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alexandr1967 [171]
3 years ago
14

If a chemical reaction occurs in a direction that has a positive change in entropy thenGroup of answer choicesthe change in enth

alpy must be negative.the reaction must be spontaneous.heat goes from the system into the surroundings.the reaction must be exothermic.the disorder of the system increases.
Chemistry
1 answer:
mezya [45]3 years ago
6 0

Answer:

The reaction must be spontaneous, the disorder of the system increases.

Explanation:

By the Second Law of Thermodynamics, a positive change in entropy is due to a net input heat, and entropy is a measure of the grade of disorder within the system. The net input heat means that resultant goes to the system from the surroundings.

By the First Law of Thermodynamics, a net input heat is due to a positive change in enthalpy.

The reaction is endothermic and spontaneous (since change in entropy is positive).

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Answer:

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Explanation:

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4 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
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Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

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The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

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+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

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