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telo118 [61]
3 years ago
6

The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the

change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l )
Chemistry
1 answer:
rewona [7]3 years ago
8 0

Answer:

The entropy decreases.

Explanation:

The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where

Δngas = n(gaseous products) - n(gaseous reactants)

  • If Δngas > 0, the entropy increases
  • If Δngas < 0, the entropy decreases.
  • If Δngas = 0, there is little or no change in the entropy.

Let's consider the following reaction.

2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)

Δngas = 0 - 3 = -3, so the entropy decreases.

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Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents
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Answer:

Explanation:

a) Mg + NiCl_2 \rightarrow MgCl_2 + Ni

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and NiCl_2 acts as oxidizing agent.

b) PCl_3 + Cl_2 \rightarrow PCl_5

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

PCl_3 acts as reducing agent and Cl_2 acts as oxidizing agent.

c) C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

C_2H_4 acts as reducing agent and O_2 acts as oxidizing agent.

d) Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and H_2SO_4 acts as oxidizing agent

e) K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

K_2S_2O_3 acts as reducing agent and I_2 acts as oxidizing agent.

f) 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and HNO_3 acts as oxidizing agent

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