Calculating Enthalpy: The Four Best Methods

symbol for enthalpy of reaction

Core Concepts

In this article, you will learn about the most important methods of calculating enthalpy of chemical reactions. After reading, you will be able to calculate a reaction’s enthalpy in a variety of common situations.

Topics Covered in Other Articles

What is Enthalpy?

Chemists and physicists define enthalpy as the heat released by a process under constant pressure. Enthalpy is a “state variable,” meaning that a system’s change in enthalpy only depends on the initial and final states of the system, rather than the particular path taken between the two states.

In chemistry, enthalpy has the most application when understanding the thermodynamics of a chemical reaction. In particular, the sign of a reaction’s change of enthalpy yields important information.

For instance, if a reaction’s change of enthalpy has a positive sign, chemists call the reaction “endothermic.” Endothermic reactions involve reactants absorbing heat from the environment. Generally, endothermic reactions are either not thermodynamically favored, or only favored at high temperatures.

By contrast, if a reaction’s change of enthalpy has a negative sign, chemists call the reaction “exothermic.” Exothermic reactions instead involve reactants releasing heat into the environment. Generally, exothermic reactions are either always thermodynamically favored, or only favored at low temperatures.

Due to the thermodynamic importance of enthalpy, chemists have developed many methods of calculating enthalpy of reaction. In this article, we will cover the four most important 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 enthalpy, depending on what information you already know about your reactants and reaction.

Calculating Enthalpy with Enthalpies of Formation

The most straightforward method of calculating the enthalpy of a reaction involves using what chemists call “enthalpies of formation.” In short, each molecule has a characteristic “enthalpy of formation,” which is essentially the change in enthalpy involved with assembling the molecule from its respective atoms. For more information about enthalpy of formation, check out this article.

If you know the enthalpies of formation for each molecule in your reaction, then you can calculate the reaction’s overall enthalpic change. To do this, you need to use an enthalpy formula 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 enthalpy of formation with its stoichiometric coefficient in the reaction equation. Second, you add the multiplied enthalpies of the products and those of the reactants. Third, you subtract the combined enthalpies of the products with that of the reactants, yielding the overall reaction enthalpy.

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 molecule’s enthalpy of formation through experimentation, specifically calorimetry. Thus, you can only use enthalpies of formation when you are dealing with familiar molecules at well-studied temperatures (such as 25°C and 37°C). Unfamiliar molecules and conditions tend to lack readily available information on their enthalpy of formation.

Example Calculation

Calculate the enthalpy of the following reaction at 25°C: right)_2

begin{gather*} {Caleft( OH right)_{2} + left( NH_{4} right)_{2} CO_{3} rightarrow CaCO_{3} + 2NH_{4} OH} end{gather*}

 ∆Hf  (kJ/mol)

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

This reaction is exothermic.

Calculating Enthalpy with Bond Enthalpies

Another frequent method of calculating a reaction’s enthalpy involves using bond enthalpies. Specifically, each chemical bond between two atoms involves some enthalpic change in its formation. To learn more about bond enthalpies, check out this article.

You will need to implement a similar method to calculate enthalpy of reaction as Hess’s Law, but with important differences.

First, you need to find the bond enthalpies of each bond formed or broken over the course of the reaction. Any bonds that remain unchanged by the reaction don’t factor into the reaction’s overall enthalpy. Second, you add all the enthalpies of the broken bonds and those of the formed bonds. Third, you subtract the added enthalpies of the broken bonds by that of the formed bonds, yielding the overall reaction enthalpy.

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 broken minus bonds formed.” This is important to point out because it essentially is the opposite: newly formed bonds are found in the products, while broken bonds are found in the reactants.

Also, using bond enthalpies works for more reactions than enthalpies of formation. With this, you can calculate reaction enthalpies for unconventional products and reactants, so long as the reaction involves familiar bonds. However, you are still limited by the familiarity of the conditions, and that of the bonds.

Example Calculation

Calculate the enthalpy of the following reaction at 25°C:

elimination reaction between 2-chlorobutane and tertbutoxide. useful example for calculating enthalpy of reaction
 ∆HBond  (kJ/mol)

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 only one of the bonds in an alkene break, you can think of it as a C=C bond breaking and a C–C bond forming.

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 simple method of calculating reaction enthalpy involves using a substance’s specific heat. Each substance has a property called “specific heat” which indicates the amount of energy to raise the substance’s 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 direct 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’t affect the reaction enthalpy. Instead, 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 air in the vessel. In this case, air serves as our reaction medium, with a specific heat of 1.01 J/(g°C). If instead 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 following formula:

begin{align*} {q=mCDelta T} end{align*}

q: Heat absorbed or released from reaction medium (in Joules)

m: Mass of the reaction medium (in grams)

C: Specific heat of the reaction medium (in J/(g°C))

∆T: Change in temperature of reaction medium (in degrees Celsius)

In most cases, the quantity of heat exchanged by the reaction medium equals the enthalpy of reaction. The only exceptions are when the conditions of “constant pressure” aren’t met. However, the pressure of the system would only change if the reaction involved gases, and there existed unequal moles of gas 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 water. The water’s temperature then increases by 13.7°C. Find the enthalpy of reaction.

begin{gather*} {NaOH + HCl rightarrow NaCl + H_{2}O} \ {q=mCDelta T = left(1000gright) left(4.18 frac{J}{gdegree C}right) left(13.7 degree Cright) = 57.2 kJ/mol} end{gather*}

Note: since we calculated the heat absorbed by the water, we need to flip the sign to get the heat given off by the reaction.

begin{gather*} {Delta H_{rxn} = -q = -57.2 kJ/mol} end{gather*}

This reaction is exothermic.

Calculating Enthalpy with Equilibrium Constants

You can also solve for reaction enthalpy using the reaction’s equilibrium constants, which are values that represent the equilibrium proportions of products and reactants. If you want to learn more about equilibrium constants, then take a look at this article.

This method represents another experimental way of calculating enthalpy. Specifically, it takes advantage of the relationship between the thermodynamic characteristics of a reaction with its equilibrium dynamics. Depending on your reaction species, equilibrium concentrations can be easily measured using spectrophotometry, yielding easy enthalpic information.

To find enthalpy using equilibrium constants, you need to measure the equilibrium concentrations of products and reactants at two different temperatures. Then, you can find the enthalpy of reaction using what chemists call the Van’t Hoff enthalpy 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: Equilibrium constant at Temperature 1

K2 : Equilibrium 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 following reaction:

begin{gather*} {A + 2B rightleftarrows C} end{gather*}

At 25°C, you observe the following equilibrium concentrations:

begin{align*} {[A] &= 0.70M} \ {[B] &= 1.4M} \ {[C] &= 0.30M} end{align*}

At 50°C, you observe the following equilibrium concentrations:

begin{align*} {[A] &= 0.44M} \ {[B] &= 0.88M} \ {[C] &= 0.54M} end{align*}

Calculate the enthalpy 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} \ {lnleft( 1.6 right) - lnleft( 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( lnleft( 1.6 right) - lnleft( 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 enthalpy of this reaction?

begin{gather*} {2Mg + O_{2} rightarrow 2MgO} end{gather*}

Problem 2

In a 10L reaction vessel exposed to UV radiation, 6.00 moles of diatomic oxygen are injected. Irradiated oxygen has the following equilibrium relationship with ozone:

begin{gather*} {3O_{2} rightleftarrows 2O_{3}} end{gather*}

At 40°C, 0.800 moles of ozone are present, while at 140°C, 0.150 moles of ozone are present.

Problem 3

The following reaction takes place in 1.00L of water:

begin{gather*} {Ag_{2}SO_{4} + 2NaCl rightarrow 2AgCl + Na_{2}SO_{4}} end{gather*}

Over time, 5.00 moles of AgCl precipitate out of solution, raising the water’s temperature by 5.6°C. What is the enthalpy of reaction?

Problem 4

What is the enthalpy of the following reaction?

 ∆HBond  (kJ/mol)

Problem 5

Rendered by

What is the change of enthalpy of this reaction?

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

Practice Problem Solutions

1: ∆Hrxn = -1202kJ/molrxn

2: ∆Hrxn = 41.9kJ/molrxn

3: ∆Hrxn = -9.36kJ/molrxn

4: ∆Hrxn = 341kJ/molrxn

5: ∆Hrxn = -577.5kJ/molrxn