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AlexFokin [52]
3 years ago
13

Which of the following processes would you predict to have an increase in entropy?

Chemistry
2 answers:
arlik [135]3 years ago
3 0

Among the choices given, the best answer is the third option. Melting ice shows an increase in entropy. Entropy is the degree of disorderliness in a system. Among the phases, solid has the most ordered structure which means it has the least entropy. The ice melting shows a phase change from solid to liquid. Liquid molecules are more disorganized than the solid thus there is an increase in entropy.

Studentka2010 [4]3 years ago
3 0

Answer: melting ice

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

1. condensation of water : Condensation process is a process where gaseous state changes to liquid state., thus decreasing randomness and decreasing entropy.

2. Freezing water : Freezing process is a process where liquid state changes to solid state, thus decreasing randomness and decreasing entropy.

3. melting ice: Fusion process is a process where solid state changes to liquid state, thus increasing randomness and increasing entropy.

4. deposition of CO_2: deposition process where gaseous state changes to solid state by escaping liquid state, thus decreasing randomness and decreasing entropy.

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<h3>Answer:</h3>

9 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>  

<u>Pre-Algebra</u>  

Order of Operations: BPEMDAS  

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Chemistry</u>  

<u>Atomic Structure</u>  

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<u>Stoichiometry</u>  

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6H₂ + O₂ → 3H₂O

[Given] 18 mol H₂

[Solve] mol H₂O

<u>Step 2: Identify Conversions</u>

[RxN] 6 mol H₂ → 3 mol H₂O

<u>Step 3: Stoich</u>

  1. [DA] Set up conversion:                                                                               \displaystyle 18 \ mol \ H_2(\frac{3 \ mol \ H_2O}{6 \ mol \ H_2})
  2. [DA] Simplify:                                                                                                 \displaystyle 18 \ mol \ H_2(\frac{1 \ mol \ H_2O}{2 \ mol \ H_2})
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