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Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

Short Answer

Expert verified
Yes, a chemical reaction can have \( \Delta U<0 \) and \( \Delta H>0 \) if the system is exothermic and simultaneously doing more work on the surroundings, which results in heat absorption from the surroundings.

Step by step solution

01

Understanding Symbols

First, understand what these symbols represent in a chemical reaction. \( \Delta U \) represents the change (increase or decrease) in the internal energy which could be due to heat transfer or work done. \( \Delta H \) represents the enthalpy change which is the heat absorbed or released in a chemical reaction at constant pressure.
02

Assessing the Feasibility of \( \Delta U0 \)

Consider a system where heat is released, signifying an exothermic process making \( \Delta U<0 \). However, let's say the work done by the system on the surroundings is significantly high. Therefore, even though the system is losing energy, it's still absorbing heat energy from the surroundings to do the work. Hence, \( \Delta H \), which is the heat absorbed at constant pressure, would be greater than zero.
03

Conclusion

To conclude, it is possible for a chemical reaction to have \( \Delta U<0 \) and \( \Delta H>0 \). Such a scenario would require the system to perform work on the surroundings while absorbing heat from the surroundings, which is plausible

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Key Concepts

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

Internal Energy Change (ΔU)
In the realm of chemistry, internal energy change, denoted as \( \Delta U \), plays a key role in understanding the dynamics of a chemical reaction. Internal energy comprises the total kinetic and potential energy in a system. Changes in internal energy can occur due to heat exchange or work done during the reaction.

  • If heat is released or work is done by the system, the internal energy decreases, thus \( \Delta U < 0 \).
  • Conversely, if heat is absorbed or work is done on the system, internal energy increases, leading to \( \Delta U > 0 \).
It's important to remember that \( \Delta U \) encapsulates all forms of energy within the system, making it a comprehensive measure of energy changes during reactions. Understanding \( \Delta U \) helps in predicting the flow of energy within a chemical process.
Enthalpy Change (ΔH)
Enthalpy change, represented as \( \Delta H \), is a vital concept in thermodynamics, specifically in chemical reactions. Enthalpy is the total heat content of a system under constant pressure conditions. It primarily reflects the heat absorbed or released when a chemical reaction occurs.

  • When \( \Delta H > 0 \), the system absorbs heat from the surroundings, denoting an endothermic process.
  • Conversely, when \( \Delta H < 0 \), it indicates heat is released into the surroundings, suggesting an exothermic reaction.
The nature of \( \Delta H \) provides insights into the energetic demands of a reaction, such as whether additional heat is required to proceed or whether it liberates heat that could be harnessed.
Exothermic Processes
Exothermic processes are chemical reactions that release energy by emitting heat into the surroundings. The term "exothermic" is derived from the Greek words 'exo', meaning "outside", and 'thermic', meaning "heat". This concept is integral to various real-world applications and chemical phenomena.

  • Exothermic reactions are characterized by a negative enthalpy change, \( \Delta H < 0 \), indicating the release of heat.
  • This concept explains why certain reactions, like combustion, feel hot as they release heat into the environment.
  • In exothermic processes, such as the formation of water from hydrogen and oxygen, the released energy plays a pivotal role in making the environment warmer.
Understanding exothermic processes aids in grasping how energy transformation occurs, and why such reactions are beneficial in fields like energy production and metallurgy.

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Most popular questions from this chapter

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

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