Question

For the following reactions at constant pressure, predict if \(\Delta H>\Delta E, \Delta H<\Delta E\), or \(\Delta H=\Delta E .\) a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

For the given reactions: - Reaction a (\(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\)): \(\Delta H = \Delta E\). - Reaction b (\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)): \(\Delta H<\Delta E\). - Reaction c (\(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)): \(\Delta H>\Delta E\).

Step-by-step solution

Reaction a

Determine the change in the number of moles for reaction a: \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\). On the left side, there are 2 moles of \(\mathrm{HF}(g)\), and on the right, there is 1 mole of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{F}_{2}(g)\). Therefore, \(\Delta n =1+1-2=0\).

Reaction a Comparisons

Because \(\Delta n=0\) for reaction a, there is no change in the number of moles, and \(\Delta V=0\). Thus, \(\Delta H=\Delta E+P \Delta V=\Delta E\).

Reaction b

Determine the change in the number of moles for reaction b: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\). On the left side, there are 1 mole of \(\mathrm{N}_{2}(g)\) and 3 moles of \(\mathrm{H}_{2}(g)\), and on the right, there are 2 moles of \(\mathrm{NH}_{3}(g)\). Therefore, \(\Delta n = 2 - (1+3)=-2\).

Reaction b Comparisons

Since \(\Delta n=-2\) for reaction b, there is a decrease in the number of moles, and \(\Delta V<0\). Thus, \(\Delta H=\Delta E+P \Delta V<\Delta E\).

Reaction c

Determine the change in the number of moles for reaction c: \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\). On the left side, there are 4 moles of \(\mathrm{NH}_{3}(g)\) and 5 moles of \(\mathrm{O}_{2}(g)\), and on the right, there are 4 moles of \(\mathrm{NO}(g)\) and 6 moles of \(\mathrm{H}_{2}\mathrm{O}(g)\). Therefore, \(\Delta n = (4+6)-(4+5)=1\).

Reaction c Comparisons

Because \(\Delta n=1\) for reaction c, there is an increase in the number of moles, and \(\Delta V>0\). Thus, \(\Delta H=\Delta E+P \Delta V>\Delta E\). In summary: - For reaction a, \(\Delta H = \Delta E\). - For reaction b, \(\Delta H<\Delta E\). - For reaction c, \(\Delta H>\Delta E\).

Key concepts

Thermodynamics

Thermodynamics is the study of energy and its transformations. In chemistry, it helps us understand the energy changes that occur during chemical reactions. One important concept in thermodynamics is enthalpy (\( \Delta H \)), which is the heat content of a system at constant pressure. Another related concept is internal energy (\( \Delta E \)), which is the total energy within a system. These values can change during reactions, giving insight into whether a reaction absorbs or releases energy.

Enthalpy is particularly useful because most chemical reactions occur at constant pressure, so it directly relates to heat exchange. The relationship between \( \Delta H \) and \( \Delta E \) is given by:

• \( \Delta H = \Delta E + P \Delta V \)

where \( P \) is the pressure and \( \Delta V \) is the change in volume. This equation shows that enthalpy depends on both internal energy changes and changes in volume, which is significant under gas law conditions.

When no volume change (\( \Delta V = 0 \)) occurs, the enthalpy change equals the internal energy change (\( \Delta H = \Delta E \)). If the volume increases, the enthalpy change is greater than the internal energy change, and if it decreases, the enthalpy change is less. This is fundamental in predicting reaction behavior.

Moles change in reactions

Understanding how moles change in reactions is crucial for predicting energy changes. The number of moles directly influences the volume of gases involved in reactions, according to the ideal gas law.

In chemical reactions, we calculate the change in the number of moles, \( \Delta n \), by subtracting the moles on the reactant side from those on the product side:

• A decrease in moles (\( \Delta n<0 \)) typically indicates a volume reduction, leading to a decrease in enthalpy relative to internal energy (\( \Delta H<\Delta E \)).
• No change in moles (\( \Delta n=0 \)) means there's no volume change, so \( \Delta H = \Delta E \)
• An increase in moles (\( \Delta n>0 \)) suggests a volume increase, resulting in an enthalpy change greater than the internal energy change (\( \Delta H>\Delta E \)).

Considering moles change helps explain why, for instance, reaction b in the original exercise shows \( \Delta H<\Delta E \), while reaction c shows \( \Delta H>\Delta E \). It's all about how gases expand or compress.

Gas laws in chemistry

Gas laws are essential in chemistry for understanding how gases behave under different conditions and how these behaviors impact chemical reactions. The ideal gas law, \( PV = nRT \), is foundational, linking pressure (\( P \)), volume (\( V \)), moles (\( n \)), the gas constant (\( R \)), and temperature (\( T \)).

In reactions involving gases, changes in pressure, volume, and temperature can influence reaction dynamics:

• If the total number of gas moles increases (\( \Delta n>0 \)), the volume tends to increase if temperature and pressure are constant, impacting the enthalpy change.
• Conversely, a decrease in moles (\( \Delta n<0 \)) often leads to a volume decrease.

By understanding these laws, you can predict how certain conditions might shift the equilibrium of reactions and the energy profiles (\( \Delta H \) and \( \Delta E \)). This knowledge explains why adjusting conditions can favor either reactants or products in a chemical equation, playing a pivotal role in reaction rates and efficiencies.