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Hot packs used by skiers produce heat via the crystallization of sodium acetate from a concentrated solution. What is the sign of \(\Delta H\) for this crystallization? Is the reaction exothermic or endothermic?

Short Answer

Expert verified
\(\Delta H\) is negative and the reaction is exothermic.

Step by step solution

01

Understand Crystallization Process

Crystallization is the process where a substance transitions from a liquid or gas phase into a solid phase.
02

Analyze Heat Production

If a reaction produces heat, it releases energy into its surroundings. This type of reaction is known as exothermic.
03

Determine \(\Delta H\)

The enthalpy change (\(\Delta H\)) of an exothermic reaction is negative because energy is released.
04

Conclusion

Since heat is produced in the crystallization of sodium acetate, \(\Delta H\) is negative and the reaction is exothermic.

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Key Concepts

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

Crystallization Process
Crystallization is a fundamental process where a substance moves from a fluid or gas state to a solid state. It is commonly used in various fields, such as chemistry and pharmaceuticals. The process involves the formation of solid crystals from a homogeneous solution.
Crystallization typically happens through nucleation, where a seed crystal forms, and then growth, where additional molecules adhere to it. This phase change is essential:
  • It often purifies the substance, as only the compound of interest crystallizes.
  • It is essential in manufacturing, especially for substances like sodium acetate.
    In the case of hot packs, sodium acetate is dissolved at high concentrations. When activated, it rapidly crystallizes, releasing heat.
Enthalpy Change
Enthalpy change, represented by \(abla H\), is a measure of the heat energy released or absorbed during a chemical process at constant pressure. It is a critical concept in thermodynamics.
Enthalpy change can be positive or negative:
  • Positive \(abla H\): Indicates endothermic reactions where the system absorbs heat.
  • Negative \(abla H\): Indicates exothermic reactions where the system releases heat.
    For the crystallization of sodium acetate, the process is exothermic. Heat production signifies energy release, making \(abla H\) negative. You can think of it like a cozy blanket on a cold day - the blanket is the solidifying sodium acetate releasing warm energy around it.
Sodium Acetate
Sodium acetate is a versatile compound with many uses, most famously in hot packs for providing warmth. In chemical terms, its formula is \(Co_3 - COONa\).
Here’s how it works in hot packs:
  • Sodium acetate is dissolved in water at a high concentration.
  • Upon activation, it crystallizes and releases heat due to its exothermic reaction.
    Sodium acetate solutions are supercooled - they remain liquid below their freezing point until they are disturbed or 'activated'. At this point, they solidify rapidly, releasing significant heat. This makes sodium acetate ideal for portable heating devices used by skiers and outdoor enthusiasts.

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Most popular questions from this chapter

An adiabatic process is one that involves no heat transfer. What is the relationship between work and the change in internal energy in an adiabatic process?

A chemist places \(1.750 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) in a bomb calorimeter with a heat capacity of \(12.05 \mathrm{~kJ} / \mathrm{K}\). The sample is burned and the temperature of the calorimeter increases by \(4.287^{\circ} \mathrm{C}\). Calculate \(\Delta E\) for the combustion of ethanol in \(\mathrm{kJ} / \mathrm{mol} .\)

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,(1) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-39.8 \mathrm{~kJ}\) (2) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) \Delta H_{\mathrm{rxn}}^{0}=-112.5 \mathrm{~kJ}\) (3) \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(\Delta H_{\mathrm{ran}}^{0}=-57.2 \mathrm{~kJ}\) (4) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) \(\Delta H_{\mathrm{rxn}}^{0}=-114.2 \mathrm{~kJ}\) (5) \(\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)\) $$\Delta H_{\mathrm{ran}}^{\circ}=54.1 \mathrm{~kJ}$$calculate the standard enthalpy of reaction for$$\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(s) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g) $$

What is the difference between the standard enthalpy of formation and the standard enthalpy of reaction?

Express Hess's law in your own words.

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