Problem 54 - General Chemistry

Question

Carbon disulfide burns in air, producing carbon dioxide and sulfur dioxide. $$ \mathrm{CS}_{2}(l)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g) ; \Delta H=-1077 \mathrm{~kJ} $$ What is \(\Delta H\) for the following equation? $$ \frac{1}{2} \mathrm{CS}_{2}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g) $$

Step-by-Step Solution

Answer

\( \Delta H = -538.5 \, \text{kJ} \) for the half reaction.

Step 1: Identify the Given Reaction

The given equation indicates that the reaction of 1 mole of CSi{2} with 3 moles of Oi{2} produces 1 mole of COi{2} and 2 moles of SOi{2}, releasing 1077 kJ of energy. This is an exothermic reaction with \( \Delta H = -1077 \, \text{kJ} \).

Step 2: Interpret the Desired Reaction

The question asks for \( \Delta H \) for the reaction \( \frac{1}{2} \mathrm{CS}_{2}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g) \).

Step 3: Adjust Reaction Scale

Notice the desired reaction contains half the moles of each substance compared to the given reaction. It is essentially half of the given reaction in terms of stoichiometry.

Step 4: Calculate \( \Delta H \) for Half Reaction

Since the desired reaction is exactly half of the given reaction, its enthalpy change \( \Delta H \) should also be half:\[\Delta H = \frac{-1077 \, \text{kJ}}{2} = -538.5 \, \text{kJ}\]

Key Concepts

Enthalpy Change

The concept of enthalpy change is fundamental in thermodynamics and chemistry. Enthalpy, represented by the symbol \( H \), refers to the heat content of a system. When a reaction occurs, the enthalpy change, \( \Delta H \), indicates whether heat is absorbed or released.
In the case of our reaction, \( \Delta H \) is \(-1077 \, \text{kJ} \), which implies that this much energy is released. Therefore, the original reaction is exothermic. For the adjusted reaction, which involves only half of the substances, we need to adjust \( \Delta H \) proportionally to reflect the change in scale. If we take half of \(-1077 \, \text{kJ} \), we get \(-538.5 \, \text{kJ} \). This new value of \( \Delta H \) represents the enthalpy change for our scaled-down reaction.

Exothermic Reaction

Exothermic reactions are reactions that release energy, usually in the form of heat, to their surroundings. This type of reaction often results in the increase of temperature in the environment where the reaction takes place.
In the original exercise, the burning of carbon disulfide is an example of an exothermic reaction. The negative sign in \( \Delta H = -1077 \, \text{kJ} \) indicates that the reaction releases energy.
Exothermic reactions are common in combustion processes, like when a fuel burns in air as in this example, and they play crucial roles in various industrial and natural processes.

• They release heat.
• They have a negative \( \Delta H \).
• They often occur spontaneously.

In identifying reactions such as the one in this exercise, it is important to recognize the signs of an exothermic event which include an increase in temperature and a negative \( \Delta H \).

Stoichiometry

Stoichiometry is the method of calculating the relative quantities of reactants and products in chemical reactions. It is based on the balanced chemical equation and relies on mole ratios. These ratios ensure that the conservation of mass is maintained in any chemical process.
In the given reaction, stoichiometry tells us the proportions in which carbon disulfide and oxygen react to form carbon dioxide and sulfur dioxide. From the balanced equation \( \mathrm{CS}_{2}(l)+3 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g) \), we see that 1 mole of \( \mathrm{CS}_{2} \) reacts with 3 moles of \( \mathrm{O}_{2} \).
The step-by-step solution adjusts these quantities to half to solve for the given reaction:

• \( \frac{1}{2} \mathrm{CS}_{2}(l) \)
• \( \frac{3}{2} \mathrm{O}_{2}(g) \)
• \( \frac{1}{2} \mathrm{CO}_{2}(g) \)
• \( \mathrm{SO}_{2}(g) \)

This scaling helps find the corresponding \( \Delta H \) while maintaining the ratio of reactants to products, essential for accurately predicting how real-world chemical reactions proceed.