# Question Video: Calculating the the Enthalpy Change for the Thermal Decomposition of Lithium Carbonate Using Enthalpies of Formation Chemistry

The group I carbonate Li₂CO₃ thermally decomposes to form lithium oxide and carbon dioxide. What is the enthalpy change of this reaction if the standard enthalpy of formation of lithium carbonate is −1,216 kJ/mol and the standard enthalpy of formation of lithium oxide is −596 kJ/mol? The standard enthalpy of formation of carbon dioxide is −394 kJ/mol.

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### Video Transcript

The group one carbonate Li2CO3 thermally decomposes to form lithium oxide and carbon dioxide. What is the enthalpy change of this reaction if the standard enthalpy of formation of lithium carbonate is negative 1,216 kilojoules per mole and the standard enthalpy of formation of lithium oxide is negative 596 kilojoules per mole? The standard enthalpy of formation of carbon dioxide is negative 394 kilojoules per mole.

It can be difficult to measure enthalpy changes for some reactions. For example, the reactions may be too slow or potentially dangerous. However, we can still calculate the enthalpy changes for these reactions by using enthalpy data from other reactions. In this problem, we are given the values of the standard enthalpy of formation of three different compounds: lithium carbonate, or Li2CO3, lithium oxide, or Li2O, and carbon dioxide, or CO2. The standard enthalpy of formation is the enthalpy change when one mole of substance forms from its constituent elements in their standard states and under standard conditions.

We can use these standard enthalpy of formation values provided. But first we will need to create a Hess cycle to show how the formation reactions are related to the decomposition reaction in the problem. To begin our Hess cycle, let’s first write a balanced chemical equation for the thermal decomposition of lithium carbonate. Lithium carbonate is a solid with the chemical formula Li2CO3. And since we are told thermal decomposition is taking place, we can add the symbol for heat over the reaction arrow.

The products of the reaction are lithium oxide and carbon dioxide. The chemical formula of lithium oxide is Li2O because two Li1+ ions are needed to bond with one O2‒ ion to form a neutral compound. Lithium oxide is a solid under standard conditions. We’re probably very familiar with the chemical formula of carbon dioxide, which is CO2, and this compound is a gas under standard conditions. As written, this chemical equation is already balanced. Let’s label this reaction as reaction one.

Now, let’s write a chemical equation to represent the formation of one mole of lithium carbonate from its constituent elements, which are lithium, carbon, and oxygen. We’ve written these elements in their standard states. And it’s important to note that oxygen is diatomic. Now we can draw a reaction arrow pointing from the elements to lithium carbonate and label this reaction as reaction two. Of course, we need to balance the equation, which means we need to write a coefficient of two in front of lithium and write a coefficient of three-halves in front of oxygen gas.

Next, let’s write an equation to represent the formation of one mole of lithium oxide and one mole of carbon dioxide. We can draw a reaction arrow from the elements to the products and label this reaction as reaction three. This chemical equation is already balanced.

Now that we’ve completed creating our Hess cycle, let’s clear some space to calculate the enthalpy change of reaction one. According to our Hess cycle, we can state that the enthalpy change of reaction one is equal to the negative value of the enthalpy change of reaction two plus the enthalpy change of reaction three. The reason that we need to change the sign of the enthalpy change of reaction two is because we need to move from the reactants to the products along our alternative pathway. Therefore, we need to reverse reaction number two.

Now we’re ready to substitute the standard enthalpies of formation that were provided into our equation. The enthalpy change of reaction two is equal to the standard enthalpy of formation of lithium carbonate, which is negative 1,216 kilojoules per mole. The enthalpy change of reaction three is equal to the sum of the enthalpies of formation of lithium oxide and carbon dioxide. After simplification, we see that the enthalpy change of reaction one equals 1,216 kilojoules per mole plus negative 990 kilojoules per mole. This gives us an answer of positive 226 kilojoules per mole.

In conclusion, the enthalpy change for the thermal decomposition of lithium carbonate is positive 226 kilojoules per mole.