Question

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix \(\mathrm{C}\) : (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)\), (b) \(\mathrm{CaCO}_{3}(s)\), (c) \(\mathrm{POCl}_{3}(l)\), (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\).

Short answer

The balanced equations for the formation of the given compounds from their elements in standard states are: (a) \(\frac{1}{2} O_2 (g) + H_2 (g) \rightarrow H_2 O_2 (g)\) with \(\Delta_{f} H^{∘} = -191.2 \thinspace kJ/mol\) (b) \(Ca (s) + C (graphite) + \frac{3}{2} O_2 (g) \rightarrow CaCO_3 (s)\) with \(\Delta_{f} H^{∘} = -1206.9 \thinspace kJ/mol\) (c) \(P (red) + \frac{3}{2} O_2 (g) + \frac{3}{2} Cl_2 (g) \rightarrow POCl_3 (l)\) with \(\Delta_{f} H^{∘} = -688.2 \thinspace kJ/mol\) (d) \(\frac{2}{2} C (graphite) + \frac{6}{2} H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow C_2H_5OH (l)\) with \(\Delta_{f} H^{∘} = -277.7 \thinspace kJ/mol\)

Step-by-step solution

Finding the balanced equations for each compound

For each compound, let's write a balanced equation that describes its formation from its constituent elements in their standard states. (a) Formation of \(\mathrm{H}_{2} \mathrm{O}_{2}(g)\): \[\frac{1}{2} O_2 (g) + H_2 (g) \rightarrow H_2 O_2 (g)\] (b) Formation of \(\mathrm{CaCO}_{3}(s)\): \[Ca (s) + C (graphite) + \frac{3}{2} O_2 (g) \rightarrow CaCO_3 (s)\] (c) Formation of \(\mathrm{POCl}_{3}(l)\): \[P (red) + \frac{3}{2} O_2 (g) + \frac{3}{2} Cl_2 (g) \rightarrow POCl_3 (l)\] (d) Formation of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\): \[\frac{2}{2} C (graphite) + \frac{6}{2} H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow C_2H_5OH (l)\] Now, we'll look up the standard enthalpy of formations for each compound in Appendix C.

Finding the standard enthalpies of formation in Appendix C

Using Appendix C or a similar reference, we can look up the standard enthalpy of formation for each compound: (a) Standard enthalpy of formation for \(\mathrm{H}_{2} \mathrm{O}_{2}(g)\): \(\Delta_{f} H^{∘} = -191.2 \thinspace kJ/mol\) (b) Standard enthalpy of formation for \(\mathrm{CaCO}_{3}(s)\): \(\Delta_{f} H^{∘} = -1206.9 \thinspace kJ/mol\) (c) Standard enthalpy of formation for \(\mathrm{POCl}_{3}(l)\): \(\Delta_{f} H^{∘} = -688.2 \thinspace kJ/mol\) (d) Standard enthalpy of formation for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\): \(\Delta_{f} H^{∘} = -277.7 \thinspace kJ/mol\) Therefore, we have found the balanced equations for the formation of the compounds from their elements in standard states and their respective standard enthalpies of formation.

Key concepts

Enthalpy of Formation

Enthalpy of formation, often denoted as \( \Delta_f H^{∘} \), is a crucial concept in thermodynamics and chemistry. It refers to the amount of energy released or absorbed when one mole of a compound forms from its elements in their standard states. The standard state of a substance is its most stable form at 1 atm pressure and at a specified temperature, usually 25°C (298 K). Enthalpy of formation can provide insights into the stability of a compound.Here’s what you need to know about enthalpies of formation:

• A negative \( \Delta_f H^{∘} \) indicates the reaction releases energy, making the formation exothermic.
• A positive \( \Delta_f H^{∘} \) signifies that energy is absorbed, indicating an endothermic formation.

In the provided exercise, looking up Appendix C shows specific enthalpies of formation for compounds like \( \text{H}_2\text{O}_2(g) \) and \( \text{CaCO}_3(s) \), helping us understand the energy changes involved in their formation processes.

Balancing Equations

Balancing chemical equations is an essential skill that ensures the law of conservation of mass is obeyed in chemical reactions. In a balanced equation, the number of atoms on the reactant side matches the number of atoms on the product side. This reflects that matter is neither created nor destroyed.Let’s review how to balance equations:

• Identify the elements in the reactants and products and count their atoms.
• Adjust the coefficients (the numbers before molecules) to have equal numbers of each type of atom on both sides.
• Use trial and error to find the smallest integer coefficients that satisfy this condition.

In the original exercise, equations were balanced for the formation of compounds like \( \text{POCl}_3(l) \) and \( \text{C}_2\text{H}_5\text{OH}(l) \). Taking \( \text{CaCO}_3(s) \) as an example: it is formed from \( \text{Ca}(s) \), \( \text{C} (\text{graphite}) \), and \( \text{O}_2(g) \). By balancing the equation, we can derive the exact proportions of reactants needed.

Standard States

Understanding standard states is fundamental when discussing enthalpy and other thermodynamic properties. A standard state is the physical state of a substance under standard conditions of 1 atm pressure and a specified temperature, typically 25°C (298 K). By convention, these provide a reference point for thermodynamic calculations.Some guidelines to recognize standard states:

• For gases, the standard state is the pure gas at 1 atm.
• For solids and liquids, the standard state is the pure substance in its most stable form at 1 atm.
• For solutions, the standard state is a concentration of 1 mol/L.

In the exercise, substances were described in their standard states, such as \( \text{O}_2(g) \) for oxygen. Establishing these states is critical for calculating properties like the enthalpy of formation, which depends on forming compounds from their standard state elements.