Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\) \(2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\), from the following data: enthalpy of combustion of ethyne, \(-1300 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of ethane, \(-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of hydrogen, \(-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

Short Answer

Expert verified
-832 kJ/mol

Step by step solution

01

Write the combustion reactions for each compound

First, write the balanced chemical equations for the combustion of ethyne, ethane, and hydrogen in oxygen to form carbon dioxide and water. The equations are: For ethyne: \(2 \text{C}_2\text{H}_2(g) + 5 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 2 \text{H}_2\text{O}(l)\) For hydrogen: \(2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l)\) For ethane: \(\text{C}_2\text{H}_6(g) + \frac{7}{2} \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(l)\)
02

Apply Hess's Law

To determine the enthalpy for the hydrogenation of ethyne to ethane, use Hess's Law, which states that the total enthalpy change for a chemical reaction is the same, no matter how the reaction is carried out in steps. Rearrange the combustion equations such that when added, they give the desired hydrogenation reaction.
03

Write the reverse combustion reaction of ethyne

Write the reverse of the combustion reaction of ethyne to find its formation reaction and change the sign of the enthalpy. Reversed reaction of ethyne: \(4 \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \rightarrow 2 \text{C}_2\text{H}_2(g) + 5 \text{O}_2(g)\), Enthalpy change: \(1300 \text{ kJ/mol}\)
04

Adjust the hydrogen combustion reaction

Multiply the hydrogen combustion reaction by 2 to provide the necessary moles of hydrogen for the hydrogenation reaction. Adjust the enthalpy change accordingly. Adjusted hydrogen reaction: \(4 \text{H}_2(g) + 2 \text{O}_2(g) \rightarrow 4 \text{H}_2\text{O}(l)\), Enthalpy change: \(2 \times (-286) = -572 \text{ kJ/mol}\)
05

Write the formation reaction of ethane

Since the combustion reaction of ethane is the desired product's reaction, keep it as is. Combustion reaction of ethane: \(\text{C}_2\text{H}_6(g) + \frac{7}{2} \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(l)\), Enthalpy change: \(-1560 \text{ kJ/mol}\)
06

Combine the adjusted combustion reactions

Add the reversed combustion reaction of ethyne and the adjusted hydrogen combustion reaction. Then subtract the combustion reaction of ethane to isolate the hydrogenation of ethyne to ethane. Sum of enthalpies: \(1300 \text{ kJ/mol} - 572 \text{ kJ/mol} - 1560 \text{ kJ/mol} = -832 \text{ kJ/mol}\)
07

Conclude with the reaction enthalpy of hydrogenation

The enthalpy of hydrogenation of ethyne to ethane is the sum of the enthalpies of the steps added together, which is \(-832 \text{ kJ/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hess's Law
When studying chemical reactions, energy changes are a fundamental component, and here, Hess's Law provides a remarkably useful tool. It's based on the principle that the total enthalpy change for a reaction is constant, regardless of the pathway or number of steps taken to achieve the reaction. In essence, if a reaction is carried out in a series of steps, the sum of the enthalpy changes for each individual step is equal to the enthalpy change if the reaction were done in a single step.

This law comes into play when we do not have direct access to the enthalpy change of a reaction. By using other known enthalpy changes that can be summed to represent the overall reaction (as in the textbook example), we can calculate the unknown enthalpy changes. It's like solving a puzzle by fitting together known pieces to reveal the missing part.

Applying Hess's Law involves writing and manipulating a series of equations such that the enthalpies of reactions we know can be combined to find the enthalpy change for the reaction we want to understand. The beauty of this law lies in its simplicity and the versatility it provides to chemists in the field of thermochemistry.
Enthalpy of Combustion
When we talk about the enthalpy of combustion, we refer to the amount of energy released when a substance completely reacts with oxygen under standard conditions to form water and carbon dioxide. It's an exothermic process, meaning it releases energy. For substances that readily combust, like the hydrocarbons in our textbook problem, we can easily measure this enthalpy change in a laboratory.

In the context of chemical thermodynamics, knowing the enthalpy of combustion allows us to determine other thermodynamic properties and to understand energy flows in chemical processes. More importantly, for calculations, combustion reactions often serve as the baseline for applying Hess's Law, where we can calculate the enthalpy of formation or other reaction types by using the enthalpy of combustion as a starting point.

In the given exercise, the enthalpy of combustion serves as the essential piece of data to help find the desired enthalpy of hydrogenation. By looking at how much energy is released when ethyne and ethane combust, we can work backwards to find out how much energy these substances would require to form from their constituent elements.
Chemical Thermodynamics
Chemical thermodynamics is the area of chemistry that deals with energy changes, particularly the study of enthalpy changes in chemical reactions. It provides us with a comprehensive framework to understand how energy is transferred and transformed. Within this domain, we consider the laws of thermodynamics that define the conservation of energy and predict the direction of energy flow.

The enthalpy changes in a reaction, such as the heat absorbed or released, offer insights into the stability of the products and reactants, as well as the spontaneity of the reaction. Chemical thermodynamics informs us not only about 'how much' energy is involved, but also about the feasibility and extent of a reaction.

Our toe-dip into thermodynamics with the enthalpy of hydrogenation is a practical application of how to predict energy changes and understand reaction behavior. By mastering these concepts, students can apply them to a wide range of phenomena, from biological systems to engineering processes, and beyond.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the masses, volumes, and moles of substances involved. In a nutshell, stoichiometry is the recipe book of chemistry that tells us how much of each ingredient we need to make a certain product and how much product we can expect to get out.

For the hydrogenation problem, stoichiometry ensures that the equations for combustion and hydrogenation are balanced, which means that the number of atoms for each element is conserved. Without a balanced stoichiometric equation, we would not be able to apply Hess's Law correctly.

Understanding stoichiometry is crucial as it forms the basis for all quantitative analysis in chemistry. When dealing with reactions, whether they are simple or complex, stoichiometry provides the foundation for precise calculations and ensures materials are used efficiently.

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Most popular questions from this chapter

Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride, \(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{~s})\), from the following data: $$ \begin{array}{ll} 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \\ & \Delta H^{\circ}=-1049 \mathrm{~kJ} \\ \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{aq}) & \Delta H^{\circ}=-74.8 \mathrm{~kJ} \\ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g}) & \Delta H^{\circ}=-185 \mathrm{~kJ} \\ \mathrm{AlCl}_{3}(\mathrm{~s}) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{aq}) & \Delta H^{\circ}=-323 \mathrm{~kJ} \end{array} $$

Which gas do you expect to have the higher molar heat capacity, \(\mathrm{NO}\) or \(\mathrm{NO}_{2}\) ? Why?

Write the thermochemical equations that give the values of the standard enthalpies of formation for (a) \(\mathrm{KClO}_{3}\) (s), potassium chlorate; (b) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH}\) (s), glycine(s); (c) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})\), alumina.

(a) Calculate the work for a system that absorbs \(150 \mathrm{~kJ}\) of heat in a process for which the increase in internal energy is \(120 \mathrm{~kJ}\). (b) Is work done on or by the system during this process?

For a certain reaction at constant pressure, \(\Delta H=-15 \mathrm{~kJ}\) and \(22 \mathrm{~kJ}\) of expansion work is done on the system. What is \(\Delta U\) for this process?

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