Problem 62 - General Chemistry

Question

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is mixed with gasoline and sold as gasohol. Use the following to calculate the grams of ethanol needed to provide \(358 \mathrm{~kJ}\) of heat. $$ \begin{gathered} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) ; \\ \Delta H=-1235 \mathrm{~kJ} \end{gathered} $$

Step-by-Step Solution

Answer

You need 13.35 grams of ethanol.

Step 1: Understand the problem

We need to determine the amount of ethanol in grams required to produce 358 kJ of heat during its combustion. The reaction enthalpy given is for the complete combustion of ethanol.

Step 2: Find the connection between moles and enthalpy

The reaction gives us that the enthalpy change (\(\Delta H\)) is \(-1235\, \text{kJ}\) for the combustion of one mole of ethanol \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). This describes the energy released per mole.

Step 3: Calculate moles of ethanol needed

To calculate the moles of ethanol needed to produce 358 kJ, we use the proportion:\[\text{Moles of ethanol} = \frac{358 \text{ kJ}}{1235 \text{ kJ/mol}}\].

Step 4: Perform the calculation for moles

Calculate the moles:\[\text{Moles of ethanol} = \frac{358}{1235} = 0.2898 \text{ mol}\].

Step 5: Convert moles to grams

Use the molar mass of ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), approximately 46.08 g/mol) to convert moles to grams:\[\text{Grams of ethanol} = 0.2898 \text{ mol} \times 46.08 \text{ g/mol}\].

Step 6: Perform calculation for grams

Calculate the grams of ethanol:\[\text{Grams of ethanol} = 13.35 \text{ g}\].

Key Concepts

Ethanol Combustion

Ethanol (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \)) is a simple alcohol that can be burned to release energy, which is what we refer to as combustion. During ethanol combustion, ethanol reacts with oxygen (\( \text{O}_2 \)) to form two by-products: carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2 \text{O} \)). The chemical equation for this process is \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (l) + 3 \mathrm{O}_2 (g) \longrightarrow 2 \mathrm{CO}_2 (g) + 3 \mathrm{H}_2 \mathrm{O} (g)\]
This equation shows that one mole of ethanol undergoes combustion with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.
Combustion processes, particularly those involving hydrocarbons like ethanol, are exothermic, meaning they release heat. Here, we know that the heat released when one mole of ethanol is burned completely is \( -1235 \text{ kJ} \), indicating a substantial amount of energy release. This property makes ethanol a useful fuel as part of gasohol, a gasoline-ethanol fuel blend.
Understanding this reaction and its dynamics lays the groundwork for exploring topics in thermochemistry.

Thermochemistry

Thermochemistry is the branch of chemistry that studies the heat energy changes of chemical reactions. In simple terms, it focuses on how energy is absorbed or released during chemical transformations. This discipline helps us understand how much heat energy is transferred, which is crucial for applications like fuel efficiency.
In ethanol combustion, thermochemistry principles reveal that the process is highly energetic. The negative sign in the enthalpy change (\( \Delta H = -1235 \text{ kJ} \)) signifies an exothermic reaction, where heat is released to the surroundings.
Studying thermochemistry allows us to predict energy changes in reactions, calculate enthalpy changes, and optimize conditions for desired outcomes. Thus, it plays a vital role in industries ranging from energy production to materials synthesis. With fuel engineering, for example, knowing how different fuel types perform energetically helps optimize blend compositions like gasohol, providing efficient combustion with reduced emissions.
Overall, thermochemistry equips us with tools to predict and control energy exchanges in chemical processes.

Enthalpy Change Calculation

Calculating enthalpy changes in a reaction offers insights into how much energy, specifically heat, is released or absorbed. To perform this calculation, we need reactions with known enthalpy changes, like the combustion of ethanol.
The given problem requires us to convert a specific amount of heat (\( 358 \text{ kJ} \)) into the number of grams of ethanol needed for that energy release.

• We start by utilizing the formula: \[\text{Moles of ethanol} = \frac{358 \text{ kJ}}{1235 \text{ kJ/mol}}\], which allows for determining the moles of ethanol required.
• Calculating this gives about \( 0.2898 \text{ mol} \) of ethanol.
• With the molar mass of ethanol (\( 46.08 \text{ g/mol} \)), we convert moles into grams using: \[\text{Grams of ethanol} = 0.2898 \text{ mol} \times 46.08 \text{ g/mol}\]
• This results in approximately \( 13.35 \text{ grams} \) of ethanol.

This calculation bridges the gap between theoretical energy changes and practical measurements, providing a pathway from energy requirements to material quantities.