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Chapter 15: Problem 53

Consider the statement: The equilibrium constant of a reacting mixture of solid \(\mathrm{NH}_{4} \mathrm{Cl}\) and gaseous \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\) is \(0.316 .\) List three important pieces of information that are missing from this statement.

Short Answer

Expert verified
The missing information from the statement includes: the temperature at which the reaction is happening, the balanced chemical equation representing the reaction, and whether the equilibrium constant is in terms of pressure or concentration.

Step by step solution

01

Missing Information 1

One piece of missing crucial information is temperature. Temperature affects the magnitude of the equilibrium constant. Therefore, we need to know the temperature at which the reaction equilibrium constant is 0.316. This would enable to understand and compare this equilibrium constant with others.
02

Missing Information 2

The consideration of the involved chemical entities is missing. Although solids such as \(\mathrm{NH}_{4} \mathrm{Cl}\) and gases like \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\) have been mentioned, the actual chemical reaction has not been provided. The stoichiometry of the reaction affects the equilibrium constant. Knowing the reaction will allow us to understand how the substances interact in order to reach the equilibrium state. A balanced chemical equation would provide clarity.
03

Missing Information 3

The third missing piece of information is about the pressure or concentration of gases. For gaseous reactions, the equilibrium constant can be expressed in terms of either partial pressures or molar concentrations. It is crucial to specify whether the equilibrium constant given is in terms of concentration (\(K_c\)) or pressure (\(K_p\)).

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Most popular questions from this chapter

Explain Le Châtelier's principle. How can this principle help us maximize the yields of reactions?

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M\). (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

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?
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