Which reaction shows that the enthalpy of formation of $H _2 S$ is $\Delta H_{ f }=-20.6 kJ / mol$?
A. $H _2(g)+ S ( s )+20.6 kJ \rightarrow H _2 S$
B. $H _2(g)+ S ( s ) \rightarrow H _2 S+20.6 kJ$
C. $2 H (g)+ S (s)+20.6 kJ \rightarrow H _2 S$
D. $H _2(g)+ S ( s ) \rightarrow H _2 S-20.6 kJ$
Answer (2)
The enthalpy of formation Δ H f represents the energy change when forming 1 mole of a substance from its elements.
For H 2 S , the formation reaction is H 2 ( g ) + S ( s ) → H 2 S ( g ) with Δ H f = − 20.6 kJ/mol .
This means H 2 ( g ) + S ( s ) → H 2 S ( g ) + 20.6 kJ , indicating energy release.
Therefore, the correct reaction is H 2 ( g ) + S ( s ) → H 2 S + 20.6 k J .
Explanation
Understanding Enthalpy of Formation The enthalpy of formation, denoted as Δ H f , represents the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states. In this case, we are looking for the reaction that correctly represents the formation of H 2 S with Δ H f = − 20.6 kJ/mol . This means that 20.6 kJ of energy is released when one mole of H 2 S is formed from its elements in their standard states.
Writing the Formation Reaction The standard state for hydrogen is H 2 ( g ) , and for sulfur, it is S ( s ) . Therefore, the formation reaction of H 2 S can be written as: H 2 ( g ) + S ( s ) → H 2 S ( g ) Since the enthalpy of formation is Δ H f = − 20.6 kJ/mol , the reaction can be written as: H 2 ( g ) + S ( s ) → H 2 S ( g ) Δ H f = − 20.6 kJ/mol This indicates that 20.6 kJ of energy is released during the formation of one mole of H 2 S . We can rewrite the reaction to explicitly include the energy term: H 2 ( g ) + S ( s ) → H 2 S ( g ) + 20.6 kJ Alternatively, we can represent it as: H 2 ( g ) + S ( s ) − ( − 20.6 kJ ) → H 2 S ( g ) H 2 ( g ) + S ( s ) + 20.6 kJ → H 2 S ( g ) However, since the enthalpy of formation is negative, the energy is released, so it should appear on the product side.
Comparing with Given Options Now, let's compare the derived reaction with the given options: Option A: H 2 ( g ) + S ( s ) + 20.6 kJ → H 2 S (This implies energy is absorbed, which is incorrect since Δ H f is negative.) Option B: H 2 ( g ) + S ( s ) → H 2 S + 20.6 kJ (This correctly shows energy is released.) Option C: 2 H ( g ) + S ( s ) + 20.6 kJ → H 2 S (Incorrect, since it uses 2 H ( g ) instead of H 2 ( g ) .) Option D: H 2 ( g ) + S ( s ) → H 2 S − 20.6 kJ (This is equivalent to H 2 ( g ) + S ( s ) + 20.6 kJ → H 2 S , which is incorrect.)
Final Answer Based on the analysis, option B correctly represents the enthalpy of formation of H 2 S as Δ H f = − 20.6 kJ/mol , showing that 20.6 kJ of energy is released when one mole of H 2 S is formed from its elements in their standard states.
Examples
Understanding enthalpy of formation is crucial in various real-world applications, such as designing chemical reactions for industrial processes. For example, when synthesizing ammonia ( N H 3 ) from nitrogen and hydrogen, knowing the enthalpy of formation helps determine the amount of heat released or absorbed during the reaction. This information is vital for optimizing reaction conditions, ensuring safety, and maximizing product yield. By carefully controlling the reaction's enthalpy change, engineers can create efficient and sustainable chemical processes.
The enthalpy of formation of H 2 S is represented by the reaction H 2 ( g ) + S ( s ) → H 2 S + 20.6 kJ , indicating that this process releases 20.6 kJ of energy. Therefore, the correct choice from the options provided is Option B .
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