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Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each.

Short Answer

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Physical equilibrium is a state involving changes in physical states of matter without a change in chemical composition, as seen while ice melts or water evaporates. Chemical equilibrium involves a state where the forward and backward reaction rates, implying a change in chemical composition, become equal - for instance in the synthesis of ammonia from nitrogen and hydrogen, or formation and decomposition of sulphur trioxide.

Step by step solution

01

Define Physical Equilibrium

In physical equilibrium, the physical states of matter change, but the chemical composition remains the same. It pertains to changes in the phase of a substance. Frequency of conversion of one physical state to another reach an equal level.
02

Examples of Physical Equilibrium

1. Melting Ice: The process where ice (solid water) converts to water (liquid water) and water converts back to ice at the same rate once the melting point is reached. 2. Evaporation of Water: It is the process where liquid water converts to water vapour (gas state) and water vapour condenses back to liquid state at the same rate once the boiling point is reached.
03

Define Chemical Equilibrium

Chemical equilibrium involves the forward and backward reaction rates becoming equal, implying a change in the chemical composition. Here, the amount of reactants converted to product and the amount of product reverting back to reactants becomes equal.
04

Examples of Chemical Equilibrium

1. Synthesis of Ammonia (Haber Process): In this process, nitrogen reacts with hydrogen to produce ammonia. This reaction can also proceed in the reverse direction, breaking down ammonia into nitrogen and hydrogen. 2. Formation of Sulphur trioxide: Sulphur dioxide reacts with oxygen to form sulphur trioxide. Sulphur trioxide can also decompose back into sulphur dioxide and oxygen.

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Key Concepts

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

Physical Equilibrium
Physical equilibrium refers to a state in a physical system where multiple phases exist together in balance without any net change in the system. At this point, the rate at which a phase change occurs in one direction is equal to the rate of the reverse change. It's important to remember that physical processes do not alter the chemical composition of the substances involved.

For example, during the melting of ice, water exists in both solid (ice) and liquid form. When the rate of ice melting equals the rate of freezing, the system reaches equilibrium. Another common example is the evaporation of water. At a certain temperature, water molecules evaporate from the liquid phase to form vapor at the same rate at which vapor molecules condense back to liquid, achieving a steady state. Therefore, in physical equilibrium situations, even though the physical form changes, the chemical identity remains intact.
Phase Changes
Phase changes involve transitions between different states of matter such as solid, liquid, and gas. A phase change is a physical process, meaning the substance itself remains chemically identical before and after the change.

Key phase changes include:
  • Melting: Solid to liquid (e.g., ice to water).
  • Freezing: Liquid to solid (e.g., water to ice).
  • Evaporation: Liquid to gas (e.g., water to steam).
  • Condensation: Gas to liquid (e.g., steam to water).
  • Sublimation: Solid to gas without passing through a liquid phase (e.g., dry ice to COâ‚‚ gas).
  • Deposition: Gas to solid without passing through a liquid phase (e.g., formation of frost).
Each of these phase changes involves an exchange of energy, typically in the form of heat. For example, melting and evaporation require energy input to break intermolecular bonds, while freezing and condensation release energy. Understanding these principles can help explain natural phenomena and everyday processes.
Reaction Rates
Reaction rates refer to how quickly or slowly a chemical reaction proceeds. Various factors influence these rates, including temperature, concentration, and the presence of catalysts. When studying chemical reactions, it's important to consider the conditions under which the reaction occurs because these conditions impact the speed at which reactants turn into products.

Increasing temperatures typically increase reaction rates because the molecules involved move faster and collide more often, leading to a higher probability of successful interactions. Similarly, higher concentrations of reactants generally lead to more frequent collisions.

Catalysts are substances that increase reaction rate without being consumed in the process. They often work by providing an alternative reaction pathway with a lower activation energy, thereby increasing the number of successful collisions. These principles are fundamental to understanding how different reactions, including those in industrial processes and biological systems, happen efficiently.
Haber Process
The Haber Process is an industrial method used to synthesize ammonia \(NH_3\) from nitrogen \(N_2\) and hydrogen \(H_2\). This process is a prime example of chemical equilibrium where both forward and reverse reactions happen at the same rate, resulting in a constant concentration of reactants and products.

The balanced chemical equation can be represented as: \[N_2 (g) + 3H_2 (g) ightleftharpoons 2NH_3 (g)\]At equilibrium, the rate of ammonia formation equals the rate of its decomposition back into nitrogen and hydrogen.

This process is conducted under high pressure and moderate temperature, using catalysts to improve efficiency. The Haber Process is essential for producing fertilizers, which are critical for agriculture, and highlights the importance of understanding chemical kinetics and equilibrium to manage industrial-scale reactions.

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Most popular questions from this chapter

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}:\) $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}\), \(1.35 \times 10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

What is the rule for writing the equilibrium constant for the overall reaction involving two or more reactions?

Pure phosgene gas \(\left(\mathrm{COCl}_{2}\right), 3.00 \times 10^{-2} \mathrm{~mol},\) was placed in a 1.50-L container. It was heated to \(800 \mathrm{~K}\), and at equilibrium the pressure of \(\mathrm{CO}\) was found to be 0.497 atm. Calculate the equilibrium constant \(K_{P}\) for the reaction $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)$$

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C},\) respectively. \()\)

Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]

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