Problem 88 - Chemistry The Molecular Nature of Matter

Question

Given the following reactions and their \(\Delta G^{\circ}\) values, \(\mathrm{COCl}_{2}(g)+4 \mathrm{NH}_{3}(g) \longrightarrow\) $$ \begin{aligned} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s)+2 \mathrm{NH}_{4} \mathrm{Cl}(s) & \Delta G^{\circ}=-332.0 \mathrm{~kJ} \\ \mathrm{COCl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{HCl}(g) \\ \Delta G^{\circ}=-141.8 \mathrm{~kJ} \end{aligned} $$ calculate the value of \(\Delta G^{\circ}\) for the reaction $$ \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{NH}_{3}(g) $$

Step-by-Step Solution

Answer

\(-473.8 \mathrm{kJ}\)

Step 1: Identify the Given Reactions

We are given two reactions with their \(\Delta G^{\circ}\) values:1. \(\mathrm{COCl}_{2}(g) + 4 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s) + 2 \mathrm{NH}_{4}\mathrm{Cl}(s)\) with \(\Delta G^{\circ} = -332.0 \mathrm{~kJ}\).2. \(\mathrm{COCl}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{HCl}(g)\) with \(\Delta G^{\circ} = -141.8 \mathrm{~kJ}\).

Step 2: Write the Target Reaction

The target reaction to find \(\Delta G^{\circ}\) is:\[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{NH}_{3}(g)\]

Step 3: Apply Hess's Law

To find the \(\Delta G^{\circ}\) for the target reaction, we need to manipulate the given reactions so that when they are combined, they will yield the target reaction. This requires adding or reversing the given reactions and adjusting their stoichiometry, as necessary, while also applying those changes to the \(\Delta G^{\circ}\) values. We use Hess's Law, which states that the change in free energy for a process is the same, no matter how the process is carried out in a sequence of steps.

Step 4: Manipulate Reaction 1

Reaction 1 is already in an appropriate direction, but we need to remove \(\mathrm{NH}_{4}\mathrm{Cl}(s)\) from the product side. This means we need to find a reaction for \(\mathrm{NH}_{4}\mathrm{Cl}(s)\) that gives us \(\mathrm{NH}_{3}(g)\). Since Reaction 1 produces two moles of \(\mathrm{NH}_{4}\mathrm{Cl}(s)\), we should aim for a reaction that decomposes two moles of it.

Step 5: Manipulate Reaction 2

In Reaction 2, \(\mathrm{COCl}_{2}(g)\) reacts with water to form \(\mathrm{CO}_{2}(g)\) and \(2 \mathrm{HCl}(g)\). We need \(\mathrm{CO}_{2}(g)\) in our target reaction, so we keep this reaction as it is. Note that \(\mathrm{COCl}_{2}(g)\) cancels out when we combine reactions 1 and 2 since it is a reactant in both.

Step 6: Combine the Reactions

After manipulation, the target reaction is obtained by combining the two reactions:\[\mathrm{COCl}_{2}(g) + 4 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s) + 2 \mathrm{NH}_{4}\mathrm{Cl}(s)\]\[\mathrm{COCl}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{HCl}(g)\]The \(\mathrm{COCl}_{2}(g)\) and \(2 \mathrm{HCl}(g)\) will cancel out from both sides, leaving us with the target reaction.

Step 7: Calculate the Free Energy Change

The free energy change for the target reaction can now be calculated by adding the \(\Delta G^{\circ}\) values of the manipulated reactions. Since we are not reversing any reactions or changing stoichiometry, we simply add the given \(\Delta G^{\circ}\) values together:\[\Delta G^{\circ}_{\text{target}} = (-332.0 \mathrm{~kJ}) + (-141.8 \mathrm{~kJ})\]\[\Delta G^{\circ}_{\text{target}} = -473.8 \mathrm{kJ}\]

Key Concepts

Hess's Law

Hess's Law is a principle in chemistry that states the total enthalpy change for a chemical reaction is the same, regardless of the pathway by which the chemical reaction takes place, as long as the initial and final conditions are the same. This law is incredibly useful when the direct measurement of enthalpy change (\( \triangle H\text{, which is the heat content of a system at constant pressure} \) is challenging or the reaction itself is not easily executable in a laboratory setting.

By using Hess's Law, one can calculate the enthalpy change of a complex reaction by breaking it down into multiple simpler reactions whose enthalpy changes are known. This principle applies analogously to other state functions like Gibbs Free Energy (\( \triangle G \)), which is dependent on both enthalpy and entropy. In essence, Hess's Law allows us to piece together known reaction data, like a puzzle, to find the energy change of a reaction that would otherwise be cumbersome to determine experimentally.

For example, if we could not directly measure the Gibbs Free Energy for the reaction \( \text{CO}\text{(NH}_2\text{)}_2\text{(s)} + \text{H}_2\text{O(l)} \longrightarrow \text{CO}_2\text{(g)} + 2 \text{NH}_3\text{(g)} \), we can use the Gibbs Free Energy changes of related reactions provided to find the answer indirectly. Just as seen in the given problem, we combine the individual \( \triangle G^{\text{o}} \) values of the known reactions to obtain the \( \triangle G \) for the desired reaction, following the rules established by Hess's Law.

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It is a foundational aspect of chemistry because it deals with energy transformations and the directionality of chemical processes.

Gibbs Free Energy is a very important concept in chemical thermodynamics as it predicts the spontaneity of a reaction at constant pressure and temperature. The equation for Gibbs Free Energy is:\[ \triangle G = \triangle H - T\triangle S \]where \( \triangle G \) is the change in free energy, \( \triangle H \) is the change in enthalpy, \( T \) is the temperature, and \( \triangle S \) is the change in entropy. A negative value of \( \triangle G \) indicates a spontaneous reaction, while a positive value suggests a non-spontaneous reaction.

Understanding Gibbs Free Energy involves breaking down each component: enthalpy indicates the total heat content of the system, while entropy measures the degree of disorder or randomness. Temperature, the driving force for reaction spontaneity, modulates the balance between enthalpy and entropy to determine the system's favorability for a reaction. Chemical thermodynamics encapsulates these concepts, ensuring that students understand the delicate interplay of factors that govern the flow of energy and the feasibility of chemical reactions.

Enthalpy of Reaction

The enthalpy of reaction, often denoted as \( \triangle H \), is the heat change that occurs during a chemical reaction at constant pressure. It's an extensive property, meaning it depends on the amount of substance reacting, akin to mass or volume. Enthalpy is a state function; it depends only on the initial and final states of the system, not on the path taken to get there.

For reactions that release heat, such as combustion reactions, the enthalpy of reaction is negative, and these are termed exothermic reactions. Conversely, reactions that absorb heat from the surroundings have a positive enthalpy change and are endothermic. In the context of our exercise, the enthalpies (or, more specifically here, the free energies) of individual reactions are added to find the enthalpy of the target reaction, which can help predict how it will behave and whether it requires energy to proceed or releases energy.

The calculation of an enthalpy change in a reaction can be done by measuring the heat exchange with the surroundings or by using Hess's Law and known enthalpies of formation. This concept is of paramount importance when considering energy requirements and releases in chemical reactions, as it directly relates to the energy profiles of reactions and their potential for work and productivity in industrial and biological applications.