Calculate the equilibrium constant ar \(25^{\circ} \mathrm{C}\) for cach of the following reactions from data available in Appendix 2A. (a) the combustion of hydrogen: \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) the oxidation of carbon monoxide: \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})\) (c) the decompostion of limestone: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

Short Answer

Expert verified
To calculate the equilibrium constants for the reactions given, use the provided \(\Delta G^\circ_{f}\) values for the respective compounds from Appendix 2A. Calculate \(\Delta G^\circ_{rxn}\) for each reaction using the formula \(\Delta G^\circ_{rxn} = \sum \Delta G^\circ_{f} \text{(products)} - \sum \Delta G^\circ_{f} \text{(reactants)}\), and then use the relationship \(\Delta G^\circ = -RT \ln K\) to find \(K\).

Step by step solution

01

- Convert \(\Delta G^\circ_{f}\) to \(K\) for the combustion of hydrogen

Use the relationship \(\Delta G^\circ = -RT \ln K\) to convert the standard free energy change for the reaction to the equilibrium constant, \(K\). For \(2 \mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}_2\mathrm{O}(\mathrm{g})\), we use the \(\Delta G^\circ_{f}\) values for \(\mathrm{H}_2\mathrm{O}(\mathrm{g})\) and \(\mathrm{O}_2(\mathrm{g})\) from Appendix 2A. Since \(\mathrm{O}_2\) is the reference form of oxygen, its \(\Delta G^\circ_{f}\) is zero. Calculate \(\Delta G^\circ_{rxn}\) using the formula \(\Delta G^\circ_{rxn} = \sum \Delta G^\circ_{f} \text{(products)} - \sum \Delta G^\circ_{f} \text{(reactants)}\). Then, use the \(\Delta G^\circ_{rxn}\) to find \(K\) at \(298 K\) (\(25^\circ C\)).
02

- Convert \(\Delta G^\circ_{f}\) to \(K\) for the oxidation of carbon monoxide

Repeat the process used in Step 1 for the reaction \(2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{g})\). Use the \(\Delta G^\circ_{f}\) values for \(\mathrm{CO}_2(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) to calculate \(\Delta G^\circ_{rxn}\), and subsequently determine \(K\).
03

- Calculate the equilibrium constant for the decomposition of limestone

For the reaction \(\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})\), use \(\Delta G^\circ_{f}\) values for \(\mathrm{CaCO}_3\), \(\mathrm{CaO}\), and \(\mathrm{CO}_2\). Both \(\mathrm{CaCO}_3\) and \(\mathrm{CaO}\) are solid and their \(\Delta G^\circ_{f}\) will be used directly. Calculate the \(\Delta G^\circ_{rxn}\) for the reaction, then convert that value to get the equilibrium constant, \(K\).

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Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics is a branch of thermodynamics that deals with the study of energy changes, particularly the energy changes during chemical reactions. It lays the foundation for understanding how energy is converted in chemical processes and how chemical equilibrium is established. One of the most fundamental concepts in chemical thermodynamics is the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.

This principle applies to chemical reactions where the energy contained in the bonds of reactants is converted during a reaction to form products, which have their own energy content. The difference in energy between reactants and products is known as enthalpy change (H). However, to fully understand chemical processes, we also need to consider entropy, a measure of system's disorder. The combination of enthalpy and entropy changes gives us the Gibbs free energy change, which is crucial for predicting reaction spontaneity.
Standard Free Energy Change
The standard free energy change (G^) represents the amount of energy available to do work when a chemical reaction occurs under standard conditions (1 bar of pressure and the specified temperature, usually 25°C or 298 K). It is a useful quantity because it provides insight into the thermodynamic favorability of a reaction.

To calculate G^ for a reaction, we consider the standard free energies of formation (G^_f) of the reactants and products. The formula G^_rxn = G^_f (products) - G^_f (reactants) gives us the change in free energy for the reaction. If G^_rxn is negative, the reaction is said to be exergonic, which means it can occur spontaneously under standard conditions. Conversely, a positive G^_rxn indicates an endergonic reaction, which is not spontaneous under standard conditions.
Gibbs Free Energy
Gibbs free energy, named after Josiah Willard Gibbs, is a thermodynamic quantity that provides comprehensive information about the direction and extent of chemical reactions. The Gibbs free energy (G) is crucial because it incorporates both enthalpy (H) and entropy (S), the two main factors that determine spontaneity. The relationship is defined by the equation G = H - T*S, where H is the enthalpy change, T is temperature in Kelvin, and S is the entropy change.

A key point is that the sign of G determines a reaction’s spontaneity. When G is negative, the reaction is spontaneous, meaning it can proceed without any external energy input. For positive G, the reaction is non-spontaneous and requires input of energy to occur. At equilibrium, G is zero, indicating that there is no net change occurring in the reaction mixture.
Reaction Spontaneity
Reaction spontaneity refers to whether a reaction will occur by itself without outside intervention. The spontaneity of a reaction is not necessarily related to the rate of the reaction – some spontaneous reactions can be very slow. The key determinants for spontaneity are the enthalpy (H) and entropy (S) changes during a reaction, which together affect the Gibbs free energy (G) of the process.

In thermodynamics, spontaneity is often discussed in terms of these energy changes: a spontaneous reaction tends to lower the free energy of the system (G < 0), while a non-spontaneous reaction increases it (G > 0). However, the actual determination of spontaneity is temperature-dependent and based on the Gibbs free energy change. Importantly, at the equilibrium point of a reaction, where the rates of the forward and reverse reactions are equal, there is no net change in the composition of the system, and the free energy change is zero (G = 0). This characteristic enables us to calculate the equilibrium constant, K, which predicts the ratio of products to reactants at this balance point.

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Most popular questions from this chapter

A 30.1-g sample of NOCl is placed into a \(200-\mathrm{mL}\) reaction vessel and heated to \(500 \mathrm{~K}\). The value of \(K\) for the decomposition of NOCl at \(500 \mathrm{~K}\) in the reaction \(2 \mathrm{NOCl}(\mathrm{g})=2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) is \(1.13 \times 10^{-3}\). (a) What are the equilibrium partial pressures of \(\mathrm{NOCl}, \mathrm{NO}\), and \(\mathrm{Cl}_{2}\) ? (b) What is the percentage decomposition of NOCl at this temperature?

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

(a) Calculate the reaction free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hl}(\mathrm{g})\) at \(700 \mathrm{~K}\) when the concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.026,0.33\), and \(1.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), respectively. For this reaction, \(K_{c}=54\) at \(700 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\) An analysis of a reaction vessel at \(2500 \mathrm{~K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{Cl}_{2}, 0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward cquilibrium?

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