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Exactly one mole of gaseous methane is oxidized at fixed volume and at \(25^{\circ} \mathrm{C}\) according to the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\). If \(212 \mathrm{~K}\) cal is liberated, what is the change in enthalpy, \(\Delta \mathrm{H}\) ?

Short Answer

Expert verified
The change in enthalpy (ΔH) for the given reaction is 887568 J.

Step by step solution

01

Convert energy units

First we need to convert the given heat energy into SI Unit. 1 kcal is equal to 1000 cal and 1 cal is equal to 4.184 Joules, so we need to multiply the given energy value by 1000 and 4.184 to get the value in Joules: Energy liberated = 212 kcal * 1000 cal/kcal * 4.184 J/cal = 887568 J
02

Applying the first law of thermodynamics

The first law of thermodynamics states that the change in internal energy(ΔU) of a system is equal to the heat(Q) added to the system minus the work(W) done by the system on its surroundings. ΔU = Q - W Since the reaction is taking place at a constant volume, the work(W) done by the system is 0. Hence, ΔU = Q = 887568 J
03

Relating enthalpy change to internal energy change

Enthalpy (H) is related to the internal energy (U) by the following equation: H = U + PV Where P and V are the pressure and volume of the system, respectively. Also, the change in enthalpy (ΔH) of the reaction can be related to the change in internal energy (ΔU) as: ΔH = ΔU + Δ(PV) But since the reaction takes place at constant volume, Δ(PV) = PΔV = 0. Therefore, ΔH = ΔU = 887568 J So, the change in enthalpy(ΔH) for the given reaction is 887568 J.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a statement of energy conservation. It posits that energy can neither be created nor destroyed, only transformed from one form to another within a closed system. In the context of chemistry and specifically in the calorimetry of chemical reactions, this law is integral for understanding how energy changes occur.

During a chemical reaction, the energy stored in chemical bonds is transformed into other forms, such as heat. If a reaction is exothermic, it releases heat into the surroundings, as is the case with the oxidation of methane in our example. According to the First Law, the total energy before and after the reaction must be the same; hence, the 'liberated' heat comes from the energy that was previously stored in methane and oxygen molecules.

Students can find this concept confusing, but a useful analogy is to think of energy like money in a bank account: you can move it around, spend it, or save it, but the total balance only changes when money is transferred in or out of the account (i.e., enter or leave the system). In our methane reaction, the 'account balance' of energy is adjusted by the amount of heat 'spent' or released.
Internal Energy
Internal energy, symbolized as U, is the total energy contained within a substance due to both the random motion of its particles and the chemical bonds between those particles. It's an extensive property, meaning it depends on the system's size or the substance amount. This concept is crucial when dealing with chemical reactions, as changes in the internal energy can indicate the reaction's direction and magnitude.

As detailed in the solution, the internal energy change (ΔU) during a constant-volume process equals the heat exchanged with the surroundings. Here, no work is done by the system, since work (usually denoted as PΔV) is associated with changes in volume, and none occur in this instance. Remembering that at constant volume, ΔU is simply the heat added or removed can be incredibly helpful for students when solving problems.

Envision internal energy as the sum of many little 'packets' of energy associated with different motions and interactions within a molecule. With the methane reaction, energy is released when the chemical bonds are reorganized, and this is quantified as the decrease in the system's internal energy.
State Functions in Chemistry
State functions are properties that only depend on the current state of the system, not on how the system arrived at that state. Temperature, pressure, volume, enthalpy (H), and internal energy (U) are all examples of state functions in chemistry.

Enthalpy is particularly important in thermochemistry because it incorporates not only the internal energy but also the product of pressure and volume. Even though enthalpy is a state function, the change in enthalpy (ΔH) during a reaction is what chemists are often most interested in. This change is an indication of the heat absorbed or released under constant pressure conditions.

To cultivate a deeper understanding, students should realize that these functions are like the coordinates on a map. You can take many different paths to reach a location (reflecting the different ways a system can undergo change), but the location itself (akin to the state function's value) doesn't tell you anything about the path you took to get there. Remembering this can make the concept of state functions, such as enthalpy change in our reaction, much easier to grasp.

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Most popular questions from this chapter

At constant volume, the heat capacity of gas differs from the heat capacity at constant pressure. Why?

Determine the quantity of heat required to convert \(10 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) to vapor at \(100^{\circ} \mathrm{C}\), For water, heat of fusion \(=80\) \(\mathrm{cal} / \mathrm{g}\), heat of vaporization \(=540 \mathrm{cal} / \mathrm{g}\), and specific heat \(=1\) \(\mathrm{cal} / \mathrm{g}-{ }^{\circ} \mathrm{C}\)

\(40 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(100 \mathrm{~g}\) of water at \(60^{\circ} \mathrm{C}\), What is the final temperature after equilibrium has been established? Heat of fusion of \(\mathrm{H}_{2} \mathrm{O}=80 \mathrm{cal} / \mathrm{g}\), specific heat = \(1 \mathrm{ca} 1 / \mathrm{g}-{ }^{\circ} \mathrm{C}\).

Using the information in the following table, determine \(\Delta \mathrm{H}^{\circ}\) for the reactions: a) \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{4}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})\) b) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightarrow 3 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) c) \(\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) $$ \begin{array}{|c|c|} \hline {\text { Heats of Formation }} \\ \hline \text { Compound } & \Delta \mathrm{H}^{\circ}(\mathrm{Kcal} / \mathrm{mole}) \\ \hline \mathrm{CO}(\mathrm{g}) & -26.4 \\ \hline \mathrm{CO}_{2}(\mathrm{~g}) & -94.1 \\ \hline \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) & -197 \\ \hline \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) & -267 \\ \hline \mathrm{FeO}(\mathrm{s}) & -63.7 \\ \hline \end{array} $$

In a hydrogen bomb, deuterium nuclei fuse to produce a helium nucleus. In the process, \(5 \mathrm{MeV}\) of energy is released per deuterium. If, \(1 / 7000\) of natural \(\mathrm{H}\) is \(\mathrm{D}\), how many liters of gasoline \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) are needed to be burned to produce the same amount of energy as derived from 1 liter of water. Assume the following: the heat of combustion of gasoline is \(5100 \mathrm{KJ} /\) mole the density of gasoline is \(.703 \mathrm{~g} / \mathrm{cm}^{3}\); \(\mathrm{leV}=1.60 \times 10^{-19} \mathrm{~J}\)

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