaporization (ΔHvap) The enthalpy modify that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for almost volatile compounds.
- Enthalpy of solution (ΔHsoln) The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent. The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent.
| Substance | ΔHvap (kJ/mol) | ΔHfus (kJ/mol) |
|---|---|---|
| argon (Ar) | 6.iii | 1.three |
| methane (CH4) | ix.two | 0.84 |
| ethanol (CH3CH2OH) | 39.3 | 7.6 |
| benzene (C6H6) | 31.0 | ten.9 |
| water (H2O) | 40.7 | 6.0 |
| mercury (Hg) | 59.0 | 2.29 |
| iron (Iron) | 340 | 14 |
The sign convention is the same for all enthalpy changes: negative if heat is released by the arrangement and positive if heat is absorbed by the system.
Enthalpy of Reaction: https://youtu.exist/z2KUaIEF9qI
Summary
For a chemical reaction, the enthalpy of reaction (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of \(ΔH_{rxn}\) are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\).
Source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.04%3A_Enthalpy_of_Reaction
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