What do the symbols \(K_{c}\) and \(K_{P}\) represent?

Short Answer

Expert verified
\(K_{c}\) is the equilibrium constant with respect to concentration while \(K_{P}\) is the equilibrium constant in terms of partial pressure.

Step by step solution

01

Understanding \(K_{c}\)

The symbol \(K_{c}\) stands for the equilibrium constant with respect to concentration. It is a measure of the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.
02

Understanding \(K_{P}\)

\(K_{P}\) also stands for the equilibrium constant, but in terms of partial pressure. It is used when dealing with gases and is a measure of the ratio of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.
03

Distinguishing between \(K_{c}\) and \(K_{P}\)

Both \(K_{c}\) and \(K_{P}\) are measures of the position of equilibrium for a chemical reaction. While the former uses concentrations (usually in mol/L), the latter uses partial pressures (usually in atmospheres), making it more appropriate for gases.

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Most popular questions from this chapter

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?

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

Outline the steps for calculating the concentrations of reacting species in an equilibrium reaction.

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}.\)

Pure NOCl gas was heated at \(240^{\circ} \mathrm{C}\) in a 1.00 - \(\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm. $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

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