Question

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

Short answer

The enthalpy change, \(\Delta H^{\circ}\), for the given reaction, \(\mathrm{FeO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g)\), is approximately -49.33 kJ.

Step-by-step solution

Our goal is to find \(\Delta H^{\circ}\) for the reaction: $$ \mathrm{FeO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) $$ #Step 2: Manipulate given reactions#

We need to reverse and multiply the given reactions to match the target reaction. Observing the given reactions, we notice that reaction 1 and reaction 3 need to be reversed based on products and reactants in the target reaction. Also, we need to manipulate them so that they can be combined in such a way that some reactants/products get canceled, and only the components of the target reaction are left. For reaction 1: Reverse the reaction and multiply it by \(\frac{1}{3}\). The revised reaction becomes: $$ \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \frac{1}{3}\mathrm{CO}(g) \hspace{1cm} \Delta H^{\circ} = 23 \frac{\mathrm{kJ}}{3} $$ For reaction 3: Reverse the reaction. The revised reaction becomes: $$ 3\mathrm{FeO}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{3}\mathrm{O}_{4}(s) + \mathrm{CO}(g) \hspace{1cm} \Delta H^{\circ} = -18 \mathrm{kJ} $$ #Step 3: Combine revised reactions#

Now, we can combine the revised reactions: $$ \begin{array}{lll} \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \frac{1}{3}\mathrm{CO}(g) & & \Delta H^{\circ} = 23 \frac{\mathrm{kJ}}{3} \\ &+& \\ \mathrm{Fe}_{3}\mathrm{O}_{4}(s) + \mathrm{CO}(g) \longrightarrow 3\mathrm{FeO}(s) + \mathrm{CO}_{2}(g) & & \Delta H^{\circ} = -18 \mathrm{kJ} \\ &+& \\ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) & & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ \hline \mathrm{FeO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) & & \Delta H^{\circ} = ? \end{array} $$ Note that \(\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\), \(\frac{1}{3}\mathrm{CO}(g)\), \(\mathrm{Fe}_{3}\mathrm{O}_{4}(s)\), \(3\mathrm{FeO}(s)\), and \(\mathrm{CO}_{2}(g)\) got canceled, leaving only the components of the target reaction. #Step 4: Calculate \(\Delta H^{\circ}\) for target reaction#

To find the \(\Delta H^{\circ}\) for the target reaction, we need to sum the enthalpy changes of the revised reactions: $$ \Delta H^{\circ}_{\text{target}} = 23 \frac{\mathrm{kJ}}{3} - 18 \mathrm{kJ} - 39 \mathrm{kJ} $$ $$ \Delta H^{\circ}_{\text{target}} = 7.67 \mathrm{kJ} - 18 \mathrm{kJ} - 39 \mathrm{kJ} $$ $$ \Delta H^{\circ}_{\text{target}} = -49.33 \mathrm{kJ} $$ Therefore, the enthalpy change, \(\Delta H^{\circ}\), for the given reaction is approximately -49.33 kJ.

Key concepts

Enthalpy Change

In chemistry, enthalpy change is a core concept when studying thermodynamics and chemical reactions. It represents the amount of heat absorbed or released during a chemical reaction at constant pressure. Enthalpy is usually denoted by the symbol \( H \), and the enthalpy change by \( \Delta H \).To determine \( \Delta H \) for a reaction, we commonly use Hess's Law, which states that the total enthalpy change for a reaction is the same, regardless of the path taken. This is because enthalpy is a state function: it depends only on the initial and final states, not on the specific process pathway.In the exercise above, we used Hess's Law to calculate the enthalpy change of the target reaction. We manipulated given chemical equations to align with the target reaction, then combined them, ensuring intermediate steps cancel out, so we derived the \( \Delta H \) for the reaction \( \mathrm{FeO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) \). The total enthalpy change of approximately -49.33 kJ indicates that the reaction is exothermic, meaning it releases heat into the surroundings.

Thermodynamics

Thermodynamics is the branch of physical science concerned with heat and its relation to other forms of energy and work. It encompasses laws and concepts that describe the behavior of energy in systems, like chemical reactions, which makes it crucial to understanding reactions' enthalpy changes.The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed. In chemical reactions, this means the energy absorbed as reactants transform into products must equal the energy released. This is fundamental when calculating and understanding the enthalpy changes using Hess's Law in thermochemical equations.In our exercise, the principle of energy conservation ensures that our calculated enthalpy change \( \Delta H = -49.33 \ \mathrm{kJ} \) remains consistent and correct, demonstrating how effectively energy transitions between different states in the chemical system of the reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, accompanied by changes in energy levels. These changes are often observed through enthalpy changes, helping us understand whether a reaction releases or absorbs heat.In this exercise, the target chemical reaction is \( \mathrm{FeO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) \). Reactions can be exothermic, releasing energy (heat), or endothermic, where they absorb energy. The given task involved determining whether this specific reaction is exothermic or endothermic.By analyzing given reactions using Hess's Law and calculating the enthalpy change, we found that the reaction releases heat, evident by the negative enthalpy change value of -49.33 kJ. Understanding these energy changes provides crucial insight into how and why chemical reactions proceed, and fuels further studies into reaction mechanisms and energy efficiencies.