Question

At \(298 \mathrm{K}, \Delta G_{\mathrm{f}}^{\mathrm{p}}[\mathrm{CO}(\mathrm{g})]=-137.2 \mathrm{kJ} / \mathrm{mol}\) and \(K_{\mathrm{p}}=\) \(6.5 \times 10^{11}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{COCl}_{2}(\mathrm{g}) . \quad\) Use these data to determine \(\Delta G_{f}\left[\mathrm{COCl}_{2}(\mathrm{g})\right],\) and compare your result with the value in Appendix D.

Short answer

The standard Gibbs free energy of formation for COCl2(g) at 298K is -207.3 kJ/mol.

Step-by-step solution

Formulate the solution based on the Gibbs free energy principle

The standard free energy change for a reaction can be calculated using the equation: \[\Delta G^{\circ} = -RT \ln K\]where \( R \) is the ideal gas constant (R = 8.314 J/mol.K), \( T \) is the temperature (T = 298K), and \( K \) is the equilibrium constant. For the reaction: \[\text{CO(g)} + \text{Cl}_2(g) \leftrightarrows \text{COCl}_2(g) \quad \quad ...(1)\]We can express the standard Gibbs free energy change (\( \Delta G^{\circ}_{\text{rxn}} \)) like this:\[\Delta G^{\circ}_{\text{rxn}} = \Delta G^{\circ}_{\text{f}}[\text{COCl}_2(g)] - \Delta G^{\circ}_{\text{f}}[CO(g)] - \Delta G^{\circ}_{\text{f}}[Cl_2(g)] ..(2)\]From the information given, we know that \(\Delta G^{\circ}_{\text{f}}[CO(g)] = -137.2 \, kJ/mol\) and because Cl2(g) is an element in its standard state, its standard Gibbs free energy of formation, \( \Delta G^{\circ}_{\text{f}}[Cl_2(g)] \), is zero. Therefore, we can simplify equation 2 as:\[\Delta G^{\circ}_{\text{rxn}} = \Delta G^{\circ}_{\text{f}}[\text{COCl}_2(g)] + 137.2 \, kJ/mol \quad \quad ...(3)\]

Calculate the Standard Gibbs Free Energy of Reaction from Equilibrium Constant

From the given equilibrium constant, we can calculate the standard free energy change for the reaction using \( \Delta G^{\circ} = -RT \ln K \). The value of Kp is given as \(6.5 \times 10^{11}\). Substituting the values of R, T, and K we get:\[\Delta G^{\circ}_{\text{rxn}} = -(8.314 \times 10^{-3} \, kJ/mol.K)(298K)\ln(6.5 \times 10^{11})\]

Solve for Gibbs Free Energy of Formation for COCl2(g)

Solving for \( \Delta G^{\circ}_{\text{rxn}} \) gives a value of -70.1kJ/mol. We can then substitute \( \Delta G^{\circ}_{\text{rxn}} \) into equation 3 and solve for \( \Delta G^{\circ}_{\text{f}}[\text{COCl}_2(g)] \): \[\Delta G^{\circ}_{\text{f}}[\text{COCl}_2(g)] = \Delta G^{\circ}_{\text{rxn}} - 137.2 \, kJ/mol = -70.1 \, kJ/mol - 137.2 \, kJ/mol = -207.3 \, kJ/mol\]

Key concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in which the forward and reverse rates of a chemical reaction are equal, resulting in no net change in the concentration of reactants and products over time. Imagine a seesaw perfectly balanced with two children of equal weight; this is the state of a chemical reaction at equilibrium. In a chemical equation, this can be represented by the double arrow \rightleftharpoons\text{CO}_{2}\text{COCl}_{2}\text{(g)}\text{(g)}\rightleftharpoons\text{) is the key link to understand the balance between reactants and products.

At equilibrium, the ratio of the concentrations of products to reactants is constant, known as the equilibrium constant, symbolized as K. This ratio is a critical aspect of the reaction and depending on its value, can tell us a lot about the system in question. For instance, a high K value implies a greater concentration of products, signifying that the reaction favors the product side, as shown in the given exercise with K_{p} = 6.5 \times 10^{11}\(, indicating a strong favor towards the production of COCl_{2}\text{(g)}\).

Standard Free Energy Change

The standard free energy change, denoted as \(\text{COCl}_{2}\text{(g)}\), is a thermodynamic property indicating the energy available to do work during a chemical process at a constant temperature and pressure, assuming that all reactants and products are in their standard states. This value can be positive or negative, indicating whether a reaction is non-spontaneous or spontaneous, respectively. In the context of the exercise, determining the \(\text{COCl}_{2}\text{(g)}\) for \(text{CO}_{2}\text{COCl}_{2}\text{(g)}\) - \(137.2 \) kJ/mol given in the problem statement.

Knowing the \(\text{COCl}_{2}\text{(g)}\) allows us to understand and predict the spontaneity and direction of a chemical reaction. When the \(\text{COCl}_{2}\text{(g)}\) is negative, as it is in our exercise, the reaction tends to occur spontaneously in the direction written.

Equilibrium Constant

The equilibrium constant, K, quantifies the ratio of product and reactant concentrations at equilibrium. For gas-phase reactions, as in our exercise, the equilibrium constant with respect to partial pressures is represented as \(K_{p} = 6.5 \times 10^{11}\), further emphasizing the considerable extent to which the products are favored. The value of K is a critical landmark in deciding the extent of the equilibrium.

Furthermore, the connection between K and the Gibbs free energy change (\(6.5 \times 10\text{COCl}_{2}\text{(g)}6.5 \times 10\text{COCl}_{2}\text{(g)}\) for this reaction. A large equilibrium constant, indicating that the products are heavily favored at equilibrium, corresponds to a large negative value of \(\Delta G^{\circ}\), reflective of a spontaneous reaction.

Thermodynamics

Thermodynamics is the branch of physical science that deals with heat and work and their relation to energy and molecules in a system. It provides the framework for understanding chemical equilibrium, free energy, and the equilibrium constant. The first and second laws of thermodynamics, in particular, govern the principles affecting chemical reactions.

In the case of the provided exercise, thermodynamics tells us that the spontaneity of a reaction at a given temperature depends on the Gibbs free energy change. If a reaction releases energy (exergonic, \(6.5 \times 10\text{COCl}_{2}\text{(g)}6.5 \times 10^{11}\) to calculate the standard Gibbs free energy change, showing us how thermodynamic principles apply to practical chemistry problems.