Problem 70 - Chemistry: The Central Science in SI Units, Global Edition

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) \(\mathrm{CH}_{3} \mathrm{OH}(l),\) (b) \(\mathrm{CaSO}_{4}(s),\) (d) \(\mathrm{P}_{4} \mathrm{O}_{6}(s),\) (c) \(\mathrm{NO}(g)\).

Step-by-Step Solution

Answer

Balanced equations and standard enthalpies of formation for compounds: a) Methanol (CH3OH): \[ C(s) + 2H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow CH_{3}OH(l) \] ΔHf° = -238.7 kJ/mol b) Calcium sulfate (CaSO4): \[ Ca(s) + S(s) + 2O_{2}(g) \rightarrow CaSO_{4}(s) \] ΔHf° = -1434.5 kJ/mol c) Nitric oxide (NO): \[ \frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow NO(g) \] ΔHf° = +90.3 kJ/mol d) Phosphorus trioxide (P4O6): \[ P_{4}(s) + 3O_{2}(g) \rightarrow P_{4}O_{6}(s) \] ΔHf° = -1640.1 kJ/mol

Step 1: Identify elements in standard states

Methanol (CH3OH) is composed of carbon, hydrogen, and oxygen. The standard states of these elements are: Carbon (C): solid Hydrogen (H2): gas Oxygen (O2): gas

Step 2: Write the balanced equation

Consider an equation showing the formation of 1 mol of CH3OH from its elements in standard states. \[ C(s) + 2H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow CH_{3}OH(l) \]

Step 3: Look up the standard enthalpy of formation in Appendix C

From Appendix C, the standard enthalpy of formation, ΔHf°, for CH3OH(l) is -238.7 kJ/mol. #b. Writing a balanced equation for the formation of CaSO4 (s) from its elements in standard states#

Step 1: Identify elements in standard states

Calcium sulfate (CaSO4) is composed of calcium, sulfur, and oxygen. The standard states of these elements are: Calcium (Ca): solid Sulfur (S): solid Oxygen (O2): gas

Step 2: Write the balanced equation

Consider an equation showing the formation of 1 mol of CaSO4 from its elements in standard states. \[ Ca(s) + S(s) + 2O_{2}(g) \rightarrow CaSO_{4}(s) \]

Step 3: Look up the standard enthalpy of formation in Appendix C

From Appendix C, the standard enthalpy of formation, ΔHf°, for CaSO4(s) is -1434.5 kJ/mol. #c. Writing a balanced equation for the formation of NO (g) from its elements in standard states#

Step 1: Identify elements in standard states

Nitric oxide (NO) is composed of nitrogen and oxygen. The standard states of these elements are: Nitrogen (N2): gas Oxygen (O2): gas

Step 2: Write the balanced equation

Consider an equation showing the formation of 1 mol of NO from its elements in standard states. \[ \frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow NO(g) \]

Step 3: Look up the standard enthalpy of formation in Appendix C

From Appendix C, the standard enthalpy of formation, ΔHf°, for NO(g) is +90.3 kJ/mol. #d. Writing a balanced equation for the formation of P4O6 (s) from its elements in standard states#

Step 1: Identify elements in standard states

Phosphorus trioxide (P4O6) is composed of phosphorus and oxygen. The standard states of these elements are: Phosphorus (P4): solid Oxygen (O2): gas

Step 2: Write the balanced equation

Consider an equation showing the formation of 1 mol of P4O6 from its elements in standard states. \[ P_{4}(s) + 3O_{2}(g) \rightarrow P_{4}O_{6}(s) \]

Step 3: Look up the standard enthalpy of formation in Appendix C

From Appendix C, the standard enthalpy of formation, ΔHf°, for P4O6(s) is -1640.1 kJ/mol.

Key Concepts

Balanced Chemical Equations

A balanced chemical equation is essential in chemistry because it ensures that the same number of each type of atom appears on both sides of the equation. This reflects the law of conservation of mass, meaning mass cannot be created or destroyed in a chemical reaction. When writing a balanced equation for the formation of a compound, we show the reaction of elements, in their standard states, to form one mole of the desired compound.

• For methanol (\(\text{CH}_3\text{OH}\)), the balanced equation becomes:\[\text{C}(\text{s}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l})\]
• In the case of Calcium Sulfate (\(\text{CaSO}_4\)), the balanced equation is:\[\text{Ca}(\text{s}) + \text{S}(\text{s}) + 2\text{O}_2(\text{g}) \rightarrow \text{CaSO}_4(\text{s})\]
• For Nitric Oxide (\(\text{NO}\)), we write:\[\frac{1}{2}\text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}(\text{g})\]
• Lastly, with Phosphorus Trioxide (\(\text{P}_4\text{O}_6\)), the equation is balanced as:\[\text{P}_4(\text{s}) + 3\text{O}_2(\text{g}) \rightarrow \text{P}_4\text{O}_6(\text{s})\]

Balancing equations is crucial in predicting the products and reactants’ quantity in reaction scenarios. This ensures experimental accuracy and theoretical calculations align.

Standard States of Elements

The standard state of an element is its most stable form under 1 bar of pressure and a specified temperature, usually room temperature (approximately 298.15 K or 25°C). For example:

• Carbon's standard state, when forming organic compounds like methanol, is as a solid (\(\text{C}(\text{s})\)).
• Nitrogen and oxygen, which are diatomic gases, have their standard states as \(\text{N}_2(\text{g})\) and \(\text{O}_2(\text{g})\), respectively.
• Calcium and sulfur have their standard states as solids: \(\text{Ca}(\text{s})\) and \(\text{S}(\text{s})\).

These standard states are important for defining reference points in thermochemistry, specifically when calculating the enthalpy changes of reactions. These provide the baseline from which the enthalpy of formation, and other thermochemical data, are derived.

Thermochemistry

Thermochemistry is the branch of chemistry that studies energy changes, particularly heat, in chemical reactions. The enthalpy of formation is a central concept in thermochemistry, defined as the energy change when one mole of a compound forms from its elements in their standard states. This value is crucial to:

• Predict reaction spontaneity and stability of compounds.
• Calculate the energy profile of chemical processes.

For example:

• The \(\Delta H_f^\circ\) for methanol (\(\text{CH}_3\text{OH}(\text{l})\)) is \(-238.7 \text{kJ/mol}\), indicating an exothermic formation.
• Calcium sulfate (\(\text{CaSO}_4(\text{s})\)) has \(-1434.5 \text{kJ/mol}\), another exothermic formation.
• Nitric oxide (\(\text{NO}(\text{g})\)), a less stable compound, shows an endothermic formation at \(+90.3 \text{kJ/mol}\).

These values are essential for engineers and chemists to design processes efficiently and safely, predicting energy requirements and impacts within chemical industries.

Formation Reactions

A formation reaction involves combining elements in their standard states to form one mole of a compound. These reactions serve as the foundation for measuring the standard enthalpy change, which is a vital part of understanding thermochemistry.

• For methanol (\(\text{CH}_3\text{OH}\)), the formation reaction combines carbon, hydrogen, and oxygen.
• Calcium sulfate (\(\text{CaSO}_4\)) is formed by reacting calcium, sulfur, and oxygen.
• Phosphorus trioxide (\(\text{P}_4\text{O}_6\)) forms from phosphorus and oxygen.
• Nitric oxide (\(\text{NO}\)) forms directly from nitrogen and oxygen.

Formation reactions are specially balanced to create precisely one mole of the product. They are central to thermodynamic calculations, as their enthalpy changes directly link to the compound's energetic profile. Understanding these reactions aids in predicting reaction behavior and energy usage across chemical sciences.