Question

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and use Appendix $\mathrm{C}$ to obtain the values of their standard enthalpies of formation: (a) ${\mathrm{H}}_2{\mathrm{O}}_2(g)$ (b) ${\mathrm{CaCO}}_3(s)$ (c) ${\mathrm{POCl}}_3(l),(\mathbf{d}){\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(l)$

Short answer

(a) Formation of H2O2 (g): $$\frac{1}{2}{\mathrm{H}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)\to {\mathrm{H}}_2{\mathrm{O}}_2(g)$$, ΔHf° = -187.8 kJ/mol. (b) Formation of CaCO3 (s): $$\mathrm{Ca}(s)+\mathrm{C}(s,graphite)+\frac{3}{2}{\mathrm{O}}_2(g)\to {\mathrm{CaCO}}_3(s)$$, ΔHf° = -1206.9 kJ/mol. (c) Formation of POCl3 (l): $$\frac{1}{4}{\mathrm{P}}_4(s,white)+\frac{3}{2}{\mathrm{Cl}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)\to {\mathrm{POCl}}_3(l)$$, ΔHf° = -542.6 kJ/mol. (d) Formation of C2H5OH (l): $$2\mathrm{C}(s,graphite)+3{\mathrm{H}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(l)$$, ΔHf° = -277.6 kJ/mol.

Step-by-step solution

a) Formation of H2O2 (g) from elements in standard states

(1) Write down the elements in standard states: H2 (g) and O2 (g). (2) Combine the elements to form H2O2 (g): H2 (g) + O2 (g) -> H2O2 (g). (3) Balance the equation: 1/2 H2 (g) + 1/2 O2 (g) -> H2O2 (g). (4) Look up the standard enthalpies of formation of H2O2 (g) in Appendix C: ΔHf° = -187.8 kJ/mol.

b) Formation of CaCO3 (s) from elements in standard states

(1) Write down the elements in standard states: Ca (s), C (s, graphite), and O2 (g). (2) Combine the elements to form CaCO3 (s): Ca (s) + C (s, graphite) + O2 (g) -> CaCO3 (s). (3) Balance the equation: Ca (s) + C (s, graphite) + 3/2 O2 (g) -> CaCO3 (s). (4) Look up the standard enthalpies of formation of CaCO3 (s) in Appendix C: ΔHf° = -1206.9 kJ/mol.

c) Formation of POCl3 (l) from elements in standard states

(1) Write down the elements in standard states: P4 (s, white), Cl2 (g), and O2 (g). (2) Combine the elements to form POCl3 (l): P4 (s, white) + Cl2 (g) + O2 (g) -> POCl3 (l). (3) Balance the equation: 1/4 P4 (s, white) + 3/2 Cl2 (g) + 1/2 O2 (g) -> POCl3 (l). (4) Look up the standard enthalpies of formation of POCl3 (l) in Appendix C: ΔHf° = -542.6 kJ/mol.

d) Formation of C2H5OH (l) from elements in standard states

(1) Write down the elements in standard states: C (s, graphite), H2 (g), and O2 (g). (2) Combine the elements to form C2H5OH (l): C (s, graphite) + H2 (g) + O2 (g) -> C2H5OH (l). (3) Balance the equation: 2 C (s,graphite) +3 H2 (g) + 1/2 O2 (g) -> C2H5OH (l). (4) Look up the standard enthalpies of formation of C2H5OH (l) in Appendix C: ΔHf° = -277.6 kJ/mol.

Key concepts

Balanced Chemical Equations

Balanced chemical equations are essential in chemistry. They represent the conversion of reactants into products while maintaining the law of conservation of mass. This means that the number of each type of atom on the reactant side must be equal to the number on the product side. Balancing an equation involves adjusting the coefficients in front of the chemical formulas until this balance is achieved.
For example, when forming ${\mathrm{H}}_2{\mathrm{O}}_2(g)$, you start with hydrogen (${\mathrm{H}}_2$) and oxygen (${\mathrm{O}}_2$) gases. The unbalanced equation is ${\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{H}}_2{\mathrm{O}}_2(\mathrm{g})$. Balancing gives: $\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{H}}_2{\mathrm{O}}_2(\mathrm{g})$.
This simple rule ensures that no atoms are created or destroyed during a chemical reaction, making it a crucial concept for all following calculations and interpretations in chemistry. Practicing balancing helps one understand how elements react and combine into compounds.

Standard States

Understanding standard states is key when discussing enthalpy of formation. The standard state of an element or compound is its form at 1 atm pressure and a specified temperature, usually 25°C (298 K). This is a reference point that allows scientists to consistently report thermodynamic quantities, like enthalpy of formation.
For instance, the standard state of graphite is considered for carbon, and diatomic ${\mathrm{O}}_2(g)$ for oxygen. These forms are used as starting points when you're calculating the formation of compounds like ${\mathrm{CaCO}}_3$ and ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$. It’s important to use these standard states to ensure the enthalpy values are precise and comparable across different reactions.
By comparing how different substances transform from their standard state to compounds like water or ethanol, chemists can determine the energy changes involved in reactions. Recognizing standard states helps in understanding both phase changes and chemical reactivity.

Thermochemistry

Thermochemistry is the study of the heat involved during chemical reactions. One of the central themes in thermochemistry is the concept of enthalpy ($H$), which measures heat content in a system. Specifically, enthalpy of formation ($\mathrm{\Delta }{H}_f^{\circ }$) deals with the enthalpy change when one mole of a compound is formed from its elements in their standard states.
This idea helps us understand whether a reaction releases heat (exothermic) or absorbs heat (endothermic). For example, when ${\mathrm{C}}_2{\mathrm{H}}_5(\mathrm{l})$ is formed from carbon graphite and hydrogen, the process is exothermic ($\mathrm{\Delta }{H}_f^{\circ }=-277.6\text{ kJ/mol}$). This negative value indicates heat is released.
Studying thermochemistry is vital not just for academic purposes but also for industrial applications. As industries need to manage energy efficiently, knowing the enthalpy changes in reactions helps them design processes economically and safely.