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Radda [10]
3 years ago
8

Which process is expected to have an increase in entropy?

Chemistry
1 answer:
olganol [36]3 years ago
4 0

Answer: Option (B) is the correct answer.

Explanation:

Degree of randomness of the molecules of a substance is known as entropy. More is the kinetic energy between the molecules of a substance more will be the degree of randomness.

Therefore, when a substance is present in a gaseous state then it has the maximum entropy. In liquid state, molecules are closer to each other so, there is less randomness between them.

On the other hand, in solid state molecules are much more closer to each other as they arr held by strong intermolecular forces of attraction. Therefore, they have very less entropy.

  • When liquid water is formed from gaseous hydrogen and oxygen molecules then gas is changing into liquid. So, there is decrease in entropy.
  • When N_{2}O_{4} decomposes then the reaction will be as follows.

            N_{2}O_{4} \rightarrow 2NO_{2}

Since, 1 mole is producing 2 moles. This means that degree of randomness is increasing as both the molecules are present in gaseous form.

  • In formation of a precipitate, aqueous solution is changing into solid state. Hence, degree of randomness is decreasing.
  • Rusting of iron also leads to the formation of solid as it forms Fe_{2}O_{3}.xH_{2}O.

Thus, we can conclude that decomposition of N_{2}O_{4} gas to NO_{2} gas is the process that is expected to have an increase in entropy.

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When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
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b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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