Question

Consider the following reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090\) For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), \(\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{~kJ} / \mathrm{mol}\) \(S^{\circ}=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=\) \(-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(b\) to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\). e. Estimate the value of \(K\) at \(500 . \mathrm{K}\). f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=2.0\) torr, and \(P_{\mathrm{HOCl}}=0.10\) torr.

Short answer

a. For the reaction, ΔG° = 5493.09 J/mol. b. The student will need to look up bond energy values and perform calculations to estimate ΔH°. c. ΔS° = (ΔH° - 5493.09 J/mol) / (298 K). The student would need to plug in the estimated ΔH° value to calculate ΔS°. d. Using the known values for Cl₂O and the calculated values for the reaction, the student will need to estimate ΔHf° and S° for HOCl. e. The student must integrate the Van't Hoff equation using the estimated ΔH° value and the initial K value of 0.090 to find K at 500 K. f. Upon converting the given pressures to atmospheres and calculating the reaction quotient, use the non-standard Gibbs free energy equation to find ΔG.

Step-by-step solution

Use the given equation to calculate ΔG°

The equation for calculating ΔG° is given as: ΔG° = -RTln(K) Where R is the gas constant (8.314 J/mol·K), T is the temperature (298 K) and K is the equilibrium constant (0.090).

Calculate ΔG°

Plug in the values into the equation: ΔG° = - (8.314 J/mol·K) * (298 K) * ln(0.090) ΔG° = 5493.09 J/mol #b. Estimate ΔH° for the reaction using bond energy values# The student will need to use the provided bond energy table to look up the bond energy values and perform the necessary calculations. #c. Estimate ΔS° for the reaction#

Use the results from parts a and b to find ΔS°

The equation relating ΔG°, ΔH°, and ΔS° is: ΔG° = ΔH° - TΔS°

Solve for ΔS°

Rearrange the equation and solve for ΔS°: ΔS° = (ΔH° - ΔG°) / T ΔS° = (ΔH° - 5493.09 J/mol) / (298 K)

Calculate ΔS°

Plug in the estimated ΔH° value from part b and perform the calculation. #d. Estimate ΔHf° and S° for HOCl# The student will use the known values for Cl₂O and the calculated values for the reaction to estimate the values for HOCl. #e. Estimate the value of K at 500 K#

Use van't Hoff equation to find K at 500 K

The Van't Hoff equation is: \(\frac{d\ln K}{dT} = \frac{\Delta H}{RT^2}\)

Integrate the equation and solve for K

Integrate the equation with respect to temperature and solve for K at 500 K, using the estimated ΔH° value from part b and the initial K value (0.090) at 298 K. #f. Calculate ΔG at 25°C for given pressures#

Use the non-standard Gibbs free energy equation

The equation for non-standard Gibbs free energy is: ΔG = ΔG° + RTln(Q)

Convert pressures to atmospheres

Convert the given pressures from torr to atmospheres, using the conversion factor 1 atm = 760 torr: P(H₂O) = 18 torr * (1 atm/760 torr) = 0.0237 atm P(Cl₂O) = 2.0 torr * (1 atm/760 torr) = 0.00263 atm P(HOCl) = 0.10 torr * (1 atm/760 torr) = 0.0001316 atm

Calculate the reaction quotient Q

The reaction quotient Q is given by: Q = (P(HOCl)²) / (P(H₂O) * P(Cl₂O))

Calculate ΔG

Plug in the values for ΔG°, R, T, and Q into the non-standard Gibbs free energy equation and calculate ΔG.

Key concepts

Gibbs Free Energy

Gibbs free energy, denoted as \(\Delta G\), is the energy associated with a chemical reaction that can be used to do work at constant temperature and pressure. It is a critical concept in understanding chemical equilibrium and spontaneity of reactions. The equation \(\Delta G = \Delta H - T\Delta S\) reveals the balance between enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and temperature (T). When calculating \(\Delta G\) for a reaction like \(\mathrm{H}_2\mathrm{O}(g) + \mathrm{Cl}_2\mathrm{O}(g) \rightleftharpoons 2\mathrm{HOCl}(g)\), we can determine whether the reaction is spontaneous by checking if \(\Delta G\) is negative. If the calculated \(\Delta G\) value is positive, the reaction is non-spontaneous under the given conditions. This concept also extends to non-standard conditions using the reaction quotient (Q) in the advanced equation \(\Delta G = \Delta G^\circ + RT\ln(Q)\), which allows us to calculate \(\Delta G\) at any point in a reaction, not just at equilibrium.

Enthalpy Change

Enthalpy change, symbolized as \(\Delta H\), is the heat change of a system at constant pressure and a measure of the total energy of a thermodynamic system. It includes energy needed to break bonds in the reactants and energy released when new bonds form in the products. When we look at a reaction such as the formation of \(\mathrm{HOCl}(g)\), the bond energy values can be used to estimate this heat change. A positive \(\Delta H\) indicates that a reaction is endothermic, meaning it absorbs heat from the surroundings. Conversely, a negative \(\Delta H\) signifies an exothermic reaction, releasing heat into the surroundings. Being able to estimate \(\Delta H\) gives us valuable insight into the energy profile of a reaction and is a crucial step in the calculation of \(\Delta S\) and \(\Delta G\).

Entropy Change

Entropy change, denoted by \(\Delta S\), represents the change in disorder or randomness of a system during a chemical reaction. An increase in entropy (\(\Delta S > 0\)) means the system becomes more disordered, such as when a solid melts into a liquid. If \(\Delta S\) is negative, the system becomes more ordered. When \(\mathrm{H}_2\mathrm{O}(g)\) and \(\mathrm{Cl}_2\mathrm{O}(g)\) react to form \(2\mathrm{HOCl}(g)\), we can estimate \(\Delta S\) using the relation between \(\Delta G\) and \(\Delta H\), as given by the equation \(\Delta G = \Delta H - T\Delta S\). Understanding the entropy change is key to predicting whether a reaction will occur spontaneously, since a higher entropy typically favors spontaneity, all other variables being equal.

van't Hoff Equation

The Van't Hoff equation describes the relationship between the equilibrium constant (K) and temperature (T). It is given by \(\dfrac{d\ln K}{dT} = \dfrac{\Delta H}{RT^2}\), which implies that the equilibrium constant is temperature dependent. By integrating this equation over a temperature range, one can estimate how the equilibrium constant varies with temperature. If the enthalpy change \(\Delta H\) is known, the Van't Hoff equation enables us to predict the direction in which K will shift when the temperature changes. For example, when estimating the equilibrium constant for the formation of \(\mathrm{HOCl}(g)\) at a different temperature, such as 500 K, this equation proves to be particularly helpful. Ultimately, it helps in understanding how the position of equilibrium and the extend of the reaction may vary with temperature.

Reaction Quotient

The reaction quotient, Q, indicates the current state of a reaction, compared to the equilibrium state. It is calculated the same way as the equilibrium constant, K, but for any set of concentrations or partial pressures, not necessarily at equilibrium. For a reaction like \(\mathrm{H}_2\mathrm{O}(g) + \mathrm{Cl}_2\mathrm{O}(g) \rightleftharpoons 2\mathrm{HOCl}(g)\), Q is expressed as \(Q = (P_\mathrm{HOCl})^2 / (P_\mathrm{H_2O} \cdot P_\mathrm{Cl_2O})\). When comparing Q to the equilibrium constant, K, we can predict the direction in which a reaction will proceed to reach equilibrium. If Q < K, the reaction will move forward, whereas if Q > K, the reaction will move in reverse. The reaction quotient is crucial in determining \(\Delta G\) under non-standard conditions, as it is part of the equation \(\Delta G = \Delta G^\circ + RT\ln(Q)\), influencing the spontaneity of the reaction at any given moment.