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Which reaction shows that the enthalpy of formation of [tex]\( H_2S \)[/tex] is [tex]\( \Delta H_f = -20.6 \)[/tex] kJ/mol?

A. [tex]\( H_2(g) + S(s) \rightarrow H_2S -20.6 \)[/tex] kJ
B. [tex]\( H_2(g) + S(s) \rightarrow H_2S +20.6 \)[/tex] kJ
C. [tex]\( H_2(g) + S(s) + 20.6 \)[/tex] kJ [tex]\(\rightarrow H_2S \)[/tex]
D. [tex]\( 2H(g) + S(s) + 20.6 \)[/tex] kJ [tex]\(\rightarrow H_2S \)[/tex]

Sagot :

GinnyAnswer
To determine which reaction correctly illustrates the enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) of hydrogen sulfide ([tex]\(H_2S\)[/tex]) as [tex]\(-20.6 \text{ kJ/mol}\)[/tex], we need to recognize that the enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

Given the options:

A. [tex]\(H_2(g) + S(s) \rightarrow H_2S -20.6 \text{ kJ}\)[/tex]
B. [tex]\(H_2(g) + S(s) \rightarrow H_2S +20.6 \text{ kJ}\)[/tex]
C. [tex]\(H_2(g) + S(s) +20.6 \text{ kJ} \rightarrow H_2S\)[/tex]
D. [tex]\(2H(g) + S(s) +20.6 \text{ kJ} \rightarrow H_2S\)[/tex]

We consider the requirements for the enthalpy of formation:
- The reactants are the elements in their standard states.
- The reaction involves the formation of one mole of the compound.
- The change in enthalpy (ΔH) must correctly represent the given enthalpy of formation.

Analyzing each option:
1. Option A: [tex]\(H_2(g) + S(s) \rightarrow H_2S -20.6 \text{ kJ}\)[/tex]
- This reaction correctly shows 1 mole each of hydrogen gas and sulfur in their standard states forming 1 mole of [tex]\(H_2S\)[/tex], with [tex]\(-20.6 \text{ kJ}\)[/tex] on the product side, indicating an exothermic reaction releasing [tex]\(20.6 \text{ kJ}\)[/tex].

2. Option B: [tex]\(H_2(g) + S(s) \rightarrow H_2S +20.6 \text{ kJ}\)[/tex]
- This option wrongly indicates that the formation of [tex]\(H_2S\)[/tex] is endothermic (absorbing [tex]\(20.6 \text{ kJ}\)[/tex], which contradicts the given [tex]\(\Delta H_f\)[/tex]).

3. Option C: [tex]\(H_2(g) + S(s) +20.6 \text{ kJ} \rightarrow H_2S\)[/tex]
- This option wrongly suggests that [tex]\(20.6 \text{ kJ}\)[/tex] needs to be added to form [tex]\(H_2S\)[/tex], indicating an endothermic process.

4. Option D: [tex]\(2H(g) + S(s) +20.6 \text{ kJ} \rightarrow H_2S\)[/tex]
- This option involves atomic hydrogen (not standard state) and suggests an endothermic reaction, which is incorrect.

As a result, the correct reaction showing the enthalpy of formation of [tex]\(H_2S\)[/tex] as [tex]\(\Delta H_f = -20.6 \text{ kJ/mol}\)[/tex] is:

A. [tex]\(H_2(g) + S(s) \rightarrow H_2S -20.6 \text{ kJ}\)[/tex]
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