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 C: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g),(\mathbf{b}) \mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{POCl}_{3}(l),(\mathbf{d}) \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\)

Short answer

The balanced equations for the formation of the given compounds from elements in their standard states, along with their standard enthalpies of formation, are: (a) \( \frac{1}{2} \) O2(g) + H2(g) \( \rightarrow \) H2O2(g) ; \( \Delta H_{f}^{\circ} \) = -136.0 kJ/mol (b) Ca(s) + C(graphite) + \( \frac{3}{2} \) O2(g) \( \rightarrow \) CaCO3(s) ; \( \Delta H_{f}^{\circ} \) = -1206.9 kJ/mol (c) \( \frac{1}{2} \)\ P4(s) + \( \frac{3}{2} \) O2(g) + \( \frac{3}{2} \) Cl2(g) \( \rightarrow \) POCl3(l) ; \( \Delta H_{f}^{\circ} \) = -945.9 kJ/mol (d) 2 C(graphite) + \( \frac{3}{2} \) H2(g) + \( \frac{1}{2} \) O2(g) \( \rightarrow \) C2H5OH(l) ; \( \Delta H_{f}^{\circ} \) = -277.7 kJ/mol

Step-by-step solution

Write balanced equations for formation from the elements

Each compound should be formed from its constituent elements, which exist in their standard states. Write down the balanced chemical equation for each reaction. (a) For H2O2(g): \( \frac{1}{2} \) O2(g) + H2(g) \( \rightarrow \) H2O2(g) (b) For CaCO3(s): Ca(s) + C(graphite) + \( \frac{3}{2} \) O2(g) \( \rightarrow \) CaCO3(s) (c) For POCl3(l): \( \frac{1}{2} \) P4(s) + \( \frac{3}{2} \) O2(g) + \( \frac{3}{2} \) Cl2(g) \( \rightarrow \) POCl3(l) (d) For C2H5OH(l): 2 C(graphite) + \( \frac{3}{2} \) H2(g) + \( \frac{1}{2} \) O2(g) \( \rightarrow \) C2H5OH(l)

Look up standard enthalpies of formation

Using Appendix C, find the standard enthalpy of formation, \( \Delta H_{f}^{\circ} \), for each substance: (a) For H2O2(g): \( \Delta H_{f}^{\circ} \) = -136.0 kJ/mol (b) For CaCO3(s): \( \Delta H_{f}^{\circ} \) = -1206.9 kJ/mol (c) For POCl3(l): \( \Delta H_{f}^{\circ} \) = -945.9 kJ/mol (d) For C2H5OH(l): \( \Delta H_{f}^{\circ} \) = -277.7 kJ/mol

Present results

After completing the steps above, we can conclude that the balanced equations for the formation of the given compounds from elements in their standard states, along with their standard enthalpies of formation, are: (a) \( \frac{1}{2} \) O2(g) + H2(g) \( \rightarrow \) H2O2(g) \( \Delta H_{f}^{\circ} \) = -136.0 kJ/mol (b) Ca(s) + C(graphite) + \( \frac{3}{2} \) O2(g) \( \rightarrow \) CaCO3(s) \( \Delta H_{f}^{\circ} \) = -1206.9 kJ/mol (c) \( \frac{1}{2} \) P4(s) + \( \frac{3}{2} \) O2(g) + \( \frac{3}{2} \) Cl2(g) \( \rightarrow \) POCl3(l) \( \Delta H_{f}^{\circ} \) = -945.9 kJ/mol (d) 2 C(graphite) + \( \frac{3}{2} \) H2(g) + \( \frac{1}{2} \) O2(g) \( \rightarrow \) C2H5OH(l) \( \Delta H_{f}^{\circ} \) = -277.7 kJ/mol

Key concepts

Balanced Chemical Equations

A balanced chemical equation represents a chemical reaction where the number of each type of atom is the same on both sides of the equation. This reflects the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. For example, when writing a balanced equation for the formation of water, \(\mathrm{H}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2\mathrm{O}(g)\), it ensures that there are two hydrogen atoms and one oxygen atom on each side. To balance equations, you might need to adjust coefficients (the numbers in front of molecules), but never change the subscripts (the small numbers within chemical formulas) as doing so would change the substance itself.
Also, while balancing, remember that fractional coefficients can be used and then multiplied by whole numbers to clear any fractions at the end. Balancing acts as a bookkeeping tool that keeps track of atoms and energy in reactions.

Standard States of Elements

Elements have specific forms and phases at a given pressure of 1 atmosphere, known as their standard states, at which thermodynamic measurements are often made and reported. For example:

• Oxygen is a diatomic gas, \(\mathrm{O}_2(g)\).
• Carbon may be found as graphite, represented as \(\mathrm{C}_{\text{graphite}}\).
• Calcium is a solid metal, denoted as \(\mathrm{Ca}(s)\).

These standard states are specifically referenced when calculating the standard enthalpy of formation for compounds, as reactions are understood to occur starting with the reactants in their standard states. Understanding the standard states of elements is crucial for writing correct reaction equations and accurately determining enthalpy changes.

Thermochemistry

Thermochemistry is the study of energy changes, particularly heat, in chemical reactions. One key component is enthalpy (\(H\)), a measure of the total heat content in a thermodynamic system. In chemical reactions, we often refer to the change in enthalpy, \(\Delta H\). For formation reactions, the standard enthalpy of formation (\(\Delta H_{f}^{\circ}\)) is especially useful, representing the heat change when one mole of a compound is formed from its elements in their standard states. It helps predict reaction behavior and feasibility. Negative \(\Delta H_{f}^{\circ}\) values indicate exothermic reactions, where heat is released, while positive values suggest endothermic reactions, where heat is absorbed.
In practical terms, this means that knowing these enthalpy values helps chemists and engineers design processes that effectively utilize or manage heat, making systems more efficient.