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Chapter 15: Problem 3

Briefly describe the importance of equilibrium in the study of chemical reactions.

Short Answer

Expert verified
Equilibrium in chemistry refers to the state when rates of forward and reverse reactions are equal, leading to no overall net effect. It's crucial in studying chemical reactions because it lets us predict the concentration of reactants and products at any point in a reaction, allowing us to manipulate a reaction's conditions to get a desired result. In addition, it helps us understand why some reactions do not go to completion, and an equilibrium state still exists.

Step by step solution

01

Define Equilibrium

In the context of chemistry, equilibrium refers to the state where the forward and reverse reactions occur at the same rate, leading to no overall net effect. It does not necessarily mean that the reactants and products are equal in quantity, but that their ratios stay constant.
02

Explain the Occurrence of Equilibrium in Reactions

Equilibrium occurs in a chemical reaction when the rate of reaction in forward direction equals the rate of reverse reaction. It means that the reactants react to form the products, and the products decompose to form the reactants at the same rate.
03

Discuss the Importance of Equilibrium in the Study of Chemical Reactions

Understanding equilibrium is essential in the study of chemical reactions because it allows us to predict the concentrations of reactants and products at any given time in a reaction. This understanding can be used to manipulate conditions of a reaction to get the desired result. For example, to maximise product output in industrial reactions. Additionally, some reactions do not go to completion, and equilibrium will still exist – an understanding of that allows us to make sense of why that happens.

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Most popular questions from this chapter

Consider the gas-phase reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture (a) at constant pressure and (b) at constant volume.

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 0.65 at \(395^{\circ} \mathrm{C}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is \(K_{\mathrm{c}}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

Consider this equilibrium system: $$ \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) $$ 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

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

In the uncatalyzed reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ at \(100^{\circ} \mathrm{C}\) the pressures of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377 \mathrm{~atm}\) and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm} .\) What would happen to these pressures if a catalyst were present?
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