Question

Given the following data: $$\begin{array}{ll}\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) & \Delta H=-393 \mathrm{~kJ} \\ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) & \Delta H=-566 \mathrm{~kJ}\end{array}$$ Calculate \(\Delta H\) for the reaction \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g)\)

Short answer

The enthalpy change, \(\Delta H\), for the reaction \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g)\) is \(+110 \mathrm{~kJ}\).

Step-by-step solution

Write the desired reaction

Write down the desired reaction: $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g)$$

Manipulate the given reactions

Our goal is to manipulate the given reactions, so that when combined, they produce the desired reaction. First, we must find a way to cancel out CO2 from the given reactions to obtain CO in the product side. Notice that by reversing the first reaction and multiplying the second reaction by 1/2, we can cancel out CO2 and obtain the desired reaction: $$\mathrm{CO}_{2}(g) \rightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) \qquad \Delta H= +393 \mathrm{~kJ}$$ $$\frac{1}{2}(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)) \qquad \frac{1}{2} \Delta H= -283 \mathrm{~kJ}$$ Now, we can combine these two manipulated reactions: $$\mathrm{CO}_{2}(g) -\hspace{0.25cm} \mathrm{CO}_{2}(g) + \mathrm{O}_2(g) + \frac{1}{2}(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)) \rightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) + \mathrm{CO}(g)$$ Simplifying the reaction, we get: $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g)$$

Calculate \(\Delta H\) for the desired reaction

Sum the corresponding enthalpy changes of the manipulated reactions: $$\Delta H_\text{desired} = +393 \mathrm{~kJ} + (-283 \mathrm{~kJ})$$ Calculate the final enthalpy change for the desired reaction: $$\Delta H_\text{desired} = 110 \mathrm{~kJ}$$ Finally, we get: $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g) \qquad \Delta H= +110 \mathrm{~kJ}$$

Key concepts

Enthalpy Change

Enthalpy change is a critical concept in understanding how energy is absorbed or released during chemical reactions. It is represented by \( \Delta H \), which indicates the amount of heat energy exchanged.

The enthalpy change can be:

• Negative (\(-\Delta H\)): This means heat is released, typically in exothermic reactions.
• Positive (\(+\Delta H\)): This implies heat is absorbed, seen in endothermic reactions.

In our exercise, we aimed to find the enthalpy change for converting carbon and oxygen into carbon monoxide. We utilized given reactions and Hess's Law to determine \( \Delta H = +110 \text{ kJ} \), showing that energy is absorbed.

Thermochemistry

Thermochemistry is the study of heat changes in chemical reactions. It combines principles of chemistry and thermodynamics to predict how energy transfer occurs.

Key points include:

• Energy Conservation: Energy cannot be created or destroyed, only transferred.
• Hess's Law: The overall enthalpy change in a reaction is the sum of enthalpy changes of individual steps.

In the exercise, the indirect method of calculating \( \Delta H \) showcases Hess's Law. By rearranging known reactions, we calculated the desired reaction’s energy change, emphasizing how thermochemistry is applied in real-world problem-solving.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. During this process, bonds are broken and formed, resulting in energy changes.

Key concepts include:

• Reactants and Products: Substances involved in the reaction.
• Balancing Equations: Ensuring the same number of atoms on each side to conserve mass.
• Reaction Pathways: Different routes can be taken, showing the concept of reaction manipulation to find desired energies.

In the context of our problem, the given reactions were manipulated to yield the target reaction. This illustrates how understanding the nature of reactions helps us predict and calculate important parameters like enthalpy changes.