Thermochemical Equations - Online Tutor, Practice Problems & Exam Prep

Stoichiometry involves the numerical relationships in balanced chemical equations, while thermochemical equations incorporate the enthalpy of reaction (ΔH). In these equations, the focus shifts from mole-to-mole comparisons to ΔH-to-mole comparisons. A thermochemical stoichiometric chart helps connect the given ΔH to moles of a compound, allowing conversions from grams or molecules to moles. This process requires a mole-to-mole comparison using coefficients from the balanced equation, ultimately enabling conversions to unknown quantities, including ΔH of unknown substances.

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 ΔH_rxn, 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 ΔH of reaction. It's our job to make a connection between your enthalpy of reaction and either moles, grams, molecules, what you have 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 direction 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 an enthalpy change of negative 375 kilojoules. So what we have to do here is we have to convert the given quantity into moles of the given.

Alright. 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 kilojoules. 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 the enthalpy involved, is negative 1204 kilojoules per mole. So we've just found our moles concerning magnesium oxide.

Step 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 12.59 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 amount of negative 375 kilojoules and using that helped us to determine the final answer of 12.59 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 and how does it differ from a regular chemical equation?

A thermochemical equation is a balanced chemical equation that includes the enthalpy change (ΔH) of the reaction. Unlike regular chemical equations that focus solely on the stoichiometric relationships between reactants and products, thermochemical equations incorporate the heat energy absorbed or released during the reaction. This ΔH value is crucial for understanding the energy dynamics of the reaction. In thermochemical equations, the focus shifts from mole-to-mole comparisons to ΔH-to-mole comparisons, allowing for the calculation of energy changes in relation to the amounts of reactants or products.

How do you use a thermochemical stoichiometric chart?

A thermochemical stoichiometric chart helps connect the given ΔH to the moles of a compound in a reaction. Start with the given ΔH and convert it to moles of the given substance using the balanced chemical equation. From there, you can convert moles of the given substance to moles of the unknown substance using the coefficients from the balanced equation. Finally, you can convert moles of the unknown substance to grams, molecules, or even a new ΔH if needed. This process involves multiple steps and requires careful attention to the stoichiometric relationships and energy changes involved.

What is the significance of ΔH in a thermochemical equation?

ΔH, or the enthalpy change, in a thermochemical equation represents the heat energy absorbed or released during a chemical reaction at constant pressure. It is a crucial parameter for understanding the energy dynamics of the reaction. A positive ΔH indicates an endothermic reaction (heat absorbed), while a negative ΔH indicates an exothermic reaction (heat released). This information is essential for predicting the feasibility and spontaneity of reactions, as well as for practical applications like designing energy-efficient processes and understanding reaction mechanisms.

How do you convert grams of a substance to ΔH in a thermochemical equation?

To convert grams of a substance to ΔH in a thermochemical equation, follow these steps: First, convert the grams of the substance to moles using its molar mass. Next, use the balanced chemical equation to find the stoichiometric relationship between the moles of the substance and the ΔH of the reaction. Finally, multiply the moles of the substance by the ΔH per mole (from the balanced equation) to find the total ΔH for the given amount of substance. This process allows you to determine the energy change associated with a specific quantity of reactant or product.

What is the role of stoichiometry in thermochemical equations?

Stoichiometry in thermochemical equations involves the quantitative relationships between reactants and products, incorporating the enthalpy change (ΔH) of the reaction. It allows for the conversion between moles, grams, and molecules of substances while accounting for the energy changes involved. By using stoichiometric coefficients from the balanced equation, you can determine the amount of energy absorbed or released for given quantities of reactants or products. This is essential for calculating energy requirements, optimizing reaction conditions, and understanding the thermodynamic properties of chemical reactions.