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Calculate \(\Delta E\) for a. the combustion of a gas that releases \(210.0 \mathrm{kJ}\) of heat to its surroundings and does \(65.5 \mathrm{kJ}\) of work on its surroundings. b. a chemical reaction that produces \(90.7 \mathrm{kJ}\) of heat but does no work on its surroundings.

Short Answer

Expert verified
Question: Determine the change in internal energy (∆E) for both situation (a) and situation (b) using the first law of thermodynamics. Given information: a. q = -210.0 kJ, w = -65.5 kJ b. q = 90.7 kJ, w = 0 kJ Answer: For situation (a), the change in internal energy (∆E) is -275.5 kJ. For situation (b), the change in internal energy (∆E) is 90.7 kJ.

Step by step solution

01

Identify given information

For each situation, make sure to have the necessary variables at hand: a. q = -210.0 kJ, w = -65.5 kJ b. q = 90.7 kJ, w = 0 kJ
02

Determine ∆E for situation (a)

Using the first law of thermodynamics equation, ∆E = q + w, plug in the values for situation a: ∆E_a = (-210.0\,\text{kJ}) + (-65.5\,\text{kJ})
03

Calculate ∆E for situation (a)

Perform the addition for situation (a): ∆E_a = -275.5\,\text{kJ} The change in internal energy for the combustion of the gas in situation (a) is -275.5 kJ.
04

Determine ∆E for situation (b)

Using the first law of thermodynamics equation, ∆E = q + w, plug in the values for situation b: ∆E_b = (90.7\,\text{kJ}) + (0\,\text{kJ})
05

Calculate ∆E for situation (b)

Perform the addition for situation (b): ∆E_b = 90.7\,\text{kJ} The change in internal energy for the chemical reaction in situation (b) is 90.7 kJ.

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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 fundamental principle in physics governing the conservation of energy. This principle asserts that energy cannot be created or destroyed, only transformed from one form to another. In simpler terms, the total energy in a closed system remains constant over time.
This principle is often expressed through an equation:
  • \( \Delta E = q + w \)
Here, \( \Delta E \) represents the change in internal energy, \( q \) is the heat exchanged with the surroundings, and \( w \) is the work done by or on the system.
In a real-world context, if a system releases heat (q is negative) or does work on the surroundings (w is negative), the internal energy of the system decreases. Conversely, when a system absorbs heat or work is done on it, its internal energy increases. This equation allows us to quantify these energy changes effectively.
Internal Energy Calculation
Internal energy represents the total energy contained within a system. Calculating changes in internal energy involves understanding how heat and work affect the energy reserves of a system.
To compute \( \Delta E \), you apply the first law of thermodynamics. Let's break it down for clarity:
  • The formula is \( \Delta E = q + w \).
  • If heat flows into the system, \( q \) is positive.
  • If the system releases heat to the surroundings, \( q \) is negative.
  • Work done on the system makes \( w \) positive.
  • Work done by the system on the surroundings makes \( w \) negative.
Thus, by knowing the values of \( q \) and \( w \), you can easily determine the change in internal energy for any process.
Combustion Reactions
Combustion is a fundamental chemical process where a substance reacts rapidly with oxygen, releasing energy mainly in the form of heat and light. This reaction is an exothermic process, meaning it releases more energy than it absorbs.
Key characteristics of combustion reactions include:
  • They usually involve hydrocarbons reacting with oxygen.
  • They produce carbon dioxide and water as common byproducts.
  • The energy released is primarily in the form of heat, making the heat term \( q \) typically negative (indicating a release of heat to the surroundings).
In the context of thermodynamics, analyzing a combustion reaction requires careful attention to how much energy is transferred as heat (q) and whether any work (w) is done during the reaction. This understanding allows us to compute the change in internal energy (ΔE), enabling us to better predict and manage energy transformations in various chemical processes.

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Most popular questions from this chapter

Hydrogen is attractive as a fuel because it has a high fuel value and produces no \(\mathrm{CO}_{2}\) $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)$$ a. One problem with hydrogen is its low fuel density. What is the fuel density (kJ/L) for \(\mathrm{H}_{2}\) given that the density of hydrogen gas is \(0.0899 \mathrm{g} / \mathrm{L} ?\) If we could liquefy hydrogen, what would the fuel density of hydrogen be, given that the density of liquid hydrogen is \(70.8 \mathrm{g} / \mathrm{L} ?\) b. One solution to the problem of hydrogen storage is to use solid, hydrogen- containing compounds that release hydrogen upon heating at low temperature. One such compound is ammonia borane, \(\mathrm{H}_{3} \mathrm{NBH}_{3}\). Calculate the enthalpy change \((\Delta H)\) for the reactions shown below given \(\Delta H_{\mathrm{f}, \mathrm{H}_{\mathrm{J}} \mathrm{NBH}_{1}}=-38.1 \mathrm{kJ} / \mathrm{mol}\) $$\begin{aligned} &\mathrm{H}_{3} \mathrm{NBH}_{3}(g) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{BH}_{3}(g)\\\ &\begin{array}{l} \mathrm{H}_{3} \mathrm{NBH}_{3}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{H}_{2} \mathrm{NBH}_{2}(g) \\ \mathrm{H}_{2} \mathrm{NBH}_{2}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{HNBH}(g) \end{array} \end{aligned}$$ c. How many kilograms of ammonia borane are needed to supply \(10.0 \mathrm{kg}\) of hydrogen?

Two solids, \(5.00 \mathrm{g}\) of \(\mathrm{NaOH}\) and \(4.20 \mathrm{g}\) of \(\mathrm{KOH}\), are added to 150 mL of water \(\left[c_{P}=75.3 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) ; T=23^{\circ} \mathrm{C}\right]\) in a calorimeter. Given \(\Delta H_{\text {soln, } \mathrm{NaOH}}=-44.3 \mathrm{kJ} / \mathrm{mol}\) and \(\Delta H_{\text {soln, } \mathrm{KOH}}=-56.0 \mathrm{kJ} / \mathrm{mol},\) what is the final temperature of the solution?

Explain how the use of \(\Delta H_{f}^{\circ}\) to calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\) is an example of Hess's law.

What is the difference between specific beat and molar beat capacity?

How are fuel values calculated from molar enthalpies of combustion?

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