Problem 64 - Chemistry The Central Science

Question

From the enthalpies of reaction $$\begin{aligned} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) & \Delta H=-221.0 \mathrm{kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{CH}_{3} \mathrm{OH}(g) & \Delta H=-402.4 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$

Step-by-Step Solution

Answer

Therefore, the enthalpy change for the target reaction is: \(\Delta H = \Delta H_2 - \Delta H'_1 = -402.4 kJ - 110.5 kJ = -512.9 kJ\)

Step 1: Define the target reaction

Our target reaction is: \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\)

Step 2: Analyze the given reactions

We are given two reactions: Reaction 1: \(2 C(s) + O_{2}(g) \longrightarrow 2 CO(g) \qquad \Delta H_1 = -221.0 kJ\) Reaction 2: \(2 C(s) + O_{2}(g) + 4 H_{2}(g) \longrightarrow 2 CH_{3}OH(g) \qquad \Delta H_2 = -402.4 kJ\)

Step 3: Modify the given reactions to obtain the target reaction

To obtain the target reaction from the given reactions, we can perform the following operations: 1. Divide Reaction 1 by 2. 2. Reverse the modified Reaction 1. 3. Subtract the modified Reaction 1 from Reaction 2. Reaction 1 (modified): \(\frac{1}{2}(2 C(s) + O_{2}(g) \longrightarrow 2 CO(g))\) \(C(s) + \frac{1}{2}O_{2}(g) \longrightarrow CO(g) \qquad \frac{\Delta H_1}{2} = -110.5 kJ\) Reverse Reaction 1 (modified): \(CO(g) \longrightarrow C(s) + \frac{1}{2}O_{2}(g) \qquad \Delta H'_1 = 110.5 kJ\) Finally, \(Reaction\, 2 - Reverse\, Reaction\, 1 (modified)\) \(2 C(s) + O_{2}(g) + 4 H_{2}(g) - [CO(g) - (C(s) + \frac{1}{2}O_{2}(g))] \longrightarrow 2 CH_{3}OH(g) - C(s) - \frac{1}{2}O_{2}(g)\) \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\)

Key Concepts

Thermochemistry

Thermochemistry is the study of energy changes that occur during chemical reactions and changes in state. In many reactions, energy is either absorbed or released, and this energy change is often linked to the concept of enthalpy, denoted as \(\Delta H\).

Enthalpy is the heat content of a system at constant pressure. When a reaction releases heat, it is exothermic and \(\Delta H\) is negative. Conversely, when a reaction absorbs heat, it is endothermic and \(\Delta H\) is positive.

In our example, comparing the given reactions reveals their heat changes. The calculated \(\Delta H\) values are critical in determining whether the reactions are exothermic or endothermic, providing insight into the energy flow during the reactions. Changes in enthalpy help predict whether reactions will occur naturally.

• Reactions that release heat (negative \(\Delta H\)) are often more spontaneous.
• The values of enthalpy changes are key for applications in various fields like environmental science and engineering.

Hess's Law

Hess's Law is a crucial principle in thermochemistry, stating that the total enthalpy change for a chemical reaction is the same, irrespective of the pathway the reaction takes. This is incredibly useful because it allows us to calculate enthalpy changes that might not be easily measurable directly.

In the original problem, to find \(\Delta H\) for the target reaction \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\), we applied Hess's Law by manipulating the given reactions. By reversing and scaling reactions, we reshaped them to derive the desired reaction while adjusting \(\Delta H\).

This exemplifies how Hess's Law simplifies complex calculations, enabling us to effectively determine the energy changes for reactions without directly measuring them.

• Hess's Law underlines the concept of conservation of energy.
• It is invaluable for complex reactions encountered in industrial and laboratory settings.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and formation of chemical bonds. They vary widely, from simple combinations to complex decomposition or substitutions.

The reactions given in the problem demonstrate how elements like carbon, oxygen, and hydrogen can form different compounds through chemical processes. Understanding the enthalpy changes helps illustrate how energy is distributed in these transformations.

In the exercise, the manipulation of equations shows the complex interplay of such reactions, underscoring the dependency on parameters like temperature and pressure.

• Reactant bonds are broken, requiring energy, and new products are formed, releasing energy.
• Chemical equilibrium plays a key role in determining the direction and extent of reactions.

These concepts connect to broader themes in chemistry, such as reaction rates and catalysis, and are foundational for understanding diverse scientific phenomena and industrial processes.