Question

In this problem, you calculate the error in assuming that \(\Delta H_{R}^{\circ}\) is independent of \(T\) for the reaction \(2 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}(s)+\mathrm{O}_{2}(g)\) The following data are given at \(25^{\circ} \mathrm{C}\) $$\begin{array}{lccc}\text { Compound } & \mathrm{CuO}(s) & \mathrm{Cu}(s) & \mathrm{O}_{2}(\mathrm{g}) \\\\\hline \Delta H_{f}^{\circ}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) & -157 & & \\ \Delta G_{f}^{\circ}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) & -130 & & \\\C_{P, m}\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) & 42.3 & 24.4 & 29.4\end{array}$$ a. From Equation (6.65), $$\int_{K_{P}\left(T_{0}\right)}^{K_{P}\left(T_{f}\right)} d \ln K_{P}=\frac{1}{R} \int_{T_{0}}^{T_{f}} \frac{\Delta H_{R}^{\circ}}{T^{2}} d T$$ To a good approximation, we can assume that the heat capacities are independent of temperature over a limited range in temperature, giving \(\Delta H_{R}^{\circ}(T)=\Delta H_{R}^{\circ}\left(T_{0}\right)+\) \(\Delta C_{P}\left(T-T_{0}\right)\) where \(\Delta C_{P}=\Sigma_{i} v_{i} C_{P, m}(i) .\) By integrat- ing Equation \((6.65),\) show that $$\begin{aligned}\ln K_{P}(T)=\ln K_{P}\left(T_{0}\right) &-\frac{\Delta H_{R}^{\circ}\left(T_{0}\right)}{R}\left(\frac{1}{T}-\frac{1}{T_{0}}\right) \\\&+\frac{T_{0} \times \Delta C_{P}}{R}\left(\frac{1}{T}-\frac{1}{T_{0}}\right) \\\&+\frac{\Delta C_{P}}{R} \ln \frac{T}{T_{0}}\end{aligned}$$ b. Using the result from part (a), calculate the equilibrium pressure of oxygen over copper and \(\mathrm{CuO}(s)\) at \(1275 \mathrm{K}\) How is this value related to \(K_{P}\) for the reaction \(2 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}(s)+\mathrm{O}_{2}(g) ?\) c. What value of the equilibrium pressure would you obtain if you assumed that \(\Delta H_{R}^{\circ}\) were constant at its value for \(298.15 \mathrm{K}\) up to \(1275 \mathrm{K} ?\)

Short answer

In this problem, it is found that the equilibrium pressure of oxygen at 1275 K is equal to Kp(1275 K). By comparing the equilibrium pressures calculated in steps 4 and 5, it is observed that there is a notable error when assuming the standard reaction enthalpy is independent of the temperature for the given reaction. This error emphasizes the importance of considering temperature effects on reaction properties for accurate predictions.

Step-by-step solution

Calculate ΔCp and ΔH°(T0) for the Reaction

For the given reaction, 2CuO(s) <=> 2Cu(s) + O2(g) We know that ΔCp = Σ(νi * Cp,m(i)) Using the given values: ΔCp = 2 * Cp(Cu) + Cp(O2) - 2 * Cp(CuO) = 2 * 24.4 + 29.4 - 2 * 42.3 ΔH°(T0) for the reaction can be calculated from the sum of the enthalpy of formation of products minus the sum of the enthalpy of formation of reactants. At 298.15 K: ΔH°(T0) = 2 * ΔHf°(Cu) + ΔHf°(O2) - 2 * ΔHf°(CuO) = 0 + 0 - 2 * (-157)

Use Equation (6.65) and Substitute Reaction Enthalpy Expression

We are given the following equation: ∫(d ln Kp) = (1/R) * ∫[(ΔH_R° * ΔCp * (T - T0))/T²]dT Incorporate the expression for the reaction enthalpy as a function of temperature ΔH_R°(T) = ΔH_R°(T0) + ΔCp * (T - T0) in the above equation. Now, integrate both sides with respect to T, and solve for ln Kp(T).

Finding Kp at T0

To find Kp at T0 = 298.15 K, we will use the following relationship of ΔGf° and Kp: ΔG_R°(T0) = -RT0 * ln Kp(T0) For the given reaction, ΔG_R°(T0) = 2 * ΔGf°(Cu) + ΔGf°(O2) - 2 * ΔGf°(CuO) Therefore, we can find Kp(T0) by using ΔG_R°(T0) value.

Calculate Equilibrium Pressure of Oxygen at 1275 K

By substituting the values of T = 1275 K and T0 = 298.15 K in the formula derived in Step 2, we can find the value of ln Kp(1275 K). Then, by exponentiating the value of ln Kp(1275 K), we can find Kp(1275 K). We know that Kp is related to the equilibrium pressure of O2 (P_O2) as: Kp = P_O2 Therefore, the equilibrium pressure of oxygen at 1275 K is equal to Kp(1275 K).

Calculate Equilibrium Pressure Assuming Constant ΔH_R°

For this step, we will calculate the equilibrium pressure at 1275 K, assuming that ΔH_R°(298.15 K) is the constant reaction enthalpy up to 1275 K. We can use the Van 't Hoff equation: ln(Kp2/Kp1) = -ΔH_R° * (1/T2 - 1/T1) / R Where T1 = 298.15 K, T2 = 1275 K, and Kp1 = Kp(T0). Now, calculating the value of Kp2 and finding the equilibrium pressure under the assumption of constant ΔH_R°. By comparing the equilibrium pressures from Step 4 and Step 5, we can observe the error in assuming that the standard reaction enthalpy is independent of temperature for the given reaction.

Key concepts

Standard Reaction Enthalpy

Standard reaction enthalpy, represented as \( \Delta H_{R}^{\circ} \), is a crucial concept in thermodynamics, referring to the total heat content, or enthalpy change, involved in a reaction under standard conditions. Standard conditions imply a pressure of 1 bar, commonly with reactants and products in their standard states. For example, calculating \( \Delta H_{R}^{\circ} \) for a reaction involves subtracting the sum of the enthalpies of the reactants from that of the products.

When exploring complex reactions, understanding how \( \Delta H_{R}^{\circ} \) varies with temperature is vital. It's not always true that this value remains constant across different temperatures. In reality, the enthalpy change can shift due to variables like heat capacity changes and phase transitions of the substances involved. Hence, assessing the accuracy of \( \Delta H_{R}^{\circ} \) at different points is important for accurate predictions of a reaction's thermodynamics.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a measure of the proportions of reactants to products at chemical equilibrium. Two common forms are \( K_c \) and \( K_p \) for molar concentration and partial pressure respectively. These constants are determined by the specific reaction at a given temperature and remain constant unless that temperature changes.

For the reaction \( 2 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}(s) + \mathrm{O}_{2}(g) \), \( K_p \) would describe the ratio of the partial pressure of oxygen to the pressures of copper and copper oxide at equilibrium. Calculating \( K_p \) at different temperatures or using different assumptions can illustrate how sensitive equilibrium is to these conditions.

Temperature Dependence of Reaction

In thermodynamics, it's fundamental to understand that the behavior of a reaction, including the position of equilibrium, can change with temperature. The temperature dependence of a reaction is often examined through the reaction coefficient \( K \) and the reaction enthalpy \( \Delta H_R^{\circ} \), indicating how reactions shift to maintain energy balance. For example, with increasing temperature, the value of the equilibrium constant \( K \) for an endothermic reaction will typically increase, favoring the production of more products.

Furthermore, the temperature dependence is key to predicting how reaction rates and extents of reactions vary in different temperature conditions, which is a cornerstone of reaction engineering and thermodynamics. This plays a critical role in industrial applications where reactions must be optimized for efficiency and yield.

Van 't Hoff Equation

The Van 't Hoff equation is quintessential for understanding the effect of temperature on the equilibrium constant \( K \). This equation, stated as \( \frac{d ln K}{d T} = \frac{\Delta H_{R}^{\circ}}{R T^2} \) for a constant reaction enthalpy, links the change in \( K \) to the absolute temperature \( T \) and the reaction enthalpy \( \Delta H_{R}^{\circ} \). It shows that as the temperature changes, the equilibrium constant will also change, in a manner that is exponential with respect to the reciprocal of temperature. Using the integrated form of this equation, as seen in the given problem, allows for the calculation of the equilibrium constant at various temperatures when the standard reaction enthalpy is known.

The integrated Van 't Hoff equation played a pivotal role in the textbook problem, allowing the student to relate the equilibrium pressure of oxygen to \( K_p \) for the reaction in question. It demonstrated the impact of temperature on reaction equilibrium and provided a vivid example of temperature-dependent thermodynamic calculations.