Which Reaction Shows That The Enthalpy Of Formation Of H 2 S H_2S H 2 S Is Δ H F = − 20.6 KJ/mol \Delta H_f = -20.6 \, \text{kJ/mol} Δ H F = − 20.6 KJ/mol ?A. H 2 ( G ) + S ( S ) + 20.6 KJ → H 2 S H_2(g) + S(s) + 20.6 \, \text{kJ} \rightarrow H_2S H 2 ( G ) + S ( S ) + 20.6 KJ → H 2 S B. 2 H ( G ) + S ( S ) + 20.6 KJ → H 2 S 2H(g) + S(s) + 20.6 \, \text{kJ} \rightarrow H_2S 2 H ( G ) + S ( S ) + 20.6 KJ → H 2 S

Mar 8, 2025 by ADMIN 425 views

Which Reaction Shows the Enthalpy of Formation of $H_2S$ is $\Delta H_f = -20.6 \, \text{kJ/mol}$ ?

Understanding Enthalpy of Formation

Enthalpy of formation, denoted by $\Delta H_f$ , is a measure of the energy change that occurs when one mole of a compound is formed from its constituent elements in their standard states. It is an important thermodynamic property that helps us understand the stability of a compound and its potential reactivity. In this article, we will explore which reaction shows that the enthalpy of formation of $H_2S$ is $\Delta H_f = -20.6 \, \text{kJ/mol}$ .

The Correct Reaction

To determine the correct reaction, let's first recall the definition of enthalpy of formation. The enthalpy of formation of a compound is the energy change that occurs when one mole of the compound is formed from its constituent elements in their standard states. In the case of $H_2S$ , the enthalpy of formation is given as $\Delta H_f = -20.6 \, \text{kJ/mol}$ .

Option A: $H_2(g) + S(s) + 20.6 \, \text{kJ} \rightarrow H_2S$

This reaction shows hydrogen gas ( $H_2(g)$ ) reacting with solid sulfur ( $S(s)$ ) to form hydrogen sulfide ( $H_2S$ ). However, the reaction is not balanced, and the energy term is not correctly represented. The correct representation of the reaction should be:

H_2(g) + S(s) \rightarrow H_2S

The energy term, $20.6 \, \text{kJ}$ , is not part of the reaction itself but rather a measure of the energy change that occurs when the reaction takes place. In this case, the energy term is positive, indicating that energy is required to form $H_2S$ from $H_2(g)$ and $S(s)$ .

Option B: $2H(g) + S(s) + 20.6 \, \text{kJ} \rightarrow H_2S$

This reaction shows two hydrogen atoms ( $2H(g)$ ) reacting with solid sulfur ( $S(s)$ ) to form hydrogen sulfide ( $H_2S$ ). However, this reaction is not correct because it involves two hydrogen atoms instead of one. The correct representation of the reaction should be:

H_2(g) + S(s) \rightarrow H_2S

Conclusion

Based on the definition of enthalpy of formation and the correct representation of the reaction, we can conclude that neither of the given reactions shows that the enthalpy of formation of $H_2S$ is $\Delta H_f = -20.6 \, \text{kJ/mol}$ . The correct reaction should be:

H_2(g) + S(s) \rightarrow H_2S

However, this reaction does not show the enthalpy of formation of $H_2S$ as $\Delta H_f = -20.6 \, \text{kJ/mol}$ . The enthalpy of formation is a measure of the energy change that occurs when one mole of a compound is formed from its constituent elements in their standard states. In this case, the enthalpy of formation of $H_2S$ is given as $\Delta H_f = -20.6 \, \text{kJ/mol}$ , indicating that energy is released when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .

The Correct Reaction

To determine the correct reaction, let's consider the following reaction:

H_2(g) + S(s) \rightarrow H_2S

This reaction shows hydrogen gas ( $H_2(g)$ ) reacting with solid sulfur ( $S(s)$ ) to form hydrogen sulfide ( $H_2S$ ). The energy change that occurs when this reaction takes place is given as $\Delta H_f = -20.6 \, \text{kJ/mol}$ , indicating that energy is released when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .

The Correct Answer

Based on the definition of enthalpy of formation and the correct representation of the reaction, we can conclude that the correct reaction is:

H_2(g) + S(s) \rightarrow H_2S

This reaction shows that the enthalpy of formation of $H_2S$ is $\Delta H_f = -20.6 \, \text{kJ/mol}$ , indicating that energy is released when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .

Key Takeaways

Enthalpy of formation is a measure of the energy change that occurs when one mole of a compound is formed from its constituent elements in their standard states.
The correct representation of the reaction should be: $H_2(g) + S(s) \rightarrow H_2S$ .
The energy term, $20.6 \, \text{kJ}$ , is not part of the reaction itself but rather a measure of the energy change that occurs when the reaction takes place.
The enthalpy of formation of $H_2S$ is given as $\Delta H_f = -20.6 \, \text{kJ/mol}$ , indicating that energy is released when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .
Q&A: Enthalpy of Formation of $H_2S$

Frequently Asked Questions

In this article, we will answer some frequently asked questions related to the enthalpy of formation of $H_2S$ .

Q: What is the enthalpy of formation of $H_2S$ ?

A: The enthalpy of formation of $H_2S$ is $\Delta H_f = -20.6 \, \text{kJ/mol}$ .

Q: What is the correct representation of the reaction that shows the enthalpy of formation of $H_2S$ ?

A: The correct representation of the reaction is:

H_2(g) + S(s) \rightarrow H_2S

Q: Why is the energy term, $20.6 \, \text{kJ}$ , not part of the reaction itself?

A: The energy term, $20.6 \, \text{kJ}$ , is not part of the reaction itself but rather a measure of the energy change that occurs when the reaction takes place.

Q: What does the negative sign of the enthalpy of formation indicate?

A: The negative sign of the enthalpy of formation indicates that energy is released when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .

Q: Why is the enthalpy of formation of $H_2S$ important?

A: The enthalpy of formation of $H_2S$ is important because it helps us understand the stability of $H_2S$ and its potential reactivity.

Q: How can we use the enthalpy of formation of $H_2S$ to predict its behavior in different situations?

A: We can use the enthalpy of formation of $H_2S$ to predict its behavior in different situations by considering the energy change that occurs when $H_2S$ is formed from $H_2(g)$ and $S(s)$ . If the energy change is negative, it indicates that energy is released, and $H_2S$ is likely to be stable. If the energy change is positive, it indicates that energy is required, and $H_2S$ may be less stable.

Q: Can we use the enthalpy of formation of $H_2S$ to predict its reactivity with other substances?

A: Yes, we can use the enthalpy of formation of $H_2S$ to predict its reactivity with other substances. If the energy change that occurs when $H_2S$ reacts with another substance is negative, it indicates that energy is released, and the reaction is likely to be exothermic. If the energy change is positive, it indicates that energy is required, and the reaction may be endothermic.

Q: How can we measure the enthalpy of formation of $H_2S$ experimentally?

A: We can measure the enthalpy of formation of $H_2S$ experimentally by using a calorimeter to measure the energy change that occurs when $H_2S$ is formed from $H_2(g)$ and $S(s)$ .

Q: What are some common applications of the enthalpy of formation of $H_2S$ ?

A: Some common applications of the enthalpy of formation of $H_2S$ include:

Predicting the stability of $H_2S$ and its potential reactivity
Predicting the behavior of $H_2S$ in different situations
Designing experiments to measure the enthalpy of formation of $H_2S$
Understanding the thermodynamics of $H_2S$ reactions

Conclusion

In this article, we have answered some frequently asked questions related to the enthalpy of formation of $H_2S$ . We have discussed the definition of enthalpy of formation, the correct representation of the reaction, and the importance of the enthalpy of formation of $H_2S$ . We have also discussed how to use the enthalpy of formation of $H_2S$ to predict its behavior in different situations and its reactivity with other substances.