Many chemical calculations are based on a law developed by the German scientist. is 1840.^[1][2] This law states that the value of the heat of reaction (the change in enthalpy, or the change in the standard enthalpy of formation of a compound) for any chemical reaction under constant pressure is equal to a constant quantity, whether the reaction takes place in one step or several steps. This means that the reaction heat depends only on the properties of the reactants and products of the reaction, that is, on the initial and final states of the reaction and is not affected by the path that the reaction takes. The importance of this law is evident in the possibility of calculating the reaction heat, also for reactions that cannot be measured by experimental methods because they occur so slowly that they cannot be studied, or side reactions occur that produce unwanted materials in addition to the required materials.

Hessian thermal law text[عدل]

The change in enthalpy

$\displaystyle \Delta H$

The occurrence during a completed process is the sum of the enthalpy changes of all successive steps of the process.

We conclude from this law that the enthalpy of the reaction does not change with the change of the course of the reaction, but depends only on the initial state and the final state of the reaction.

For example[عدل]

Graphite can be combusted directly, producing carbon dioxide (1), or through several steps, first producing carbon monoxide (2), (3). Then it produces carbon dioxide. So the enthalpy of the overall reaction is Δ_RH is the same in both cases.

$\displaystyle \beginmatrix(1)\;&\mathrmC(Graphite)&+&\mathrmO_2(g) &\rightarrow&\mathrmCO_2(g). ) &\;\;\Delta_RH_1=-393\,\mathrm kJ\cdot mol^-1 \;\;\\\\(2)\;&\ mathrm C(Graphite) &+&\frac 12\mathrm O_2(g) &\rightarrow &\mathrm CO(g) &\;\;\Delta _RH_2=-111\,\mathrm kJ\cdot mol^-1 \\\\(3)&\mathrm CO(g) &+&\frac 1 2\mathrm O_2(g) &\rightarrow &\mathrm CO_2(g) &\;\;\Delta_RH_3=-282\ ,\mathrm kJ\cdot mol^-1\endmatrix$

The enthalpy of the complete reaction in both cases is equal and is -393 kJ/mol. The negative sign of entropy means that this reaction is exothermic.

mathematical formula[عدل]

Hess’s law makes it easy to calculate the enthalpy change (ΔH) during a reaction when it cannot be measured directly. We calculate it with several simple mathematical operations using the relevant reaction equation, and we also use some known enthalpy values ​​that were set before.

According to the previous example, a complete reaction can be split, such as the combustion of graphite (carbon) to obtain carbon dioxide. Hess’s law says that the total enthalpy change for a reaction is equal to the enthalpy changes for each step of the reaction. That is, the ΔH of one of the reaction steps can be calculated by knowing the difference in the heat of formation of a chemical compound (product) and the heat of formation of the substances involved in the reaction:

$\displaystyle \Delta H_reaction^\ominus =\sum \Delta H_\mathrm f \,(products)^\ominus -\sum \Delta H_\mathrm f \ ,(reactants)^\ominus$

Where is the tag ^o Mean values ​​are at standard conditions for the material (see standard enthalpy of formation).

Its relationship to entropy and free energy[عدل]

Hess’s law can be formulated to include entropy changes and Gibbs free energy, which are also functions of the state of a system. An example of such an application is the Borewell thermodynamic cycle, which takes advantage of easy measurements of chemical equilibrium and redox potentials to determine the Gibbs free energy that cannot be measured directly. Further relate the ΔG values^o From the Bordwell cycle with values ​​of enthalpy change ΔH^o Which we get from Hess’s law we can set the entropy change also mathematically.

For Gibbs free energy we get:

$\displaystyle \Delta G_reaction^\ominus =\sum \Delta G_\mathrm f \,(products)^\ominus -\sum \Delta G_\mathrm f \ ,(reactants)^\ominus$

The situation is different for entropy as entropy can be measured and its absolute values ​​set (see standard molar entropy). We use the absolute entropy values ​​of the input and output materials of the reaction:

$\displaystyle \Delta S_reaction^\ominus =\sum S_(products)^\ominus -\sum S_(reactants)^\ominus$

Another example[عدل]

Hess’s law enables us to calculate the enthalpy change (ΔH) of a reaction even if it cannot be determined empirically. The method includes performing some simple calculations on the reaction equation with the use of known values ​​of the standard enthalpies of formation for the input and output materials of the reaction.

Combining several chemical equations, a result of the reaction can be obtained. If the enthalpy change for each equation is known, we get the outcome of the enthalpy change for the overall reaction. If the total enthalpy change is negative, the sign (ΔH_net < 0), the reaction is exothermic. If ΔH is positive, the reaction is endothermic. Entropy plays an important role in knowing whether a chemical reaction proceeds automatically or not, as some reactions in which the enthalpy change is positive, and yet the reaction proceeds automatically by itself.

Hess’s law says that enthalpy changes can be summed. That is, ΔH for one of the reactions can be calculated as the difference between the standard enthalpy of formation of the reaction products minus the standard enthalpy of formation of the materials involved in the reaction, according to the equation: .

$\displaystyle \Delta H_reaction^\ominus =\sum \Delta H_\mathrm f \,(products)^\ominus -\sum \Delta H_\mathrm f \ ,(reactants)^\ominus$

where the symbol indicates ^o

on the standard case.

Now we want to calculate ΔH_f For interaction:

• (2B (s) + (3/2) O₂ (g) → B₂O₃ (s

The equation says we want to add boron (solid) to oxygen (gas) in the quantities available to get boron oxide (solid).

This chemical equation can be the result of the following reactions:

• B₂O₃ (s) + 3H₂O(g) → 3O₂ (g) + B₂H₆ (g) (ΔH = 2035 kJ/mol)
• (H₂O (l) → H₂O (g) (ΔH = 44 kJ/mol
• (H₂ (g) + (1/2)O₂ (g) → H₂O (l) (ΔH = -286 kJ/mol
• (2B (s) + 3H₂ (g) → B₂H₆ (g) (ΔH = 36 kJ/mol

The signs of enthalpy change for the first, second, and fourth reactions indicate that the reaction is an endothermic reaction. The third reaction with a negative sign (ΔH = -286 kJ/mol) indicates an exothermic reaction. Noting the phase of the materials, whether they are solid, liquid or gas (phase transitions are associated with phase transition enthalpy).

We do some multiplication and reverse the reaction equations (so the enthalpy change sign is reversed), and we get:

• B₂H₆ (g) + 3O₂ (g) → B₂O₃ (s) + 3H₂O (g) (ΔH = -2035 kJ/mol)
• (3H₂Or (g) → 3H₂O (l) (ΔH = -132 kJ/mol
• (3H₂Or (l) → 3H₂ (g) + (3/2) O₂ (g) (ΔH = 858 kJ/mol
• (2B (s) + 3H₂ (g) → B₂H₆ (g) (ΔH = 36 kJ/mol

And we add the four equations with crossing out the similar amounts on both sides of the equation, and we get:

• (2B (s) + (3/2) O₂ (g) → B₂O₃ (s) (ΔH = -1273 kJ/mol

This result says that the formation of boron oxide from the reaction of boron and oxygen is an exothermic reaction (that is, it is supposed to proceed spontaneously), and a heat of 1273 kJ / mol spreads from it.)

See also[عدل]

• chemical energy (chemistry)
• Standard enthalpy of formation
• Sublimation enthalpy
• Heater
• reaction equation
• Benson’s theory
• entropy
• crystallization
• Calorimetry
• Masaar
• adiabatic process
• Thermodynamic databases for pure substances
• standard molar entropy

• Physics portal
• Physical chemistry portal
• Chemistry portal

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