Question

Write thermochemical equations for the decomposition of one mole of the following compounds into the elements in their stable states at \(25^{\circ} \mathrm{C}\) and 1 atm. (a) ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) (b) sodium fluoride \((s)\) (c) magnesium sulfate \((s)\) (d) ammonium nitrate (s)

Short answer

In summary, we have written the balanced thermochemical equations for the decomposition of the given compounds into their respective elements in their stable states at \(25^{\circ}\mathrm{C}\) and 1 atm: a) Ethyl alcohol: $$\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) \rightarrow 2\mathrm{C}(s,\,graphite\,)+ 3\mathrm{H}_{2}(g) + \dfrac{1}{2}\mathrm{O}_{2}(g)$$ b) Sodium fluoride: $$\mathrm{NaF}(s) \rightarrow \mathrm{Na}(s) + \dfrac{1}{2}\mathrm{F}_{2}(g)$$ c) Magnesium sulfate: $$\mathrm{MgSO}_{4}(s) \rightarrow \mathrm{Mg}(s) + \dfrac{1}{8}\mathrm{S}_{8}(s,\,crystalline\,)+ 2\mathrm{O}_{2}(g)$$ d) Ammonium nitrate: $$\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \rightarrow \dfrac{3}{2}\mathrm{N}_{2}(g) + 2\mathrm{H}_{2}(g) + \dfrac{1}{2}\mathrm{O}_{2}(g)$$

Step-by-step solution

(a) Decomposition of ethyl alcohol

Ethyl alcohol, or ethanol, has the chemical formula \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l)\). The elements present in it are carbon, hydrogen, and oxygen. At \(25^{\circ}\mathrm{C}\) and 1 atm, carbon is in the solid state and exists as graphite, hydrogen is in gaseous state as \(\mathrm{H}_{2}\), and oxygen is in gaseous state as \(\mathrm{O}_{2}\). The balanced thermochemical equation is: $$\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) \rightarrow 2\mathrm{C}(s,\,graphite\,)+ 3\mathrm{H}_{2}(g) + \dfrac{1}{2}\mathrm{O}_{2}(g)$$

(b) Decomposition of sodium fluoride

Sodium fluoride has the chemical formula \(\mathrm{NaF}(s)\). The elements present are sodium and fluorine. At \(25^{\circ}\mathrm{C}\) and 1 atm, sodium is in the solid state and fluorine is in gaseous state as \(\mathrm{F}_{2}\). The balanced thermochemical equation is: $$\mathrm{NaF}(s) \rightarrow \mathrm{Na}(s) + \dfrac{1}{2}\mathrm{F}_{2}(g)$$

(c) Decomposition of magnesium sulfate

Magnesium sulfate has the chemical formula \(\mathrm{MgSO}_{4}(s)\). The elements present are magnesium, sulfur, and oxygen. At \(25^{\circ}\mathrm{C}\) and 1 atm, magnesium is in the solid state, sulfur is in the solid state and exists as crystalline \(\mathrm{S}_{8}\), and oxygen is in gaseous state as \(\mathrm{O}_{2}\). The balanced thermochemical equation is: $$\mathrm{MgSO}_{4}(s) \rightarrow \mathrm{Mg}(s) + \dfrac{1}{8}\mathrm{S}_{8}(s,\,crystalline\,)+ 2\mathrm{O}_{2}(g)$$

(d) Decomposition of ammonium nitrate

Ammonium nitrate has the chemical formula \(\mathrm{NH}_{4}\mathrm{NO}_{3}(s)\). The elements present are nitrogen and hydrogen. At \(25^{\circ}\mathrm{C}\) and 1 atm, nitrogen is in gaseous state as \(\mathrm{N}_{2}\) and hydrogen is in gaseous state as \(\mathrm{H}_{2}\). The balanced thermochemical equation is: $$\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \rightarrow \dfrac{3}{2}\mathrm{N}_{2}(g) + 2\mathrm{H}_{2}(g) + \dfrac{1}{2}\mathrm{O}_{2}(g)$$

Key concepts

Chemical Decomposition

Understanding the concept of chemical decomposition is crucial when tackling any chemistry assignment. It refers to a type of chemical reaction where a single compound breaks down into two or more simpler substances, often elements or simpler compounds. This process is typically the result of heating, but it can also occur due to electric current, light, or even the presence of a catalyst.

In the exercise provided, decomposition reactions are used to break down complex molecules like ethanol and magnesium sulfate into their elemental constituents under standard conditions, which means they are at a temperature of 25 degrees Celsius and a pressure of 1 atmosphere. The products of these reactions are the elements in their most stable form at these conditions, such as solids like graphite from carbon or gases like hydrogen (H2) and oxygen (O2).

It's important to note that the equations we are creating are thermochemical equations, which means they not only represent the stoichiometry of the reaction but also implicitly include the thermal conditions—such as temperature and physical states of the substances—during the reaction. When you're given a compound like ethyl alcohol (C2H5OH(l)), think about what elements make it up and in what state they would exist under standard conditions. Then, write an equation that shows these elements separately, balanced correctly according to the conservation of mass.

Standard State Conditions

Let's delve deeper into the significance of the standard state conditions. These are a set of reference conditions for physical and chemical properties used to provide a basis for comparing different substances. They are important in chemistry because they are the 'default' conditions at which the properties of substances are often described, and against which changes, such as reactions, are measured.

In the context of our exercises, the standard state conditions mentioned are a temperature of 25 degrees Celsius (298.15 Kelvin) and a pressure of 1 atmosphere. Under these conditions, when we talk about the decomposition of a compound, we are considering the products that would exist in their most stable forms at this temperature and pressure. For example, carbon is most stable as graphite, while hydrogen and oxygen are most stable as diatomic molecules, H2 and O2, respectively.

Understanding and utilizing standard state conditions allow chemists to predict and calculate the properties of chemical reactions accurately. It is essential when writing thermochemical equations to specify the standard states to avoid ambiguity, as the standard states indicate the phase (solid, liquid, gas) of the elements or compounds at the specified conditions.

Chemical Reaction Balancing

Moving onto chemical reaction balancing, which is the foundation of stoichiometry in chemistry. The primary rule here is the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a closed system. When writing chemical equations, it me an s that you must have the same number of atoms of each element on both the reactant and product sides of the equation.

In each of the decomposition reactions presented in the exercise, we start with our compound and determine the elements that make it up. We then write products that reflect those elements in their most stable forms under standard state conditions, just as we've discussed. The next step is ensuring that the equation is balanced. For instance, in ethyl alcohol (C2H5OH), we have two carbons, six hydrogens, and one oxygen. The equation must reflect the formation of two carbon atoms in graphite, three diatomic hydrogen molecules (each with two hydrogen atoms), and a half molecule of oxygen (since O2 has two oxygen atoms).

It can be easy to overlook balancing especially when dealing with more complex molecules with multiple atoms of each element. Breaking it down into a step-wise approach can be helpful: list all the elements, tally the atoms of each in the reactants and products, and adjust coefficients in front of molecules to balance the counts on either side of the reaction arrow. Patience and practice are key, as some trial and error may be required to achieve a balanced equation. This meticulous approach ensures accuracy in the depiction and understanding of the chemical process taking place.