Thermochemical Equations: Study with Video Lessons, Practice Problems & Examples

Stoichiometry involves the numerical relationships in balanced chemical equations, while thermochemical equations incorporate the enthalpy of reaction (ΔH_rxn). In these equations, the focus shifts from mole-to-mole comparisons to ΔH-to-mole comparisons. By using a thermochemical stoichiometric chart, one can convert given ΔH to moles of a compound, then make a mole-to-mole comparison using coefficients from the balanced equation to find unknown quantities. This process highlights the connection between enthalpy and various forms of measurement, such as grams or molecules, essential for understanding thermochemistry.

Thermochemical Equations

Now recall that stoichiometry deals with the numerical relationship between compounds in a balanced chemical equation. Now with a thermochemical equation, we're going to deal with chemical reactions that include an enthalpy of reaction, which is ΔHrxn, so delta H of reaction. Here with thermochemical equations, we're going to be introduced to our thermochemical stoichiometric chart. Now here, the chart uses the given quantity of a compound to determine the unknown quantity of another compound. So here with the thermochemical equation, we have our balanced chemical equation, and to the side of it, you'll see your delta H of reaction. It's our job to make a connection between your enthalpy of reaction and either moles, grams, molecules, what have you, in terms of the chemical reaction. So in a thermochemical equation, what we're trying to do is not a mole-to-mole comparison; we're trying to do a delta H to mole comparison. And that's the key difference with our thermochemical equation. Now that we've seen this, let's move onwards and talk about more, with thermochemical equations.

Thermochemical Equations

Now, if you've seen my stoichiometric chart under stoichiometry, or solution chemistry, or gas stoichiometry, this should be pretty familiar to you. But if this is the first time you're seeing this stoichiometric chart, let's go through it. Alright. So what we need to realize here is that with thermochemical equations, it's pretty common to be given the ΔH of our chemical reaction. So we'll start out with ΔH of given. And like I said before, in stoichiometry we're used to doing a multiple comparison which can still happen here, but the more important thing is that we establish a connection between the ΔH of reaction and one of the moles for one of the compounds within our chemical, reaction. So, here we go from ΔH of given to moles of given. And what we can say here is that besides going from ΔH of given to moles of given, we can go from grams of given to moles of given, or we can go from ions, atoms, formula units, or molecules of given to moles of given. Once we get there, we have to go to moles of our unknown. To do this requires a leap of faith in a sense, because you're going from an area where you know information to an area where you know nothing at all. So, we call this the jump. When you make this jump, in order to do it correctly you have to do a mole to mole comparison and use the coefficients in the balanced equation. From this point, you are at moles of unknown, and from here you can go in any way you want. You can go from moles of unknown to ions, atoms, formula units, or molecules, you can go to grams, or you can go to a new ΔH of unknown. So, just realize with the thermochemical equation, we'll have a balanced chemical equation, which introduces the idea of stoichiometry, but for it to be a thermochemical equation, we'll also have the ΔH of reaction present.

Thermochemical Equations Example 1

Here it says, consider the following thermochemical reaction. We have 2 moles of magnesium solid reacting with 1 mole of oxygen gas to produce 2 moles of magnesium oxide solid. It gives us an enthalpy of reaction equal to negative 1204 kilojoules. Here we're asked how many grams of magnesium oxide are produced during enthalpy change of negative 375 kilojoules. So what we have to do here is we have to convert the given quantity into moles of given. Alright. So they're giving us negative 375 kilojoules of energy, and what we need to do is establish a relationship between magnesium oxide and this value of negative 375. Well according to my balanced equation, for step 2, it says we need to do a mole to mole comparison to convert moles of given into moles of unknown. Here because it's a thermochemical equation, it's going to be moles of given relating to delta h of reaction. So we're going to say for every 2 moles of magnesium oxide, the energy involved or enthalpy involved is negative 1204 kilojoules per mole. So we've just found our moles in terms of magnesium oxide.

Steps 3 says, if necessary convert the moles into desired units. Here they want grams not moles, so we're going to do one more step and say for every one mole of magnesium oxide, the mass of magnesium oxide, 1 magnesium, is 24.31 grams according to the periodic table. One oxygen is 16 grams. Multiplying and adding those numbers together gives us 40.31 grams. Moles cancel out and now I'm going to have 25.11 grams of magnesium oxide. Step 4 isn't needed here because, in step 4, if we have to calculate more than one final amount then we must compare them to determine the theoretical yield. Here we're only given one given the amount of negative 375 kilojoules and using that helped us to determine the final answer of 25.11 magnesium oxide, grams of magnesium oxide.

Nitromethane (CH₃NO₂), sometimes used as a fuel for drag racing, burns according to the following reaction: 

4 CH₃NO₂ (l) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (g) + 4 NO₂ (g)      ∆Hº = – 2441.6 kJ

How much heat is released by burning 125.0 g of nitromethane (MW:61.044 g/mol)? 

• A. 1.250 x 10³ kJ
• B. -1.250 x 10³ kJ
• C. 2.440 x 10³ kJ
• D. -2.440 x 10³ kJ

Consider the following reaction: 

2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂­O (g)  ∆Hº = – 6278 kJ

What volume of benzene (C₆H₆, d = 0.880 g/mL, molar mass = 78.11 g/mol) is necessary to evolve 5.19 x 10⁹ kJ of heat? 

• A. 1.47 x 10⁸ mL
• B. 5.19 x 10⁹ mL
• C. 4.37 x 10⁷ mL
• D. 9.51 x 10⁹ mL

The creation of liquid methanol is accomplished by the hydrogenation of carbon monoxide: 

CO (g) + 2 H₂ (g) → CH₃OH (l)       ∆Hº = – 128.1 kJ

How much heat (in kJ) is released when 125.0 g CO reacts with 2.32 x 10² g H₂?

• A. 275 kJ
• B. -572 kJ
• C. -128 kJ
• D. 281 kJ

Here’s what students ask on this topic:

What is a thermochemical equation?

A thermochemical equation is a balanced chemical equation that includes the enthalpy change (ΔH) of the reaction. This ΔH value represents the heat absorbed or released during the reaction. Unlike standard stoichiometric equations that focus on mole-to-mole comparisons, thermochemical equations emphasize the relationship between ΔH and the moles of reactants or products. For example, in the combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ, the ΔH value indicates that 890 kJ of energy is released per mole of CH₄ combusted.

How do you use a thermochemical stoichiometric chart?

A thermochemical stoichiometric chart helps convert given quantities (like ΔH, grams, or molecules) to moles and then to other desired quantities. Start with the given ΔH and convert it to moles of the given substance. From there, use the balanced chemical equation to perform a mole-to-mole comparison to find the moles of the unknown substance. Finally, convert the moles of the unknown to the desired unit (grams, molecules, etc.). This process ensures accurate calculations involving enthalpy changes in chemical reactions.

What is the difference between a thermochemical equation and a regular stoichiometric equation?

The main difference between a thermochemical equation and a regular stoichiometric equation is the inclusion of the enthalpy change (ΔH) in thermochemical equations. While regular stoichiometric equations focus on the mole-to-mole relationships between reactants and products, thermochemical equations emphasize the relationship between ΔH and the moles of substances involved. This allows for the calculation of energy changes during chemical reactions, which is crucial for understanding thermochemistry.

How do you calculate the enthalpy change (ΔH) for a reaction using a thermochemical equation?

To calculate the enthalpy change (ΔH) for a reaction using a thermochemical equation, follow these steps: 1) Write the balanced chemical equation with the ΔH value. 2) Identify the moles of reactants and products involved. 3) Use the stoichiometric coefficients to relate the moles of substances to the given ΔH. For example, if the equation is CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ, and you have 2 moles of CH₄, the total ΔH would be 2 × -890 kJ = -1780 kJ.

Why is it important to include ΔH in a thermochemical equation?

Including ΔH in a thermochemical equation is important because it provides information about the energy changes during a chemical reaction. This helps in understanding whether the reaction is exothermic (releases heat) or endothermic (absorbs heat). Knowing the ΔH value is crucial for applications in energy management, industrial processes, and predicting reaction behavior. It also allows for accurate calculations involving the conversion of energy to moles, grams, or other units, which is essential for practical and theoretical chemistry.