Problem 67 - Chemistry: Atoms First

Question

Calculate the standard molar enthalpy of formation of \(\mathrm{NO}(g)\) from the following data: \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=66.4 \mathrm{kJ}\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=-114.1 \mathrm{kJ}\)

Step-by-Step Solution

Answer

\(\Delta H^{\circ}_{\text{formation}} \text{ of } \mathrm{NO}(g) = -47.7 \mathrm{kJ/mol}\)

Step 1 - Write down the given reactions and their enthalpies

We are given the following two reactions: 1. \( \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=66.4 \mathrm{kJ} \) 2. \( 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=-114.1 \mathrm{kJ} \) We will use these reactions to calculate the standard molar enthalpy of formation of NO(g).

Step 2 - Derive the reaction for the formation of \( \mathrm{NO}(g) \) from its elements in their standard states

The standard molar enthalpy of formation of a compound is defined as the change in enthalpy when 1 mole of the compound is formed from its elements in their standard states. For NO(g), the reaction is: \[\frac{1}{2} \mathrm{N}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)\]

Step 3 - Manipulate the given reactions to derive the formation reaction of NO(g)

To get the formation reaction for NO(g), we need 1 mole of NO(g) on the products side and the elements N and O in their standard states on the reactants side. We take half of the first reaction: \[\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)\] For which the enthalpy will be \(\frac{66.4 \mathrm{kJ}}{2} = 33.2 \mathrm{kJ}\). Then we reverse the second reaction to obtain NO(g) from NO2(g): \[2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g)\] Which will have an enthalpy of \(114.1 \mathrm{kJ}\) (reversing the reaction changes the sign of \(\Delta H^{\circ}\)). Now, we subtract the modified first reaction from the second reversed reaction to cancel out the NO2 and O2: \[2 \mathrm{NO}_{2}(g) - (\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g)) = 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) - (\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g))\]

Step 4 - Combine the enthalpies of the modified reactions

By Hess's law, the enthalpy change for the overall reaction can be found by adding the enthalpy changes of the individual steps: \[\Delta H^{\circ}_{\text{formation}} = (2 \times 33.2 \mathrm{kJ}) - 114.1 \mathrm{kJ}\] We must multiply the enthalpy for the half-reaction by 2 to account for the two moles of NO2(g), since we reversed the second equation for two moles of NO2(g). Calculation gives: \[\Delta H^{\circ}_{\text{formation}} = 66.4 \mathrm{kJ} - 114.1 \mathrm{kJ} = -47.7 \mathrm{kJ}\]

Step 5 - Adjust enthalpy for the desired number of moles

Since we have calculated the enthalpy change for the formation of 1 mole of NO(g), no further adjustment is necessary. The standard molar enthalpy of formation of NO(g) is therefore -47.7 kJ per mole.

Key Concepts

Enthalpy Change Calculation

Understanding the calculation of enthalpy change is crucial in chemical thermodynamics, as it provides insight into the energy changes during chemical reactions. One aspect of this is the standard molar enthalpy of formation, which is the heat change that occurs when one mole of a substance is formed from its elements in their standard states.

To calculate the enthalpy change, you should first write down the related chemical equations along with their associated enthalpy changes. Then, using stoichiometry, you manipulate the equations to reflect the formation of the desired substance from its constituent elements. Throughout this process, follow the law of conservation of energy which implies that the total energy of the system remains constant, thus allowing the use of Hess's law for calculations.

Hess's Law

Hess's law is a powerful tool in chemical thermodynamics, which states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to carry out the reaction. This principle is based on the fact that enthalpy is a state function, which means its value depends only on the current state of the system, not the path taken to reach that state.

For example, if a reaction can occur directly or through several steps, the total enthalpy change will be identical in both cases. In practice, we use Hess's law to calculate unknown enthalpy changes by adding or subtracting known enthalpy changes of intermediate steps formulated in stoichiometric balance.

Chemical Thermodynamics

Chemical thermodynamics deals with the study of energy and work in the context of chemical reactions and changes in physical states. It's concerned with the heat and energy transfer associated with these processes. The first law of thermodynamics, also known as the law of energy conservation, establishes that energy cannot be created or destroyed in an isolated system.

The concept of enthalpy plays a central role in chemical thermodynamics because it reflects the heat content within a system at constant pressure. Understanding the enthalpy changes, often calculated using Hess's law, allows chemists to predict whether a reaction will release heat (exothermic) or absorb heat (endothermic) and how that impacts the reaction's spontaneity, which is further analyzed using Gibbs free energy.

Stoichiometry

Stoichiometry is a section of chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It is built on the balanced chemical equation, which represents the conservation of mass. Through stoichiometry, you can predict the amount of reactants needed or the amount of products formed.

For enthalpy calculations, stoichiometry is used to determine the correct proportions of each substance involved in a reaction or a series of reactions. This ensures that when using Hess's law, the enthalpies can be accurately combined, added, or subtracted to compute the energy change for the overall reaction. A clear understanding of stoichiometry is critical when dealing with complex reactions to achieve balanced chemical equations, which serve as the foundation for further thermodynamics calculations.