Problem 41 - Chemistry: The Central Science in SI Units, Global Edition

Question

Atomic hydrogen (H) is used in welding (AHW). The atoms recombine to hydrogen molecules with a large release of heat according to the following reaction: $$ 2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g) $$ (a) Using the thermodynamic data in Appendix C, calculate the enthalpy change for this reaction per mole of \(\mathrm{H}_{2}\). (b) Which has the higher enthalpy under these conditions, \(2 \mathrm{H}(g)\) or \(\mathrm{H}_{2}(g) ?\)

Step-by-Step Solution

Answer

The enthalpy change for the given reaction is \(-436 kJ/mol\), and under these conditions, 2 H(g) has a higher enthalpy than H₂(g).

Step 1: Find the enthalpy change for the reaction

First, we need to find the enthalpy change for the reaction using the thermodynamic data provided in Appendix C. The enthalpy change for the reaction can be written as: \(ΔH_{reaction} = H_{products} - H_{reactants}\) Using the thermodynamic data provided, we can find the enthalpy of formation for each species involved in the reaction: \(H_{H_{2}(g)} = 0 kJ/mol\) (by definition, the standard enthalpy formation of an element in its standard state is zero) \(H_{H(g)} = 218 kJ/mol\)

Step 2: Calculate the enthalpy change

Now, we can substitute the enthalpies of the species into the equation for the enthalpy change of the reaction: \(ΔH_{reaction} = (1 \times H_{H_{2}(g)}) - (2 \times H_{H(g)})\) \(ΔH_{reaction} = (1 \times 0) - (2 \times 218)\) \(ΔH_{reaction} = -436 kJ/mol\)

Step 3: Compare the enthalpies to determine which species has higher enthalpy

Now that we have the enthalpy change, we can determine which species - atomic hydrogen (H) or molecular hydrogen (H₂) - has the higher enthalpy under the given conditions. Since \(ΔH_{reaction} = H_{products} - H_{reactants}\), a negative enthalpy change means that the reactants have a higher enthalpy than the products. In this case, the reactants are 2 H(g) and the products are H₂(g). Therefore, under these conditions, 2 H(g) has a higher enthalpy than H₂(g). The answers are: (a) \(-436 kJ/mol\) (b) 2 H(g) has the higher enthalpy

Key Concepts

Thermodynamic Data

Understanding thermodynamic data is crucial when studying chemical reactions and their energy changes. Thermodynamic data includes a variety of values like heat capacity, entropy, and most importantly for this context, enthalpy. This data is often compiled in tables or appendices of textbooks to provide standard values needed for calculations.

When calculating the enthalpy change for a reaction, we rely heavily on the standard enthalpy of formation values. These values are measured under standard conditions, typically at 1 atm pressure and 298 K temperature. By using these standard values, calculations become more straightforward and are consistent across different reactions.

In the case of the reaction where atomic hydrogen \(2 \ \mathrm{H}(g)\) recombines into molecular hydrogen \(\mathrm{H}_{2}(g)\), using thermodynamic data allows us to accurately determine how much heat is released during the process. This makes the calculations more precise and helps in understanding the energy dynamics of the reaction.

Enthalpy of Formation

The enthalpy of formation is a specific type of enthalpy change. It's defined as the heat change when one mole of a compound is formed from its elements in their standard state. For most elements in their normal state, such as \(\mathrm{H}_{2}(g)\), this value is zero, as there is no formation needed from another elemental form.

When calculating reaction enthalpy using the enthalpy of formation, the general formula used is \(\Delta H_{reaction} = H_{products} - H_{reactants}\). In the given chemical reaction, \(\mathrm{H}_{2}(g)\) is the product and has an enthalpy of zero. Atomic hydrogen \(\mathrm{H}(g)\), on the other hand, has a formation enthalpy of 218 \text{kJ/mol}\.

By plugging these values into our equation, we find the enthalpy change for the reaction:

• The enthalpy of the products (\mathrm{H}_{2}(g)) is 0 \text{kJ/mol}\.
• Twice the enthalpy of the reactants (2 \(\mathrm{H}(g)\)) is 2 \times 218 \text{kJ/mol} = 436 \text{kJ/mol}\.

The \(\Delta H_{reaction}\) thus comes out to be \-436 \text{kJ/mol}\, indicating an exothermic reaction where energy is released.

Atomic Hydrogen

Atomic hydrogen is simply hydrogen in its singular atomic form, represented as \(\mathrm{H}(g)\). It's different from the more stable and commonly found molecular form, \(\mathrm{H}_{2}(g)\).

While atomic hydrogen is not stable and tends to recombine to form molecular hydrogen, it plays a significant role in reactions like welding, where the recombination releases a considerable amount of heat. This heat discharge is valuable for fueling the reaction or process.

The chemical reaction of atomic hydrogen forming molecular hydrogen is highly exothermic. This means it releases energy, which is why it is effectively utilized in welding applications. The high energy of atomic hydrogen compared to \(\mathrm{H}_{2}(g)\) can be quantified using thermodynamic data, illustrating its higher relative enthalpy. This underlines why understanding the states of hydrogen and their enthalpy is crucial in practical applications and chemical thermodynamics.