Problem 75 - General Chemistry

Question

Compounds with carbon-carbon double bonds, such as ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\), add hydrogen in a reaction called hydrogenation. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$ Calculate the enthalpy change for this reaction, using the following combustion data: $$ \begin{gathered} \mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-1411 \mathrm{~kJ} \end{gathered} $$ $$ \begin{gathered} \mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-1560 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-286 \mathrm{~kJ} \end{gathered} $$

Step-by-Step Solution

Answer

The enthalpy change for the reaction is \(-137 \text{ kJ/mol}\).

Step 1: Writing Hess's Law Equation

To find the enthalpy change for the hydrogenation reaction \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2}\mathrm{H}_{6}(g)\), we use Hess's Law. We can find the enthalpy change by combining the given combustion reactions: \(\Delta H_1\) for combustion of ethylene and \(\Delta H_2\) for combustion of ethane, and subtracting the formation of water from hydrogen \(\Delta H_3\).

Step 2: Arrange Equations

The combustion reaction for ethylene is \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l)\) with \(\Delta H = -1411 \text{ kJ}\).The combustion reaction for ethane is \(\mathrm{C}_{2}\mathrm{H}_{6}(g) + \frac{7}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l)\) with \(\Delta H = -1560 \text{ kJ}\).The combustion reaction for hydrogen is \(\mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) with \(\Delta H = -286 \text{ kJ}\).

Step 3: Reverse the Reactions

To calculate the enthalpy change for hydrogenation, reverse the combustion of \(\mathrm{C}_{2}\mathrm{H}_{6}\):\(2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2}\mathrm{H}_{6}(g) + \frac{7}{2} \mathrm{O}_{2}(g), \Delta H = +1560 \text{ kJ}\).We need \(\mathrm{C}_{2}\mathrm{H}_{4}\) and \(\mathrm{H}_{2}\), so we will use reactions of these as they are.

Step 4: Apply Hess's Law and Calculate

Add the reversed ethane reaction to the ethylene and hydrogen reactions:\[(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l), -1411 \text{ kJ})\]\[(2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2}\mathrm{H}_{6}(g) + \frac{7}{2} \mathrm{O}_{2}(g), +1560 \text{ kJ})\]\[(\mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l), -286 \text{ kJ})\]Combine to yield: \[\mathrm{C}_{2}\mathrm{H}_{4}(g) + \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2}\mathrm{H}_{6}(g)\]The enthalpy change \(\Delta H\) is calculated as:\[\Delta H = -1411 + 1560 - 286 = -137 \text{ kJ}\].

Step 5: Final Result: Enthalpy Change for Hydrogenation

The enthalpy change, \(\Delta H\), for the hydrogenation of ethylene to form ethane under the provided conditions is \(-137 \text{ kJ/mol}\).

Key Concepts

Enthalpy Change

Enthalpy change, often represented as \(\Delta H\), is a measure of heat energy change in a reaction at constant pressure. It indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). For the hydrogenation reaction of ethylene to ethane, we calculate \(\Delta H\) by using Hess's Law.
This law states that the total enthalpy change for a reaction is the same, regardless of the pathway taken. This principle allows us to find \(\Delta H\) using other reactions like combustion.
We use known enthalpy changes from combustion reactions of the substances involved. Different chemical reactions have different \(\Delta H\) values depending on the bonds broken and formed during the reaction.

• For a reaction where heat is released (exothermic), \(\Delta H\) is negative.
• Conversely, in endothermic reactions, \(\Delta H\) is positive.

Understanding enthalpy change helps in predicting the energy requirement or release, crucial for industrial chemical processes.

Hydrogenation Reaction

A hydrogenation reaction involves the addition of hydrogen (\(\mathrm{H}_2\)) to another compound. In this specific context, the reaction is between ethylene (\(\mathrm{C}_2\mathrm{H}_4\)) and hydrogen gas to form ethane (\(\mathrm{C}_2\mathrm{H}_6\)).
This process usually requires a catalyst and is highly exothermic, meaning it releases heat.
Hydrogenation reactions are fundamental in various industries for producing saturated compounds from unsaturated ones.

• The reaction of ethylene and hydrogen is an example of an alkene reacting with hydrogen.
• This reaction is commonly used in the food industry to convert oils into solid fats, known as hydrogenated oils.
• The energy change associated with this reaction helps in understanding its feasibility and the conditions needed, such as catalyst selection and temperature control.

The enthalpy change calculated through Hess's Law gives us insights into the heat exchange involved in such reactions, which is crucial for energy management in chemical industries.

Combustion Reactions

Combustion reactions involve a substance reacting with oxygen gas (\(\mathrm{O}_2\)) and are typically exothermic, releasing energy in the form of heat and light. They are vital for various applications from generating energy to synthesizing important chemicals.
In the case of the given exercise, combustion reactions of ethylene, ethane, and hydrogen provide the necessary data to calculate the enthalpy change for the hydrogenation of ethylene.

• The combustion of ethylene to produce carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)) releases \(-1411\) kJ/mol.
• For ethane, the combustion releases \(-1560\) kJ/mol, and for hydrogen, it releases \(-286\) kJ/mol.
• These values are integral for applying Hess's Law to derive the \(\Delta H\) for secondary reactions, like hydrogenation.

Understanding combustion reactions allows us to utilize these reactions to assess and manipulate energy changes needed for synthesis and other chemical processes.