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Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)

Short Answer

Expert verified
The values of \(K_{c}\) for the reactions are (a) 0.011, (b) \(2.2 \times 10^{5}\), and (c) 0.429.

Step by step solution

01

Solution to part (a)

For the equation \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), \(\Delta n = 2 - 1 = 1\). Therefore, \(K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{2.9 \times 10^{-2}}{(0.0821 \times 303)^1} = 0.011\).
02

Solution to part (b)

For the equation \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\), \(\Delta n = 2 - 3 = -1\). Also, the temperature needs to be converted from Celsius to Kelvin, thus \(T = 184 + 273 = 457K\). Therefore, \(K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{1.48 \times 10^{4}}{(0.0821 \times 457)^{-1}} = 2.2 \times 10^{5}\).
03

Solution to part (c)

For the equation \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), \(\Delta n = 3 - 3 = 0\). Therefore, \(K_{p}= K_c\) as \((RT)^{\Delta n} = 1\), hence \(K_{c} = 0.429\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a reversible reaction where the forward and backward reactions occur at the same rate. This means that the concentrations of reactants and products remain constant over time. In equations, chemical equilibrium is often represented by the equilibrium constant, either as \( K_c \) for concentrations or \( K_p \) for partial pressures, especially in gas reactions.
It's important to note that equilibrium does not mean that the reactants and products are equal in concentration. It simply implies that their rates of formation are equal, making the system stable. Understanding equilibrium is crucial in predicting how changes in conditions like temperature or pressure affect a system. By manipulating these factors, you can shift the equilibrium in favor of either reactants or products, known as Le Chatelier’s Principle.
Gas Phase Reactions
Gas phase reactions involve chemical processes where all reactants and products are in the gaseous state. They are significant because gases freely mix and expand, making the reactions often quicker than those in other phases.
The behavior of gases is described by the ideal gas law, \( PV = nRT \), where pressure \( P \), volume \( V \), moles \( n \), the gas constant \( R \), and temperature \( T \) are key components. Understanding these principles helps in calculating changes during reactions like pressure and volume shifts.
  • When converting \( K_p \) (pressure-based constant) to \( K_c \) (concentration-based constant) in gas reactions, the change in the number of moles \( \Delta n \) determines the conversion factor \((RT)^{\Delta n}\).
Converting between these constants allows chemists to better understand and predict the outcomes of gas reactions under different conditions.
Thermodynamics in Chemistry
Thermodynamics in chemistry deals with energy changes in chemical reactions. It helps explain why reactions occur and whether they will proceed spontaneously.
  • One key thermodynamic term is the free energy change, \( \Delta G \), indicating the spontaneity of a reaction.
  • Another is enthalpy, \( \Delta H \), representing heat exchange, and entropy, \( \Delta S \), representing disorder.
In gas reactions, thermodynamics also involves understanding how temperature and pressure affect the equilibrium state.
The relationship between \( K_p \) and \( K_c \) highlights the thermodynamic principles at play. As temperature influences these constants, adjusting temperature can lead to different thermodynamically favored conditions. This makes thermodynamics an essential tool for chemists in designing and controlling chemical processes efficiently.

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Most popular questions from this chapter

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

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