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 molar concentrations of the reactants, each raised to a power equal to its stoichiometric coefficient. For example, for the reaction \(N_2 + 3H_2 \leftrightarrow 2NH_3\), the rate of the forward reaction is \(k_{forward}[N_2][H_2]^3\) and the rate of the backward reaction is \(k_{backward}[NH_3]^2\).

Step by step solution

01

Understanding the Law of Mass Action

The Law of Mass Action, stated by Cato Guldberg and Peter Waage, explains that the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants, each raised to a power corresponding to its stoichiometric coefficient in the balanced chemical equation.
02

Mathematical Representation

The mathematical expression for the Law of Mass Action can be stated as follows for a generic chemical reaction: \(aA + bB \leftrightarrow cC + dD\). The rate of forward reaction: \(Rate_{forward} = k_{forward} [A]^a [B]^b\) and the rate of backward reaction: \(Rate_{backward} = k_{backward} [C]^c [D]^d\) where \(k_{forward}\) and \(k_{backward}\) are rate constants, and [A], [B], [C], [D] are molar concentrations of the reactants and products.
03

Example

For a reaction, \(N_2 + 3H_2 \leftrightarrow 2NH_3\), if the concentrations of \(N_2\), \(H_2\), and \(NH_3\) are denoted as [\(N_2\)], [\(H_2\)], and [\(NH_3\)] respectively, then according to the Law of Mass Action, the rate of forward reaction = \(k_{forward}[N_2][H_2]^3\) and rate of backward reaction = \(k_{backward}[NH_3]^2\).

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Key Concepts

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

Chemical Reaction Rates
Chemical reaction rates are fundamental to studying how reactions proceed over time. Simply put, the reaction rate is a measure of how quickly a reactant is converted into a product. According to the Law of Mass Action, this rate is influenced by the concentrations of the reactants.

For instance, if you increase the concentration of a reactant in a chemical reaction, the rate at which the products are formed will generally increase. This happens because there are more reactant molecules that can potentially collide and react with each other. The Law of Mass Action quantitatively expresses this relationship by stating that the rate of a reaction is proportional to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

The constants of proportionality in this relationship are the rate constants (\(k_{forward}\) for the forward reaction and \(k_{backward}\) for the reverse reaction). These constants provide crucial information about the reaction's dynamics and are influenced by various factors such as temperature and the presence of catalysts.
Reaction Equilibrium
When a reaction proceeds in both the forward and reverse directions, it can reach a state of reaction equilibrium. This is the point where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time.

The Law of Mass Action is also used to express the equilibrium condition through the equilibrium constant, \(K_{eq}\). For the general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is \(K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\). It's important to understand that while the concentrations of the reactants and products remain constant at equilibrium, the actual values may vary depending on the reaction and the conditions present.

Exploring Equilibrium in Dynamic Reactions

Even at equilibrium, reactants and products are continuously being formed and decomposed; it's an ongoing dynamic process. The equilibrium constant sheds light on the ratio of product to reactant concentrations at equilibrium, and is a valuable parameter for predicting the extent of a reaction and determining optimal conditions for industrial processes.
Stoichiometry
Stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), and it provides a quantitative relationship between reactants and products in a chemical reaction. It enables chemists to predict how much product will form from a certain amount of reactants or how much reactant is needed to create a certain amount of product.

The stoichiometric coefficients in a balanced chemical equation represent the molar ratios in which substances react and are formed. These coefficients are central to calculations involving the Law of Mass Action. When we apply the law to a balanced equation, it tells us that not only do the amounts of reactants affect the rates and equilibrium, but the proportions in which they react are just as important.

Practical Use of Stoichiometry

Stoichiometry is not just a theoretical concept; it has practical applications in laboratory experiments, industrial production, and environmental monitoring. Accurate stoichiometric calculations ensure that reactions are carried out efficiently, resources are conserved, and that we understand the environmental impact of chemical reactions. In essence, grasping stoichiometry is essential for any student or professional involved in the field of chemistry.

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Most popular questions from this chapter

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\)

At \(1024^{\circ} \mathrm{C},\) the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is 0.49 atm: $$4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g)$$ (a) What is \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) that will decompose if 0.16 mole of it is placed in a \(2.0-\mathrm{L}\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would the fraction be if a 1.0 mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)$$ at the same temperature.

At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)$$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right]\)

Consider the following equilibrium system involving \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (sulfuryl dichloride): $$\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)$$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system; (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system; (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant.

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