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Write an equilibrium constant, \(K_{c},\) for the formation from its gaseous elements of \((a) 1\) mol \(\mathrm{HF}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g}) ;(\mathrm{c}) 2 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) 1 \mathrm{mol} \mathrm{ClF}_{3}(1)\)

Short Answer

Expert verified
The equilibrium constants are as follows: (a) \(K_{c} = \frac{[HF]^2}{[H_2][F_2]}\) (b) \(K_{c} = \frac{[NH_3]^2}{[N_2][H_2]^3}\) (c) \(K_{c} = \frac{[N_{2O}]^2}{[N_2][O_2]}\) (d) \( K_{c} = \frac{[ClF_{3}]^2}{[Cl_{2}][F_{2}]^3}\)

Step by step solution

01

Write equation for formation of HF(g)

The formation reaction for hydrofluoric acid, \(HF(g)\), can be written as:\[ H_{2(g)} + F_{2(g)} \rightarrow 2HF_{(g)}\]
02

Establish the equilibrium constant expression for HF(g)

The equilibrium constant expression for \(HF(g)\) is the product of the concentrations of the products divided by the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient, as follows:\[K_{c} = \frac{[HF]^2}{[H_2][F_2]}\]
03

Write equation for formation of NH3(g)

The formation reaction for ammonia, \(NH_{3(g)}\), is:\[N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}\]
04

Establish the equilibrium constant expression for NH3(g)

The equilibrium constant for this reaction is:\[K_{c} = \frac{[NH_3]^2}{[N_2][H_2]^3}\]
05

Write equation for the formation of N2O(g)

The formation reaction for nitrous oxide, \(N_{2O(g)}\), is:\[N_2(g) + O_2(g) \rightarrow 2N_{2O(g)}\]
06

Establish the equilibrium constant expression for N2O(g)

The equilibrium constant for this reaction is:\[K_{c} = \frac{[N_{2O}]^2}{[N_2][O_2]}\]
07

Write equation for the formation of ClF3(g)

The formation reaction for chlorotrifluoride, \(ClF_{3(g)}\), is:\[Cl_{2(g)} + 3F_{2(g)} \rightarrow 2ClF_{3(g)}]\]
08

Establish the equilibrium constant expression for ClF3(g)

The equilibrium constant for this reaction is:\[ K_{c} = \frac{[ClF_{3}]^2}{[Cl_{2}][F_{2}]^3}\]

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium represents a state in which the rate of the forward reaction equals the rate of the reverse reaction, making the concentrations of reactants and products constant over time. This does not mean that the reactants and products are equal in concentration; rather, their ratios remain stable due to ongoing dynamic processes. In a chemical equilibrium, these processes happen simultaneously, allowing a reaction to balance between reactants turning into products and products reverting back into reactants.

To understand an equilibrium constant, denoted as \(K_{c}\), consider it as a snapshot of this delicate balance, providing insights into the extent of a reaction at equilibrium. The value of \(K_{c}\) is calculated using the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. A large \(K_{c}\) indicates a reaction favoring products, while a small \(K_{c}\) suggests a preference for reactants.
  • Equilibrium is dynamic, with continuous motion at the molecular level.
  • \(K_{c}\) provides a measure of a reaction's progress at equilibrium.
  • Knowing \(K_{c}\) helps predict how changing conditions might affect a reaction.
Reaction Stoichiometry
Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. Understanding this concept is crucial as it determines the proportions in which different substances react and form products.

When writing balanced chemical equations, the stoichiometric coefficients indicate the relative amount of each substance involved. These coefficients are essential when calculating the equilibrium constant, since they dictate the exponents to which concentrations are raised in the \(K_{c}\) expression.

By using stoichiometry:
  • You can predict the amount of products generated from given reactants.
  • Determine limiting reactants to know the maximum yield possible.
  • Understand how much of each reactant is needed to consume completely.
For example, in the formation of ammonia \( (NH_3) \), the balanced equation \( N_2 + 3H_2 \rightarrow 2NH_3 \) shows that one mole of \( N_2 \) requires three moles of \( H_2 \), producing two moles of \( NH_3 \).
Gaseous Reactions
Gaseous reactions involve reactants and products in the gas phase and are described by equations that account for the behavior and interactions of these gases. An essential aspect to understand here is how volume, pressure, and temperature can affect gaseous reactions at equilibrium.

The equilibrium constant for gaseous reactions, \( K_c \), can vary with temperature and is based on concentrations (moles per unit volume). In many cases, reactions involving gases might also be expressed in terms of partial pressures, known as \( K_p \).

Concepts related to gaseous reactions:
  • Gases can be compressed and expanded significantly, affecting equilibrium position.
  • When temperature changes, the reaction may shift to absorb or release heat.
  • Ideal gas law equations often come in handy to relate different parameters involved.
Knowing how these factors interplay helps in predicting the direction of a reaction shift when external conditions change, according to Le Chatelier's principle.
Formation Reactions
Formation reactions describe how a compound forms from its elements, typically in their standard states. These reactions are fundamental in studying the thermodynamics and dynamics of chemical processes.

For every formation reaction, a corresponding chemical equation, like \( H_{2(g)} + F_{2(g)} \rightarrow 2HF_{(g)} \), provides the blueprint for calculating its equilibrium constant \( K_c \), crucial for understanding the reaction's feasibility and extent.

Key points about formation reactions:
  • They allow calculation of standard enthalpy changes, essential for thermodynamic studies.
  • Used to derive equilibrium constants, offering insight into the behavior of reactions.
  • Integral for synthesis processes, especially in industrial chemistry.
Grasping formation reactions helps in predicting how different elements might combine and provides the foundation for constructing more complex chemical networks.

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Most popular questions from this chapter

The two common chlorides of phosphorus, \(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5},\) both important in the production of other phosphorus compounds, coexist in equilibrium through the reaction $$ \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{g}) $$ At \(250^{\circ} \mathrm{C},\) an equilibrium mixture in a \(2.50 \mathrm{L}\) flask contains \(0.105 \mathrm{g} \mathrm{PCl}_{5}, 0.220 \mathrm{g} \mathrm{PCl}_{3},\) and \(2.12 \mathrm{g} \mathrm{Cl}_{2}\) What are the values of (a) \(K_{c}\) and (b) \(K_{\mathrm{p}}\) for this reaction at \(250^{\circ} \mathrm{C} ?\)

Starting with \(0.3500 \mathrm{mol} \mathrm{CO}(\mathrm{g})\) and \(0.05500 \mathrm{mol}\) \(\mathrm{COCl}_{2}(\mathrm{g})\) in a \(3.050 \mathrm{L}\) flask at \(668 \mathrm{K},\) how many moles of \(\mathrm{Cl}_{2}(\mathrm{g})\) will be present at equilibrium? $$\begin{aligned} \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}} &=1.2 \times 10^{3} \mathrm{at} \ 668 \mathrm{K} \end{aligned}$$

Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go to completion. Explain this fact in terms of Le Châtelier's principle.

The following two equilibrium reactions can be written for aqueous carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq}) & K_{1} \\ \mathrm{HCO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) & K_{2} \end{array} $$ For each reaction write the equilibrium constant expression. By using Le Châtelier's principle we may naively predict that by adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the system, the concentration of \(\mathrm{CO}_{3}^{2-}\) would increase. What we observe is that after adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the equilibrium mixture, an increase in the concentration of \(\mathrm{CO}_{3}^{2-}\) occurs when \(\left[\mathrm{CO}_{3}^{2-}\right] \ll \mathrm{K}_{2}\) however, the concentration of \(\mathrm{CO}_{3}^{2-}\) will decrease when \(\left[\mathrm{CO}_{3}^{2-}\right] \gg K_{2} .\) Show that this is true by considering the ratio of \(\left[\mathrm{H}^{+}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) before and after adding a small amount of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the solution, and by using that ratio to calculate the \(\left[\mathrm{CO}_{3}^{2-}\right]\)

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

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