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Hydrogen fluoride decomposes according to the following equation: $$ 2 \mathrm{HF}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) $$ The value of \(K_{c}\) at room temperature is \(1.0 \times 10^{-95}\). From the magnitude of \(K_{c}\), do you think the decomposition occurs to any great extent at room temperature? If an equilibrium mixture in a \(1.0\) - \(\mathrm{L}\) vessel contains \(1.0 \mathrm{~mol} \mathrm{HF}\), what is the amount of \(\mathrm{H}_{2}\) formed? Does this result agree with what you expect from the magnitude of \(K_{c}\) ?

Short Answer

Expert verified
The reaction does not occur to a great extent; only \(1.0 \times 10^{-47.5}\) mol \(\mathrm{H}_{2}\) is formed.

Step by step solution

01

Understand the Reaction and Equilibrium Constant

The reaction given is the decomposition of hydrogen fluoride: \(2 \mathrm{HF}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\). The equilibrium constant \(K_c = 1.0 \times 10^{-95}\) is extremely small. This indicates that, at equilibrium, the concentration of products \(\mathrm{H}_2\) and \(\mathrm{F}_2\) is much lower compared to the concentration of the reactant \(\mathrm{HF}\).
02

Define the Equilibrium Expression

The equilibrium constant \(K_c\) for the decomposition is expressed as:\[ K_c = \frac{[\mathrm{H}_2][\mathrm{F}_2]}{[\mathrm{HF}]^2} \]
03

Assign Initial and Change in Concentrations

Initially, we have \([\mathrm{HF}] = 1.0\) mol/L and \([\mathrm{H}_2] = [\mathrm{F}_2] = 0\). Let \(x\) be the amount of \([\mathrm{H}_2]\) formed at equilibrium, then \([\mathrm{F}_2]\) will also be \(x\) and \([\mathrm{HF}] = 1.0 - 2x\).
04

Write the Equilibrium Concentrations

At equilibrium, the concentrations are:\([\mathrm{HF}] = 1.0 - 2x\), \([\mathrm{H}_2] = x\), and \([\mathrm{F}_2] = x\).
05

Set Up the Equation Using \(K_c\)

Substitute the equilibrium concentrations into the expression for \(K_c\):\[ K_c = \frac{x^2}{(1.0 - 2x)^2} = 1.0 \times 10^{-95} \].
06

Solve for \(x\)

Because \(K_c\) is extremely small, it implies \(x\) is also very small and \(1.0 - 2x \approx 1.0\). Thus, we can simplify the equation:\[ x^2 \approx 1.0 \times 10^{-95} \]\[ x \approx \sqrt{1.0 \times 10^{-95}} \approx 1.0 \times 10^{-47.5} \].
07

Interpret the Result

The small value of \(x\) (\(1.0 \times 10^{-47.5}\)) means that hardly any \(\mathrm{H}_2\) (or \(\mathrm{F}_2\)) is formed at equilibrium, confirming that the reaction does not proceed to any significant extent. This agrees with the small \(K_c\) value.

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Key Concepts

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

Equilibrium Constant
In the study of chemical reactions, the equilibrium constant, denoted as \(K_c\), plays a crucial role. It is a numerical value that indicates the ratio of concentrations of products to reactants at equilibrium, and is specific to a particular reaction at a given temperature. For the decomposition of hydrogen fluoride (HF) into hydrogen (Hâ‚‚) and fluorine (Fâ‚‚), the provided equilibrium constant \(K_c = 1.0 \times 10^{-95}\) is extremely small.
This value reveals that, at equilibrium, the concentration of HF is substantially greater than that of Hâ‚‚ and Fâ‚‚. Therefore, the reaction hardly proceeds forward at room temperature. A small \(K_c\) value suggests the reaction strongly favors the reactants.
Understanding \(K_c\) gives a direct insight into how much of each species is present at equilibrium, highlighting whether the forward or reverse reaction is predominant.
Decomposition Reaction
Decomposition reactions involve the breakdown of a compound into simpler substances or elements. In our equation, \(2 \mathrm{HF}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{F}_{2}(g)\) represents the decomposition of hydrogen fluoride into hydrogen gas and fluorine gas.
This type of reaction is endothermic, meaning it requires energy input to break the bonds in HF, allowing it to decompose into Hâ‚‚ and Fâ‚‚.
Decomposition reactions are often affected by temperature, catalysts, and pressure. Here, the marked low tendency of HF to decompose (as indicated by the extremely small \(K_c\)) suggests that even under ambient temperature conditions, this breakdown happens to a very limited extent.
  • Decomposition can be reversible, as seen in this reaction where Hâ‚‚ and Fâ‚‚ could recombine.
  • Reversibility points to the existence of chemical equilibrium.
Understanding how decomposition behaves under different conditions is critical for predicting the behavior of chemical systems.
Equilibrium Concentration
Equilibrium concentration refers to the amount of reactants and products present in a reaction mixture at equilibrium. For the HF decomposition reaction, the concentration changes as the system reaches equilibrium must be understood through an ICE (Initial, Change, Equilibrium) table setup.
At the start, we have an initial concentration of HF as 1.0 mol/L, while Hâ‚‚ and Fâ‚‚ start at 0 mol/L each. As the reaction progresses towards equilibrium, a small change (denoted as \(x\)) occurs due to the reaction's direction. This change results in an equilibrium concentration of HF as \(1.0 - 2x\), while Hâ‚‚ and Fâ‚‚ both become \(x\).
Given the very small value of \(K_c\), the \(x\) representing Hâ‚‚ and Fâ‚‚ concentrations at equilibrium is negligible, emphasizing the reaction's reluctance to proceed. These calculations are fundamental in predicting the quantities of substances in a balanced system.
Equilibrium Expression
An equilibrium expression outlines how the concentrations of reactants and products relate in a chemical equilibrium state. For the decomposition of HF, the equilibrium expression is given by:\[K_c = \frac{[\mathrm{H}_2][\mathrm{F}_2]}{[\mathrm{HF}]^2}\]This formula shows that \(K_c\) is the product of the concentrations of Hâ‚‚ and Fâ‚‚, divided by the square of the concentration of HF.
By substituting equilibrium concentrations into this equation, scientists can deduce the balance between products and reactants.
For this reaction, the extremely low \(K_c\) value indicates that the numerator \([\mathrm{H}_2][\mathrm{F}_2]\) is exceedingly small compared to the denominator \([\mathrm{HF}]^2\), reflecting minimal forward reaction progress.
  • Equilibrium expressions are essential for quantifying reaction extents and comparing different reactions.
  • They help in predicting how changes in conditions affect a system's equilibrium.
Overall, the equilibrium expression serves as a valuable tool for analyzing and understanding chemical equilibria in various reactions.

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Most popular questions from this chapter

Consider the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\). Draw a graph illustrating the changes of concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) as equilibrium is approached. Describe how the rates of the forward and reverse reactions change as the mixture approaches dynamic equilibrium. Why is this called a dynamic equilibrium?

You place \(0.600\) mol of nitrogen, \(\mathrm{N}_{2}\), and \(1.800 \mathrm{~mol}\) of hydrogen, \(\mathrm{H}_{2}\), into a reaction vessel at \(450^{\circ} \mathrm{C}\) and \(10.0 \mathrm{~atm}\). The reaction is $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ What is the composition of the equilibrium mixture if you obtain \(0.048\) mol of ammonia, \(\mathrm{NH}_{3}\), from it?

At \(850^{\circ} \mathrm{C}\) and \(1.000\) atm pressure, a gaseous mixture of carbon monoxide and carbon dioxide in equilibrium with solid carbon is \(90.55 \%\) CO by mass. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Calculate \(K_{c}\) for this reaction at \(850^{\circ} \mathrm{C}\).

In the contact process, sulfuric acid is manufactured by first oxidizing \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\), which is then reacted with water. The reaction of \(\mathrm{SO}_{2}\) with \(\mathrm{O}_{2}\) is $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ A \(2.000\) - \(\mathrm{L}\) flask was filled with \(0.0400 \mathrm{~mol} \mathrm{SO}_{2}\) and \(0.0200\) mol \(\mathrm{O}_{2}\). At equilibrium at \(900 \mathrm{~K}\), the flask contained \(0.0296\) \(\mathrm{mol} \mathrm{SO}_{3}\). How many moles of each substance were in the flask at equilibrium?

Calculate the composition of the gaseous mixture obtained when \(1.25 \mathrm{~mol}\) of carbon dioxide is exposed to hot carbon at \(800^{\circ} \mathrm{C}\) in a \(1.25\) -L vessel. The equilibrium constant \(K_{c}\) at \(800^{\circ} \mathrm{C}\) is \(14.0\) for the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$

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