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If there is no change in concentrations, why is the equilibrium state considered dynamic?

Short Answer

Expert verified
Equilibrium is dynamic because the forward and reverse reactions continue at equal rates, even though concentrations remain constant.

Step by step solution

01

Understanding Equilibrium

Equilibrium in a chemical reaction is the state where the rates of the forward and reverse reactions are equal. This means that there is no net change in the concentrations of reactants and products over time.
02

Analyzing Concentrations at Equilibrium

Even though the concentrations of reactants and products remain constant at equilibrium, individual molecules of reactants are still converting to products and vice versa. This indicates that reactions are still happening.
03

Defining Dynamic Equilibrium

Because the reactions continue to occur at equal rates but with no net change in concentrations, the system is said to be in a state of dynamic equilibrium. The word 'dynamic' reflects the continuous activity at the molecular level.

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Key Concepts

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

chemical equilibrium
The concept of chemical equilibrium is essential in understanding how reactions behave over time. In a chemical reaction, equilibrium is the point at which the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of the reactants and products remain constant, though they are not necessarily equal. At equilibrium, a reaction has reached a stable state where the quantities of all chemical components do not change.

However, it’s important to recognize that equilibrium doesn't mean the reactions have stopped. The reactions are still happening on a molecular level, but because they occur at the same rate in both directions, the overall concentrations of the chemicals involved don't change.

In summary, chemical equilibrium is characterized by constant concentrations of all reactants and products and a balance between the forward and reverse reaction rates.
reaction rates
Reaction rates are crucial for understanding dynamic equilibrium. The rate of a reaction is determined by how quickly reactants are converted into products. At dynamic equilibrium, the forward reaction rate (reactants to products) and the reverse reaction rate (products to reactants) are equal.

To break it down:
  • Forward Reaction Rate: This is the speed at which reactants turn into products.
  • Reverse Reaction Rate: This is the speed at which products turn back into reactants.
When these two rates are equal, no net change in concentrations occurs, but the individual molecular reactions continue to proceed, signifying continuous activity. Think of it like a busy intersection where the number of cars entering and exiting is the same, making the traffic appear constant and unchanging.

Understanding these rates helps explain why we use the term dynamic equilibriumβ€”the reactions are ongoing, but there is no overall shift in chemical amounts.
concentration stability
Concentration stability is a key concept to grasp when studying dynamic equilibrium. At dynamic equilibrium, the concentrations of reactants and products do not change over time. This stability indicates that, although reactions are still occurring, they do so at rates that balance each other out.

Consider the following:
  • Constant Concentrations: The amounts of reactants and products remain fixed.
  • Continuous Reactions: Molecules are still reacting with each other, but because the forward and reverse reaction rates are equal, the overall concentrations don't shift.
This stability can be visualized by imagining a leaky bucket being filled with water at the same rate it leaks. Although water continuously enters and exits the bucket, the water level stays the same.

Therefore, even though we don't see a concentration change, the system is dynamic, not static. This steady state where concentration remains constant yet reactions continue is what makes dynamic equilibrium so fascinating and important in chemical studies.

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Most popular questions from this chapter

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

Compound A decomposes according to the equation $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g) $$ A sealed 1.00-L container initially contains \(1.75 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{A}(g)\) \(1.25 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{B}(g),\) and \(6.50 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{C}(g)\) at \(100^{\circ} \mathrm{C}\) At equilibrium, [A] is \(2.15 \times 10^{-3} \mathrm{M}\). Find [B] and [C].

Isolation of Group \(8 \mathrm{~B}(10)\) elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) .\) The metal oxide is reduced by the \(\mathrm{H}_{2}\) in water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\) to impure \(\mathrm{Ni}: \mathrm{NiO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g) .\) The \(\mathrm{CO}\) in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, \(\mathrm{Ni}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g),\) which is sub- sequently decomposed to the metal. (a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of \(\mathrm{Ni}_{3} \mathrm{~S}_{2}\) to \(\mathrm{Ni}(\mathrm{CO})_{4}\). (b) Show that the overall \(Q_{\mathrm{c}}\) is the product of the \(Q_{c}\) 's for the individual reactions.

Balance each reaction and write its reaction quotient, \(Q_{c}\) (a) \(\mathrm{NO}_{2} \mathrm{Cl}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{POCl}_{3}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, \(Q_{c}\) (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\operatorname{SnO}_{2}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \operatorname{Sn}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(l)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)\)

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