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Chapter 38: Problem 4

In general, how does the rate of the forward reaction \((\mathrm{Y} \rightarrow \mathrm{Z})\) at equilibrium compare to the rate of the reverse reaction \((\mathrm{Z} \rightarrow \mathrm{Y})\) at equilibrium?

Short Answer

Expert verified
At equilibrium, the rate of the forward reaction (Y to Z) is equal to the rate of the reverse reaction (Z to Y). Both occur at the same rate.

Step by step solution

01

Define Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of the starting (reactant) and end (product) molecules do not change over time. This is because the rate at which the reactants produce the products (forward reaction) is the same as the rate at which the products produce the reactants (reverse reaction)
02

Apply the Concept to the Forward and Reverse Reactions

In this given scenario, Y converts to Z (forward reaction) and Z converts back to Y (reverse reaction). Both these reactions occur at the same rate.

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Most popular questions from this chapter

Consider a situation in which the reaction quotient, \(Q_{\mathrm{c}}\), for a given reaction is larger than the equilibrium constant, \(K_{\mathrm{c}}\) a) How must the value of \(\mathrm{Q}_{\mathrm{c}}\) change to reach equilibrium? b) Describe how the concentrations of reactants and/or the concentration of products must change to reach equlibrium.

Assume that the system is at equilibrium as determined in CTQ \(13 .\) What will happen to the number of moles of \(\mathrm{PCl}_{3}\) present if some \(\mathrm{Cl}_{2}\) gas is suddenly injected into the box? Explain your reasoning.

Assume that the system is at equilibrium as determined in CTQ \(13 .\) What will happen to the number of moles of \(\mathrm{PCl}_{3}\) present if some \(\mathrm{PCl}_{5}\) gas is suddenly injected into the box? Explain your reasoning.

Explain why the reaction quotient is useful.
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