Question

From the following data, calculate \(\Delta H_{R, 391.4 \mathrm{K}}^{\circ}\) for the reaction \(\mathrm{CH}_{3} \mathrm{COOH}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)\) $$\frac{\Delta H_{R}^{\circ}\left(\mathbf{k J} \mathbf{~ m o l}^{-1}\right)}{\mathrm{CH}_{3} \mathrm{COOH}(l)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)-871.5}$$ $$\begin{array}{ll} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & 40.656 \\ \mathrm{CH}_{3} \mathrm{COOH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(g) & 24.4 \end{array}$$ Values for \(\Delta H_{R}^{\circ}\) for the first two reactions are at \(298.15 \mathrm{K}\) and for the third reaction at \(391.4 \mathrm{K}\) $$\begin{array}{lccccc} \text { Substance } & \mathrm{CH}_{3} \mathrm{COOH}(l) & \mathrm{O}_{2}(g) & \mathrm{CO}_{2}(g) & \mathrm{H}_{2} \mathrm{O}(l) & \mathrm{H}_{2} \mathrm{O}(g) \\ \hline C_{P, m} / R & 14.9 & 3.53 & 4.46 & 9.055 & 4.038 \end{array}$$

Short answer

The combined enthalpy change for the reaction at 391.4 K is calculated as follows: \(\Delta H_{R, 391.4 \mathrm{K}}^{\circ} = 24.4 + 14.9(391.4 - 298.15) + 0 + 2[40.656 + 9.055(391.4 - 298.15)] -871.5 + 4.46(391.4 - 298.15)\) \(\Delta H_{R, 391.4\mathrm{K}}^{\circ} = -827.23 \mathrm{~kJ~mol^{-1}}\)

Step-by-step solution

Rewrite the given reaction with the required state changes.

The reaction can be rewritten as follows by introducing the state changes for the reactants and products: $$\mathrm{CH}_{3} \mathrm{COOH}(l) \longrightarrow \mathrm{CH}_{3}\mathrm{COOH}(g) \rightarroweqno{(1)}$$ $$2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{2}(g) \rightarroweqno{(2)}$$ $$2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(g) \rightarroweqno{(3)}$$ $$\mathrm{CH}_{3} \mathrm{COOH}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) + 2 \mathrm{CO}_{2}(g) \rightarroweqno{(4)}$$

Find \(\Delta H^{\circ}\) for each reaction at 298.15 K.

Using the given data, let's find the standard enthalpy change for each reaction at 298.15 K: For reaction (1): \(\Delta H_{1}^{\circ} = 24.4 \mathrm{~kJ~mol^{-1}}\) (given) For reaction (2): \(\Delta H_{2}^{\circ} = 0 \mathrm{~kJ~mol^{-1}}\) (since there's no change in state) For reaction (3): \(\Delta H_{3}^{\circ} = 40.656 \mathrm{~kJ~mol^{-1}}\) (given) For reaction (4): \(\Delta H_{4}^{\circ} = -871.5 \mathrm{~kJ~mol^{-1}}\) (given)

Calculate the combined enthalpy change at 391.4 K.

We can use the heat capacities, \(C_{P,m}\), to find the temperature-dependent enthalpy change from 298.15 K to 391.4 K using the following formula: $$\Delta H = \Delta H^{\circ} + \int_{298.15}^{391.4} C_{P,m}dT$$ Since the heat capacities are constant values, we can simplify: $$\Delta H = \Delta H^{\circ} + C_{P,m}\left(T_{2} - T_{1}\right)$$ Let's apply this to each reaction: $$\Delta H_{1}(391.4) = 24.4 + 14.9(391.4 - 298.15)$$ $$\Delta H_{2}(391.4) = 0$$ $$\Delta H_{3}(391.4) = 40.656 + 9.055(391.4 - 298.15)$$ $$\Delta H_{4}(391.4) = -871.5 + 4.46(391.4 - 298.15)$$

Calculate the combined enthalpy change at 391.4 K with stoichiometry.

Now we can find the combined enthalpy change for the reaction at 391.4 K using stoichiometry: $$\Delta H_{R, 391.4 \mathrm{K}}^{\circ} = \Delta H_{1}(391.4) + \Delta H_{2}(391.4) + 2\Delta H_{3}(391.4) + \Delta H_{4}(391.4)$$ Calculate each term and find the overall enthalpy change.

Key concepts

Enthalpy Change

Enthalpy change, often represented as \( \Delta H \), is a crucial concept in physical chemistry. It refers to the heat absorbed or released during a chemical reaction at constant pressure. When a reaction releases heat, \( \Delta H \) is negative, indicating an exothermic process. Conversely, when a reaction absorbs heat, \( \Delta H \) is positive, indicating an endothermic process.
To understand enthalpy change, consider the reaction of acetic acid with oxygen as given. Each phase transition—from liquid to gas for acetic acid or water—affects the process’s total enthalpy due to the heat absorbed or released in these changes. Enthalpy change provides insight into the energy profile of reactions, crucial for predicting reaction feasibility and energy requirements.
Using Hess's Law, the enthalpy change of a complex reaction is the sum of enthalpy changes of individual steps. This principle is invaluable in calculating enthalpy changes when direct measurement is challenging.

Standard Enthalpy of Reaction

The standard enthalpy of reaction, \( \Delta H^{\circ}_{R} \), is the enthalpy change when a reaction occurs under standard conditions: 1 atm pressure and 298.15 K (25°C). It is an important constant that helps in comparing the energy changes of different reactions under uniform conditions.
In exercises like this, calculating \( \Delta H^{\circ}_{R} \) at a temperature other than 298.15 K, like 391.4 K, involves using the standard enthalpy along with corrections for temperature changes. This allows for accurate predictions of enthalpy at the desired temperature, providing essential data for thermochemical calculations in various conditions.
The reactions' enthalpy changes are often available in data tables or are calculated using bond enthalpies. Hess's Law again plays a key role, allowing us to use known reactions to find unknown enthalpy changes.

Heat Capacity

Heat capacity, symbolized as \C_{P, m}\ and expressed in units of energy per mole per temperature change (e.g., kJ mol\(^{-1} \) K\(^{-1} \)), quantifies the amount of heat required to raise the temperature of a substance by one degree Kelvin at constant pressure. In the context of chemical reactions, substances have distinct molar heat capacities which influence the total enthalpy change between different temperatures.
In this exercise, the molar heat capacities are employed to adjust the standard enthalpy change from a baseline temperature (298.15 K) to 391.4 K using calorimetry calculations. This approach involves integrating heat capacity across the temperature range of interest, which in practice simplifies to multiplying the heat capacity by the difference in temperature, providing a vital link between standard conditions and the actual conditions of the reaction.
Heat capacity variations must be considered when designing experiments and industrial processes because they determine how much thermal energy is needed or released, impacting energy efficiency and safety.

Temperature Dependence of Reaction Enthalpy

The temperature dependence of reaction enthalpy is a key aspect of thermodynamics. While we often calculate enthalpy changes at standard conditions, real-world reactions often occur at temperatures quite different from 298.15 K. This necessitates adjusting \( \Delta H^{\circ} \) to reflect conditions relevant to the process at hand by integrating heat capacity data over the temperature range.
The temperature dependence is determined using the relation: \[ \Delta H = \Delta H^{\circ} + C_{P,m}(T_2 - T_1) \] By doing this, we adjust the enthalpy to account for additional heat gained or released as the components’ temperatures change from initial (\( T_1 \)) to final (\( T_2 \)) states.
This principle is not only important academically but has practical applications in industries where reactions are optimized for efficiency at various temperatures, such as in fuels, pharmaceuticals, and materials manufacturing. Understanding how reaction enthalpy varies with temperature is essential for designing effective and sustainable chemical processes.