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The standard heats of combustion \(\left(\Delta H^{\circ}\right)\) per mole of 1,3-butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) ;\) butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) ;\) and \(\mathrm{H}_{2}(\mathrm{g})\) are \(-2540.2,-2877.6,\) and \(-285.8 \mathrm{kJ},\) respectively. Use these data to calculate the heat of hydrogenation of 1,3-butadiene to butane. $$\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) \quad \Delta H^{\circ}=?$$ [Hint: Write equations for the combustion reactions. In each combustion, the products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\left.\mathrm{H}_{2} \mathrm{O}(1) .\right]\)

Short Answer

Expert verified
The heat of hydrogenation of 1,3-butadiene to butane is \(-119.6 kJ\).

Step by step solution

01

Write the combustion reactions

The combustion reactions of the reactants and product are as follows:\n\n1,3-Butadiene:\n\[C_{4}H_{6}(g) + 6O_{2}(g) \rightarrow 4CO_{2}(g) + 3H_{2}O(l)\] \[\Delta H_{1}^{\circ} = -2540.2 kJ\]\n\nHydrogen:\n\[H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)\] \[\Delta H_{2}^{\circ} = -285.8 kJ\]\n\nButane:\n\[C_{4}H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_{2}O(l)\] \[\Delta H_{3}^{\circ} = -2877.6 kJ\]
02

Apply Hess's Law

Hess's Law states that the total enthalpy change of a reaction is the same irrespective of the different reaction routes taken, provided the initial and final conditions are the same. In this problem, the hydrogenation reaction of 1,3-butadiene to butane will be calculated by rearranging the combustion reactions. The combustion reactions of 1,3-butadiene and hydrogen have to be summed to give the combustion reaction of butane. Therefore: \[\Delta H_{total}^{\circ} = \Delta H_{3}^{\circ} - \Delta H_{1}^{\circ} - 2\Delta H_{2}^{\circ}\]
03

Substitute the values

Substitute the given standard heats of combustion into the formula: \[\Delta H_{total}^{\circ} = -2877.6 kJ - (-2540.2 kJ) - 2(-285.8 kJ)\]
04

Calculate the heat of hydrogenation

Calculate the total enthalpy change to find the heat of hydrogenation: \[ \Delta H_{total}^{\circ} = -2877.6 kJ + 2540.2 kJ + 2 * 285.8 kJ = -119.6 kJ \]

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Key Concepts

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

Hess's Law
Hess's Law is a foundational principle in thermochemistry that simplifies the way we calculate enthalpy changes in chemical reactions. It states that the total enthalpy change of a reaction is the same regardless of the path taken, as long as the initial and final conditions remain constant. This means that we can combine known reactions to find unknown enthalpy changes, making complex calculations more manageable.
By using Hess's Law, we can rearrange given chemical equations to match the desired reaction pathway. The sum of the enthalpies of these rearranged reactions gives the total enthalpy change. This approach is particularly useful when direct measurement of a reaction's enthalpy is challenging or impossible.
Heats of Combustion
The heat of combustion is the energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. This is typically expressed in kilojoules per mole (kJ/mol).
In the context of our problem, the heats of combustion for 1,3-butadiene, hydrogen, and butane are provided. These values allow us to calculate the heat of hydrogenation by comparing the energy changes in the combustion of reactants and products.
Understanding the heats of combustion is crucial, as it provides insight into the energy efficiency of fuels and the stability of various compounds. By comparing these values, chemists can predict how much energy can be harnessed from different substances.
Enthalpy Change
Enthalpy change, represented as \(\Delta H\), indicates the heat absorbed or released by a system during a chemical reaction at constant pressure. In this exercise, we're specifically interested in the enthalpy of hydrogenation.
The enthalpy change can tell us whether a reaction is exothermic (releases heat, \(\Delta H < 0\)) or endothermic (absorbs heat, \(\Delta H > 0\)). For the hydrogenation of 1,3-butadiene to butane, the calculated enthalpy change was \(-119.6 \, \text{kJ/mol}\), meaning it's exothermic.
Calculating enthalpy changes enables scientists and engineers to understand the energy dynamics of reactions, crucial for designing energy-efficient processes.
Hydrogenation Reaction
A hydrogenation reaction involves the addition of hydrogen ( H_2 ) to other compounds, often to saturate organic molecules by converting double bonds to single bonds. This process is widely used in industries to modify the properties of oils and fats.
In the example of 1,3-butadiene to butane, hydrogenation results in more stable and saturated hydrocarbons. Calculating the heat of hydrogenation involves determining how much energy is released during the conversion of the double bonds in 1,3-butadiene to single bonds in butane.
Understanding hydrogenation is important not only for industrial applications but also for studying reaction pathways and optimizing energy use in chemical manufacturing.

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Most popular questions from this chapter

The heat of solution of \(\mathrm{NaOH}(\mathrm{s})\) in water is \(-41.6 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) When \(\mathrm{NaOH}(\mathrm{s})\) is dissolved in water the solution temperature (a) increases; (b) decreases; (c) remains constant; (d) either increases or decreases, depending on how much NaOH is dissolved.

Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

The heat of solution of \(\mathrm{KI}(\mathrm{s})\) in water is \(+20.3 \mathrm{kJ} / \mathrm{mol}\) KI. If a quantity of KI is added to sufficient water at \(23.5^{\circ} \mathrm{C}\) in a Styrofoam cup to produce \(150.0 \mathrm{mL}\) of 2.50 M KI, what will be the final temperature? (Assume a density of \(1.30 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(2.7 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1} \text {for } 2.50 \mathrm{M} \mathrm{KI} .\right)\)

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