Problem 86 - Chemistry and Chemical Reactivity

Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell),\) is an efficient oxygen scavenger. It is sometimes added to steam boilers to remove traces of oxygen that can cause corrosion in these systems. Combustion of hydrazine gives the following information: $$\begin{aligned}&\mathrm{N}_{2} \mathrm{H}_{4}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\\\&\Delta H_{\mathrm{rxn}}^{\circ}=-534.3 \mathrm{kJ}\end{aligned}$$ (a) Is the reaction product- or reactant-favored? (b) Use the value for \(\Delta H_{\mathrm{rxn}}^{\circ}\) with the enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) to calculate the molar enthalpy of formation of \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\).

Step-by-Step Solution

Answer

(a) The reaction is product-favored. (b) The molar enthalpy of formation of \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\) is \(50.7\, \mathrm{kJ/mol}\).

Step 1: Analyzing Reaction Enthalpy for Favorability

The standard enthalpy change, \(\Delta H_{\mathrm{rxn}}^{\circ}\), for the reaction is given as \(-534.3\, \mathrm{kJ}\). A negative \(\Delta H_{\mathrm{rxn}}^{\circ}\) indicates that the reaction is exothermic, meaning it releases heat. This suggests that the reaction is product-favored because energy is released when moving from reactants to products.

Step 2: Understanding Reaction Stoichiometry

The balanced chemical reaction is \(\mathrm{N}_{2} \mathrm{H}_{4} (\ell) + \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{N}_{2} (\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O} (\mathrm{g})\). This indicates that one mole of \(\mathrm{N}_{2} \mathrm{H}_{4} (\ell)\) reacts with one mole of \(\mathrm{O}_{2} (\mathrm{g})\) to produce one mole of \(\mathrm{N}_{2} (\mathrm{g})\) and two moles of \(\mathrm{H}_{2} \mathrm{O} (\mathrm{g})\).

Step 3: Using Enthalpy of Formation Data

The given \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction is \(-534.3 \mathrm{kJ/mol}\). We need the standard molar enthalpies of formation for the reactants and products. For \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H_{f}^{\circ} = -241.8\, \mathrm{kJ/mol}\). \(\Delta H_{f}^{\circ}\) for \(\mathrm{N}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) are both zero in their standard states.

Step 4: Applying Hess's Law and Enthalpy Calculations

Using Hess's Law, we know: \[\Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ} (\text{products}) - \sum \Delta H_{f}^{\circ} (\text{reactants})\]Substitute the known values:\[-534.3 = [0 + 2(-241.8)] - [\Delta H_{f}^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)) + 0]\]Solving the equation:\[-534.3 = -483.6 - \Delta H_{f}^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(\ell))\]\[\Delta H_{f}^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)) = -483.6 + 534.3\]\[\Delta H_{f}^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)) = 50.7\, \mathrm{kJ/mol}\]

Step 5: Conclusion

The reaction is product-favored due to the negative enthalpy change, and the molar enthalpy of formation for \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\) is \(50.7\, \mathrm{kJ/mol}\).

Key Concepts

Combustion Reactions

Combustion reactions are powerful processes where a substance combines with oxygen to release energy in the form of heat and light. In the combustion of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\), the reaction with oxygen \(\mathrm{O}_{2}(\mathrm{g})\) produces nitrogen gas \(\mathrm{N}_{2}(\mathrm{g})\) and water vapor \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). A key sign of a combustion reaction is the exothermic release of energy. This reaction releases \(-534.3 \; \mathrm{kJ/mol}\), confirming its exothermic nature.
One of the fundamental reasons combustion reactions are vital is they can transform chemical potential energy into usable energy. You'll see this process in everyday phenomena like burning wood or fuel.

• Common Features: Reaction with \(\mathrm{O}_{2}\) and energy release.
• Results: Often produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\).

Hess's Law

Hess's Law is a powerful tool in thermochemistry that allows us to calculate the total enthalpy change for a reaction by adding up the enthalpy changes of individual steps that lead to the overall reaction.
It states that the total enthalpy change in a chemical reaction is the same, no matter what multiple steps or different pathways we take to get there. This concept applies even if intermediate substances are formed that lead to the final products.
For the combustion of hydrazine, Hess's Law helps us determine the enthalpy of formation for \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\). Here's how you can use it:

• Use the equation: \[\Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{f}^{\circ} (\text{products}) - \sum \Delta H_{f}^{\circ} (\text{reactants})\]
• Substitute known enthalpy of formation values to solve for unknowns.

This allows chemists to calculate reactions that are difficult to observe directly.

Enthalpy of Formation

The enthalpy of formation refers to the heat change associated with the formation of a compound from its elements, with all substances in their standard states.
For a compound like hydrazine, finding the enthalpy of formation involves studying the energy required to form it from nitrogen and hydrogen in their gaseous states.
In a standard state, the enthalpy of formation for elements like \(\mathrm{N}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is zero. For this exercise, using the enthalpy of reaction and Hess's Law, we can determine that the enthalpy of formation for hydrazine \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)\) is \(50.7\, \mathrm{kJ/mol}\).

• Importance: Helps calculate unknowns in reactions.
• Used For: Understanding energy requirements of forming new compounds.

Exothermic Reactions

Exothermic reactions are processes that release energy, typically as heat, into the surrounding environment. When a chemical reaction has a negative enthalpy change like the combustion of hydrazine \((-534.3 \; \mathrm{kJ/mol})\), it is exothermic.
These reactions are characterized by the following features:

• Heat Release: Feel warmer to the touch due to heat release.
• Product-Favored: Energy is often released when new bonds in the products are more stable than those in the reactants.

In practical terms, exothermic reactions are used extensively for heating and energy production, making them essential in power plants and even everyday products like hand warmers. The study of these reactions helps us optimize energy usage across various applications.