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Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{CO}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{CO} 2(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of hydrogen atoms or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), (c) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) and \(0.5 \mathrm{~mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C}\), (d) \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(300^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
(a) 1 mole of CO2(g) has higher enthalpy than 1 mole of CO2(s). (b) 2 moles of hydrogen atoms have higher enthalpy than 1 mole of H2. (c) 1 mole of H2(g) and 0.5 mole of O2(g) have higher enthalpy than 1 mole of H2O(g). (d) 1 mole of N2(g) at 300°C has higher enthalpy than 1 mole of N2(g) at 100°C.

Step by step solution

01

Case (a)

In this case, we have 1 mole of CO2 either in solid or gaseous state. Since gases have more enthalpy than solids due to the increased kinetic energy of the particles, 1 mole of CO2(g) will have a higher enthalpy than 1 mole of CO2(s).
02

Case (b)

In this case, we have 2 moles of hydrogen atoms versus 1 mole of H2 molecules. Hydrogen atoms have higher energy than H2 molecules because H2 molecules are formed with a strong bond, which releases energy. So 2 moles of hydrogen atoms will have a higher enthalpy than 1 mole of H2.
03

Case (c)

In this case, we have 1 mole of H2(g) and 0.5 mole of O2(g) versus 1 mole of H2O(g). When 1 mole of H2(g) combines with 0.5 mole of O2(g), it forms 1 mole of H2O(g) in an exothermic reaction, releasing heat energy. Hence, 1 mole of H2(g) and 0.5 mole of O2(g) will have a higher enthalpy than 1 mole of H2O(g), as the enthalpy was released when the substances combined.
04

Case (d)

In this case, we have 1 mole of N2(g) at 100°C and 1 mole of N2(g) at 300°C. The enthalpy of a system increases with increasing temperature due to the increased kinetic energy of the particles. Therefore, 1 mole of N2(g) at 300°C will have a higher enthalpy than 1 mole of N2(g) at 100°C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It describes how energy changes from one form to another. In the context of chemistry, thermodynamics often focuses on enthalpy, a measure of heat content in a system. Changes in enthalpy can indicate whether a process absorbs or releases energy. For example, when a gas becomes a solid, it takes less energy to keep the molecules moving, reducing enthalpy. The amount of energy involved gives insights into how substances will react or change states. Understanding these principles helps predict changes in energy when chemicals interact.
  • Enthalpy relates to the heat content in a system.
  • Thermodynamics helps explain energy transformations.
  • Energy changes occur during state changes and chemical reactions.
Grasping thermodynamics is crucial for understanding how energy affects matter in various physical and chemical processes.
Exploring Chemical Reactions
Chemical reactions involve the rearrangement of atoms to transform substances. They may either release or absorb energy, which is evident through changes in enthalpy. When substances react, bonds between atoms break and new bonds form. The energy changes associated with these processes lead to different enthalpy levels. In the exercise example, forming water from hydrogen and oxygen is an exothermic reaction because it releases energy, thus resulting in a lower enthalpy for the water than for the separate reactants.
  • Chemical reactions transform substances through atom rearrangement.
  • Exothermic reactions release energy; endothermic reactions absorb energy.
  • Enthalpy change indicates the energy absorbed or released.
By understanding chemical reactions, you can predict energy changes that occur when substances interact.
States of Matter and Enthalpy
The state of matter (solid, liquid, gas) significantly impacts the enthalpy of a substance. Each state has different energy levels tied to the movement of particles. Gas particles move faster, needing more energy, and thus have higher enthalpy compared to solids and liquids. In the solid state, molecules vibrate in place, requiring less energy, resulting in lower enthalpy. For example, CO2(g) has higher enthalpy than CO2(s) because gaseous particles move more freely.
  • Gases have higher kinetic energy and enthalpy than solids or liquids.
  • State of matter affects how molecules move and how much energy is needed.
  • Enthalpy differences explain why states of matter affect energy content.
Comprehending how states of matter influence enthalpy helps predict energy needs or releases during physical transitions.
Kinetic Energy and Temperature
Kinetic energy refers to the energy an object possesses due to its motion. In substances, it's tied to the temperature, as warmer substances have particles moving faster. Higher temperatures mean more kinetic energy, leading to higher enthalpy. In the exercise, N2 at 300°C has more enthalpy than N2 at 100°C due to increased kinetic energy from higher temperatures.
  • Kinetic energy is related to the motion of particles.
  • Temperature changes affect how fast particles move, impacting energy.
  • Higher temperatures correlate with increased kinetic energy and enthalpy.
Understanding kinetic energy helps explain how temperature affects the energy states of matter, providing insights into enthalpy changes.

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Most popular questions from this chapter

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

During a normal breath, our lungs expand about \(0.50 \mathrm{~L}\) against an external pressure of \(1.0\) atm. How much work is involved in this process (in J)?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J}\). (b) A system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system.

(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose (table sugar), \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s), \Delta H_{f}^{\circ}\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\).

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