Calculating Enthalpy: The Four Best Methods | ChemTalk

Core Concepts

In this article, you will learn about the most authoritative methods of calculating heat content of chemical reactions. After reading, you will be able to calculate a reaction ’ second heat content in a variety of coarse situations .

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What is Enthalpy?

Chemists and physicists define heat content as the heat released by a procedure under constant press. Enthalpy is a “ submit variable, ” mean that a arrangement ’ south change in heat content only depends on the initial and final states of the organization, rather than the particular way taken between the two states. In chemistry, heat content has the most application when understanding the thermodynamics of a chemical reaction. In particular, the sign of a reaction ’ s change of heat content yields crucial information. For case, if a chemical reaction ’ s change of heat content has a positive bless, chemists call the reaction “ endothermic. ” Endothermic reactions involve reactants absorbing heat from the environment. Generally, endothermic reactions are either not thermodynamically favored, or entirely favored at high temperatures.

By contrast, if a reaction ’ s change of heat content has a minus sign, chemists call the reaction “ exothermic. ” Exothermic reactions rather involve reactants releasing heating system into the environment. Generally, exothermic reactions are either constantly thermodynamically favored, or alone favored at moo temperatures. due to the thermodynamic importance of heat content, chemists have developed many methods of calculating heat content of reaction. In this article, we will cover the four most authoritative methods, each using different data or resources :

  • Bond Enthalpies
  • Enthalpies of Formation
  • Specific Heat
  • Reaction Equilibrium

As you will see, different situations require different methods to get heat content, depending on what data you already know about your reactants and reaction .

Calculating Enthalpy with Enthalpies of Formation

The most straightforward method acting of calculating the heat content of a reaction involves using what chemists call “ enthalpies of formation. ” In short, each molecule has a feature “ heat content of formation, ” which is basically the change in heat content involved with assembling the molecule from its respective atoms. For more information about heat content of formation, check out this article. If you know the enthalpies of formation for each atom in your reaction, then you can calculate the reaction ’ s overall enthalpic change. To do this, you need to use an heat content convention known as Hess ’ s law : \begin{gather*} {\Delta H_{rxn} = \sum \Delta H_{f,products} - \sum \Delta H_{f,reactants}} \end{gather*} first, you multiply each molecule ’ s heat content of constitution with its stoichiometric coefficient in the reaction equation. Second, you add the breed enthalpies of the products and those of the reactants. Third, you subtract the blend enthalpies of the products with that of the reactants, yielding the overall reaction heat content. \begin{gather*} {aA + bB \rightarrow cC + dD} \\ {\Delta H_{rxn} = \left( \left( c*\Delta H_{f,C}\right) + \left( d*\Delta H_{f,D}\right) \right) - \left \left( a*\Delta H_{f,A}\right) + \left( b*\Delta H_{f,B}\right) \right)} \end{gather*} Chemists only know a atom ’ s heat content of constitution through experiment, specifically calorimetry. frankincense, you can alone use enthalpies of formation when you are dealing with companion molecules at well-studied temperatures ( such as 25°C and 37°C ). unfamiliar molecules and conditions tend to lack readily available data on their heat content of formation .

Example Calculation

Calculate the heat content of the follow reaction at 25°C : \right ) _2 \begin{gather*} {Ca\left( OH \right)_{2} + \left( NH_{4} \right)_{2} CO_{3} \rightarrow CaCO_{3} + 2NH_{4} OH} \end{gather*}

  ∆Hf  (kJ/mol)
Ca(OH)2 -1003.
(NH4)2CO3 -412.1
CaCO3 -1207
NH4OH -362.5

\begin{align*} {\Delta H_{rxn} &= \left( \Delta H_{f,CaCO_{3}} + \left( 2\Delta H_{f,NH_{4}OH}\right) \right) - \left( \Delta H_{f,Ca\left( OH \right)_{2}} + \Delta H_{f,\left( NH_{4} \right)_{2} CO_{3}} \right)} \\ {\Delta H_{rxn} &= \left( -1207 + \left( 2*-362.3\right) \right) - \left( -1003 + -412.1 \right) = -516.9 kJ/mol} \end{align*} This reaction is exothermic .

Calculating Enthalpy with Bond Enthalpies

Another patronize method of calculating a reaction ’ second heat content involves using adhesiveness enthalpies. specifically, each chemical bond between two atoms involves some enthalpic change in its constitution. To learn more about bond enthalpies, check out this article. You will need to implement a exchangeable method acting to calculate heat content of reaction as Hess ’ s Law, but with important differences. foremost, you need to find the alliance enthalpies of each bond formed or broken over the course of the chemical reaction. Any bonds that remain unaltered by the reaction don ’ deoxythymidine monophosphate factor into the reaction ’ s overall heat content. Second, you add all the enthalpies of the break dance bonds and those of the form bonds. Third, you subtract the total enthalpies of the broken bonds by that of the form bonds, yielding the overall reaction heat content. \begin{gather*} {\Delta H_{rxn} = \sum \Delta H_{Broken} - \sum \Delta H_{Formed}} \end{gather*} Unlike Hess ’ s Law, this method doesn ’ t involve “ products minus reactants, ” but rather “ bonds break subtraction bonds formed. ” This is significant to point out because it basically is the opposite : newly formed bonds are found in the products, while break bonds are found in the reactants. besides, using bond enthalpies works for more reactions than enthalpies of formation. With this, you can calculate reaction enthalpies for improper products and reactants, so long as the chemical reaction involves conversant bonds. however, you are however limited by the acquaintance of the conditions, and that of the bonds .

Example Calculation

Calculate the heat content of the follow chemical reaction at 25°C :elimination reaction between 2-chlorobutane and tertbutoxide. useful example for calculating enthalpy of reaction

  ∆HBond  (kJ/mol)
C-H 412
C-C 348
C-Cl 338
C=C 612
O-H 463

\begin{gather*} {\Delta H_{rxn} = \left( \Delta H_{C-H} + \Delta H_{C-C} + \Delta H_{C-Cl} \right) - \left( \Delta H_{C=C} + \Delta H_{O-H} \right)} \end{gather*} note : When entirely one of the bonds in an alkene break, you can think of it as a C=C bind demote and a C–C adhesiveness form. \begin{align*}  {\Delta H_{rxn} &= \left( 412 + 348 +338 \right) - \left( 612 + 443 \right) = 43 kJ/mol} \end{align*} This reaction is endothermic .

Calculating Enthalpy with Specific Heat

Another dim-witted method of calculating chemical reaction heat content involves using a kernel ’ mho specific heat. Each substance has a place called “ specific hotness ” which indicates the total of energy to raise the means ’ mho temperature. If you ’ d like to learn more about specific heat, check out this article. This approach involves measuring the heat exchange by a reaction, making it far more send and experimental than the first two. Since chemical reactions by definition involve some substances changing into others, the specific heats of the products and reactants don ’ triiodothyronine affect the reaction heat content. alternatively, we can look at the specific heat of the reaction environment. If we isolate a reaction into a vessel, we can measure the heat given off by the reaction by monitoring the temperature change of the vent in the vessel. In this event, air travel serves as our reaction medium, with a specific heat of 1.01 J/ ( g°C ). If alternatively we observe a reaction in an aqueous solution, water serves as our medium, with a specific heat of 4.18 J/ ( g°C ). To find heat, we need to use the come convention : \begin{align*} {q=mC\Delta T} \end{align*} q : Heat absorbed or released from reaction medium ( in Joules ) meter : mass of the reaction medium ( in grams ) hundred : specific hotness of the reaction metier ( in J/ ( g°C ) )

∆T : change in temperature of reaction medium ( in degrees Celsius ) In most cases, the quantity of heat exchanged by the chemical reaction medium equals the heat content of reaction. The only exceptions are when the conditions of “ changeless pressure ” aren ’ thymine met. however, the pressure of the system would only change if the reaction involved gases, and there existed inadequate moles of gasoline between the products and reactants .

Example Calculation

One mole each of NaOH and HCl react to form NaCl and H2O in 1.00L ( 1000g ) of urine. The urine ’ sulfur temperature then increases by 13.7°C. Find the heat content of reaction. \begin{gather*} {NaOH + HCl \rightarrow NaCl + H_{2}O} \\ {q=mC\Delta T = \left(1000g\right) \left(4.18 \frac{J}{g\degree C}\right) \left(13.7 \degree C\right) = 57.2 kJ/mol} \end{gather*} eminence : since we calculated the heat absorbed by the body of water, we need to flip the signboard to get the heat given off by the chemical reaction. \begin{gather*} {\Delta H_{rxn} = -q = -57.2 kJ/mol} \end{gather*} This reaction is exothermic .

Calculating Enthalpy with Equilibrium Constants

You can besides solve for reaction heat content using the chemical reaction ’ second equilibrium constants, which are values that represent the balance proportions of products and reactants. If you want to learn more about chemical equilibrium constants, then take a look at this article. This method represents another experimental way of calculating heat content. specifically, it takes advantage of the relationship between the thermodynamic characteristics of a reaction with its equilibrium dynamics. Depending on your chemical reaction species, equilibrium concentrations can be easily measured using spectrophotometry, yielding easy enthalpic information. To find heat content using balance constants, you need to measure the balance concentrations of products and reactants at two unlike temperatures. then, you can find the heat content of reaction using what chemists call the Van ’ t Hoff heat content equation : \begin{gather*} {lnK_{2} - lnK_{1} = \left( \frac{-\Delta H}{R} \right) \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)} \end{gather*} K1 : balance constant at temperature 1 K2 : balance constant at temperature 2 T1 : temperature 1 ( in degrees Kelvin ) T2 : temperature 2 ( in degrees Kelvin ) r : Ideal Gas Constant ( 8.314 J/molK )

Example Calculation

You have a flask 1.0M of reactant A and 2.0M of reactant B, and you observe the play along reaction : \begin{gather*} {A + 2B \rightarrow C} \end{gather*} At 25°C, you observe the come chemical equilibrium concentrations : \begin{align*} {[A] &= 0.70M} \\ {[B] &= 1.4M} \\ {[C] &= 0.30M} \end{align*} At 50°C, you observe the comply equilibrium concentrations : \begin{align*} {[A] &= 0.44M} \\ {[B] &= 0.88M} \\ {[C] &= 0.54M} \end{align*} Calculate the heat content of reaction. \begin{gather*} {T_{1} = 298K} \\ {T_{2} = 323K} \\ {K_{1} = \frac{0.30M}{0.70M \left( 1.4M \right)^2} = 0.39} \\ {K_{1} = \frac{0.54M}{0.44M \left( 88M \right)^2} = 1.6} \\ {ln\left( 1.6 \right) - ln\left( 0.39 \right) = \left( \frac{-\Delta H_{rxn}}{8.314} \right) \left( \frac{1}{323} - \frac{1}{298} \right)} \\ {\Delta H_{rxn} = \frac{-8.314 \left( ln\left( 1.6 \right) - ln\left( 0.39 \right) \right)}{\frac{1}{323} - \frac{1}{298}} = 9.1 J/mol} \end{gather*} This reaction is endothermic .

Practice Problems for Calculating Enthalpy

Problem 1 Rendered by What is the change in heat content of this reaction ? \begin{gather*} {2Mg + O_{2} \rightarrow 2MgO} \end{gather*} Problem 2 Rendered by What is the change of heat content of this reaction ?

\begin{gather*} {CH_{4} + O_{2} \rightleftarrows CO_{2} + H_{2}O} \end{gather*}

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