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Discuss this statement: "No true chemical equilibrium can exist unless reactant molecules are constantly changing into product molecules, and vice versa."

Short Answer

Expert verified
Chemical equilibrium is dynamic; reactants and products interconvert continuously.

Step by step solution

01

Define Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the forward and reverse reactions occur at the same rate. As a result, the concentrations of reactants and products remain constant over time.
02

Dynamic Nature of Equilibrium

Equilibrium is dynamic, meaning that reactant molecules are continuously converting into product molecules and vice versa. Even though the macroscopic properties (like concentration) remain constant, the microscopic process of interconversion is ongoing.
03

Explain Reaction Dynamics at Equilibrium

At equilibrium, the rate of the forward reaction (reactants to products) is equal to the rate of the reverse reaction (products to reactants). As such, there is no net change in the concentrations of reactants and products, even though reactions are continuously happening.
04

Addressing the Statement

The statement correctly highlights the necessity for continuous interconversion of reactants and products for maintaining a true chemical equilibrium. It underscores the dynamic aspect at the molecular level, despite apparent stability at the macroscopic level.

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Key Concepts

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

Dynamic Equilibrium
The concept of dynamic equilibrium is central to understanding chemical equilibrium. In a chemical reaction, equilibrium is not a static state. It is dynamic in nature. This means that, on the microscopic level, individual molecules are always shifting and changing. Reactant molecules are turning into product molecules, and at the same time, products are turning back into reactants.

Imagine a busy highway with cars moving in both directions. Even if you stand at one spot, the number of cars traveling in each direction appears steady. Similar to this scenario, in a dynamic equilibrium, though the chemical concentrations remain unchanging, the reactions at the molecular level continue to "dance around." It is this constant back and forth action that keeps the system in balance.
  • Molecular motion continues endlessly.
  • Balance is dynamic, not still.

Understanding dynamics is key to recognizing that while equilibrium maintains a constant overall appearance, it's the constant activity of molecules that makes this steadiness possible.
Reaction Rates
In any chemical reaction, rates measure how quickly reactants turn into products. At the outset of a reaction, the rate at which reactants convert into products is typically high. As the reaction proceeds, this rate diminishes as reactants are consumed. In contrast, the reverse reaction (products converting back into reactants) starts slow but builds as products accumulate.

Ultimately, in a state of chemical equilibrium, these rates become equal. The forward reaction rate (reactants to products) matches the reverse reaction rate (products back to reactants). When this balance is achieved, the amount of each substance remains constant over time because losses in one direction are compensated by gains in the other.
  • Rates are measures of speed.
  • Equal rates at equilibrium.

This equal race to and fro ensures that while changes continuously happen, their net effect on concentrations is zero, epitomizing the magical balance of equilibrium.
Forward and Reverse Reactions
In a chemical equilibrium, both forward and reverse reactions are constantly occurring. The forward reaction involves reactants converting to products, while the reverse reaction deals with products transforming back into reactants.

In a closed system, this constant conversion between forward and reverse reactions creates a cycle. Imagine a teeter-totter perfectly balanced on its fulcrum. The forward reaction pushes it one way, while the reverse reaction pushes it the opposite. Neither side is allowed to dominate, ensuring the teeter-totter (system) remains balanced. It is in this way that a system achieves equilibrium.
  • Forward converts reactants to products.
  • Reverse turns products back into reactants.
  • Balance maintains the system's stability.

Thus, equilibrium is truly a matter of balance, with forward and reverse reactions acting as invisible forces keeping the system poised and immobile to the observer, yet deeply active beneath the surface.

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Most popular questions from this chapter

Chemists carried out a study of the high temperature reaction of sulfur dioxide with oxygen in which a sealed reactor initially contained \(0.0076-\mathrm{M} \mathrm{SO}_{2}, 0.0036-\mathrm{M} \mathrm{O}_{2}\), and no \(\mathrm{SO}_{3}\). After equilibrium was achieved, the \(\mathrm{SO}_{2}\) concentration decreased to \(0.0032 \mathrm{M}\). Calculate \(K_{\mathrm{c}}\) at this temperature for $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$

Consider the system $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \\ \Delta_{\mathrm{r}} H^{\circ} &=-1530.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ (a) How will the amount of ammonia at equilibrium be affected by (i) removing \(\mathrm{O}_{2}(\mathrm{~g})\) without changing the total gas volume? (ii) adding \(\mathrm{N}_{2}(\mathrm{~g})\) without changing the total gas volume? (iii) adding water without changing the total gas volume? (iv) expanding the container? (v) increasing the temperature? (b) Which of these changes (i to v) increases the value of \(K ?\) Which decreases it?

Two molecules of A react to form one molecule of \(\mathrm{B},\) as in the reaction $$ 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) $$ Three experiments are done at different temperatures and equilibrium concentrations are measured. For each experiment, calculate the equilibrium constant, \(K_{\mathrm{c}^{*}}\) (a) \([\mathrm{A}]=0.74 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.74 \mathrm{~mol} / \mathrm{L}\) $$ \begin{array}{l} \text { (b) }[\mathrm{A}]=2.0 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=2.0 \mathrm{~mol} / \mathrm{L} \\ \text { (c) }[\mathrm{A}]=0.01 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.01 \mathrm{~mol} / \mathrm{L} \end{array} $$ What can you conclude about this statement: "If the concentrations of reactants and products are equal, then the equilibrium constant is always \(1.0 . "\)

Write the equilibrium constant expression for each reaction. (a) The oxidation of ammonia with \(\mathrm{ClF}_{3}\) in a rocket motor $$ \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{ClF}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{HF}(\mathrm{g})+\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) $$ (b) The simultaneous oxidation and reduction of a chlorite ion $$ \begin{aligned} 3 \mathrm{ClO}_{2}^{-}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\ \text { (c) } \mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \rightleftharpoons & \\\ & \mathrm{IO}_{6}^{5-}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$

The equilibrium constants for dissolving silver sulfate and silver sulfide in water are \(1.7 \times 10^{-5}\) and \(6 \times 10^{-30}\), respectively. (a) Write the balanced dissociation reaction equation and the associated equilibrium constant expression for each process. (b) Which compound is more soluble? Explain your answer. (c) Which compound is less soluble? Explain your answer.

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