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How do chemists define a state of chemical equilibrium?

Short Answer

Expert verified
A state of chemical equilibrium is defined as a point in a reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products over time. The system is in a dynamic state, with reactions continuing to occur at equal rates, maintaining constant concentrations of both reactants and products.

Step by step solution

01

Introduction to Chemical Reactions

In a chemical reaction, reactants are converted into products. A forward reaction is when reactants are transformed into products, while a reverse reaction is when products are converted back into reactants. When a reaction takes place, the concentrations of reactants and products change over time.
02

Dynamic Equilibrium

Sometimes, after a reaction has started, the products formed also begin to react and convert back into the original reactants. This is known as the reverse reaction. When the rates of both the forward and reverse reactions become equal, the system has reached a state called dynamic equilibrium.
03

No Net Change in Concentration

At the state of chemical equilibrium, the concentrations of the reactants and the products remain constant over time. This does not mean that the reaction has stopped. The forward and reverse reactions are still occurring, but their rates are equal, resulting in no net change in the concentration of reactants and products.
04

Balanced Chemical Equation

Chemists use balanced chemical equations to represent chemical reactions, including those that involve equilibrium. In a balanced equation, the number of atoms of each element on both sides of the equation remains the same. For equilibrium reactions, the balanced chemical equation is often written with a double arrow symbol (⇌) to represent the forward and reverse reactions occurring simultaneously.
05

Equilibrium Constant (K)

To quantitatively describe the state of chemical equilibrium, chemists use a value called the equilibrium constant (K). K is the ratio of the concentration of products to the concentration of reactants at equilibrium, each raised to the power of their stoichiometric coefficients. A larger value of K indicates that the equilibrium favors the formation of products, while a smaller value of K shows that the equilibrium favors the formation of reactants. In conclusion, chemists define a state of chemical equilibrium as a point in a reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products over time. At equilibrium, the system is in a dynamic state, meaning the reactions continue to occur, but their rates are equal, and the concentrations of reactants and products remain constant.

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Key Concepts

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

dynamic equilibrium
In the realm of chemical reactions, dynamic equilibrium is a key concept that often occurs in reversible reactions. Imagine a system where a chemical reaction can proceed in both directions: forwards and backwards. At dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

Even though the reactions continue to happen, the overall concentrations of reactants and products do not change over time. This balance does not mean that the chemical activity has ceased, but rather, it's like a state of balance where everything is moving at the right pace, so it seems still.

Dynamic equilibrium is essential for processes such as:
  • Synthesis of ammonia in the Haber process
  • Biochemical processes in cells
  • Certain atmospheric reactions
By understanding dynamic equilibrium, chemists can predict the outcome of reactions and manipulate conditions to favor the production of desired products.
forward and reverse reactions
In reversible chemical reactions, both forward and reverse reactions occur. Initially, the forward reaction typically dominates, converting reactants into products. However, as products accumulate, they can begin to transform back into reactants, marking the start of the reverse reaction.

A good understanding of both forward and reverse reactions allows chemists to:
  • Determine how a system reaches equilibrium
  • Adjust conditions to shift equilibria
  • Optimize the yield of desired products
One classic example is the conversion of nitrogen and hydrogen into ammonia. This process involves both forward and reverse reactions. By manipulating temperature, pressure, and concentration, chemists guide the reaction towards maximizing ammonia production, despite the ongoing reverse reaction.
equilibrium constant
The equilibrium constant, denoted as \( K \), is a crucial parameter that provides insight into the balance of a chemical reaction at equilibrium. Essentially, \( K \) is the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation at equilibrium.

The value of \( K \) helps predict the extent of a reaction:
  • \( K > 1 \) implies a greater concentration of products than reactants, favoring product formation.
  • \( K < 1 \) indicates a greater concentration of reactants, suggesting the reaction favors reactants.
By calculating the equilibrium constant, chemists can better understand reaction dynamics and predict which direction a reaction will likely progress under given conditions. This tool is vital in fields ranging from industrial synthesis to environmental science.

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Most popular questions from this chapter

Write the equilibrium expression for each of the following reactions. a. \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(C_{2} H_{4}(g)+C l_{2}(g) \rightleftharpoons C_{2} H_{4} C l_{2}(g)\).

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide. $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 \mathrm{M},\) and \([\mathrm{NO}]=4.7 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K\) for the reaction.

What is the effect on the equilibrium position if an endothermic reaction is performed at a higher temperature? Does the net amount of product increase or decrease? Does the value of the equilibrium constant change if the temperature is increased?

Suppose the reaction system $$\mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ has already reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional \(\mathrm{UO}_{2}(s)\) is added to the system. b. 5.0 mol of \(\mathrm{Xe}(g)\) is added to the system. c. The reaction is performed in a glass reaction vessel; HF(g) attacks and reacts with glass. d. Water vapor is removed. e. The size of the reaction vessel is increased.

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

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