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What is the law of mass action?

Short Answer

Expert verified
The law of mass action states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

Step by step solution

01

Introduction to the Law of Mass Action

The law of mass action is a principle in chemistry that relates the rate of a reaction to the concentration of the reactants. It states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to the number of moles in the balanced chemical equation.
02

Understanding Reaction Rate Expression

In a reaction where reactants A and B form products, for example: \[ aA + bB ightarrow cC + dD \]The rate of the reaction can be expressed as:\[ ext{Rate} = k[A]^m[B]^n \]where \(k\) is the rate constant, \(m\) and \(n\) are the stoichiometric coefficients of reactants A and B respectively in the balanced equation.
03

Application of the Law

To apply the law of mass action, we use the concentration terms of each reactant involved in the reaction. The stoichiometric coefficients in the balanced reaction become the exponents for these concentration terms in the rate equation.
04

Example Reaction

Consider a specific example: \(2 ext{NO} + ext{O}_2 ightarrow 2 ext{NO}_2\). The law of mass action gives the rate expression as:\[ ext{Rate} = k[ ext{NO}]^2[ ext{O}_2]^1 \]Here, each concentration is raised to the power of its stoichiometric coefficient in the chemical equation.

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Key Concepts

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

Reaction Rate
The reaction rate is a fundamental concept in chemistry that describes how quickly a chemical reaction proceeds. It's the speed at which reactants are converted into products in a given reaction. In mathematical terms, the reaction rate is often expressed as the change in concentration of a reactant or product over time. The formula for this is:
\[\text{Rate} = \frac{-\Delta [A]}{\Delta t} = \frac{\Delta [B]}{\Delta t}\]This formula shows the decrease in concentration of reactant \([A]\) or the increase in concentration of product \([B]\) over time \(\Delta t\). Understanding reaction rates helps chemists control processes like manufacturing and medication delivery.
  • The rate is affected by factors like temperature, pressure, and concentration.
  • A higher concentration of reactants usually leads to a faster reaction rate.
Stoichiometric Coefficients
Stoichiometric coefficients are numbers placed in front of compounds in a balanced chemical equation. They indicate the ratio of moles of each substance involved in the reaction. These coefficients are vital for writing the rate law of a reaction.
In the reaction \(aA + bB \rightarrow cC + dD\), the stoichiometric coefficients are \(a\), \(b\), \(c\), and \(d\).
  • They determine how much of each reactant is needed and how much product is formed.
  • In rate laws, these coefficients become exponents that show how reactant concentration affects the rate.
For example, if \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\), then \([\text{NO}]\) is squared in the rate expression because of its coefficient of 2.
Chemical Kinetics
Chemical kinetics studies how fast chemical reactions occur and what factors affect these rates. It's a branch of chemistry focused on understanding the speed of reactions and their mechanisms.
This field is important because it helps in:
  • Predicting reaction speeds under different conditions.
  • Designing systems to optimize the efficiency of chemical processes.
  • Understanding how energy is transferred in reactions.
Chemical kinetic studies include exploring the dependence of reaction rates on temperature, pressure, and catalyst use. By examining these effects, scientists can optimize reactions for industrial applications and environmental considerations.
Concentration of Reactants
The concentration of reactants plays a key role in determining the rate of a chemical reaction. According to the law of mass action, the rate of a reaction is directly proportional to the concentration of the reactants, each raised to a power as indicated by its stoichiometric coefficient.
For instance, in a reaction where \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\), the rate law is expressed as:
\[\text{Rate} = k[\text{NO}]^2[\text{O}_2]^1\]Here, the concentration of \([\text{NO}]\) is squared, indicating its significant influence on the rate.
  • Higher concentration usually means a higher reaction rate.
  • Changes in reactant concentration can be used to control the speed of chemical processes.
  • Understanding this concept is vital in fields like pharmaceuticals and environmental science.

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Most popular questions from this chapter

The equilibrium constant \(K_{P}\) for the reaction: $$ \mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177,0.223,\) and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Photosynthesis can be represented by: $$\begin{aligned} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=2801 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of \(\mathrm{CO}_{2}\) is increased, (b) \(\mathrm{O}_{2}\) is removed from the mixture, (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased.

Consider the reaction between \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in a closed container: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) $$ Initially, \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(x \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(x\) and \(P\), the total pressure. (b) How does the expression in part (a) help you predict the shift in equilibrium due to an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

Ammonium carbamate \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes as follows: $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ Starting with only the solid, it is found that when the system reaches equilibrium at \(40^{\circ} \mathrm{C},\) the total gas pressure \(\left(\mathrm{NH}_{3}\right.\) and \(\mathrm{CO}_{2}\) ) is 0.363 atm. Calculate the equilibrium constant \(K_{P}\).

List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

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