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Chapter 5: Problem 42

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) 1 \(\mathrm{mol} \mathrm{CO}_{2}(s)\) or 1 \(\mathrm{mol} \mathrm{CO}_{2}(g)\) at the same temperature, ( b) 2 \(\mathrm{mol}\) of hydrogen atoms or 1 \(\mathrm{mol}\) of \(\mathrm{H}_{2},(\mathbf{c}) 1 \mathrm{mol} \mathrm{H}_{2}(g)\) and 0.5 \(\mathrm{mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{mol} \mathrm{N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\) .

Short Answer

Expert verified
(a) 1 mol of CO鈧�(g) has higher enthalpy than 1 mol of CO鈧�(s). (b) 2 mol of hydrogen atoms have higher enthalpy than 1 mol of H鈧�. (c) 1 mol H鈧�(g) and 0.5 mol O鈧�(g) have higher enthalpy than 1 mol H鈧侽(g) at 25掳C. (d) 1 mol of N鈧�(g) at 300掳C has a higher enthalpy than 1 mol of N鈧�(g) at 100掳C.

Step by step solution

01

Case (a): 1 mol CO鈧�(s) or 1 mol CO鈧�(g) at the same temperature

When comparing a solid and a gas at the same temperature, the gas usually has higher enthalpy. The reason is that, in the gas state, particles have higher kinetic energy due to their increased freedom of movement. Therefore, 1 mol of CO鈧�(g) has higher enthalpy than 1 mol of CO鈧�(s).
02

Case (b): 2 mol of hydrogen atoms or 1 mol of H鈧�

Here, we can consider the fact that a covalent bond exists in H鈧�, which requires energy to break. In other words, forming a bond in H鈧� releases energy, making the molecule's enthalpy lower. Therefore, 2 mol of hydrogen atoms have higher enthalpy than 1 mol of H鈧�.
03

Case (c): 1 mol H鈧�(g) and 0.5 mol O鈧�(g) at 25掳C or 1 mol H鈧侽(g) at 25掳C

In this case, we can use enthalpy of formation concept. The enthalpy of formation of H鈧侽(g) is negative, meaning that its formation releases energy. Conversely, breaking the bonds in H鈧侽(g) requires energy. Therefore, 1 mol H鈧�(g) and 0.5 mol O鈧�(g) have higher enthalpy since they have not yet formed into H鈧侽(g).
04

Case (d): 1 mol N鈧�(g) at 100掳C or 1 mol N鈧�(g) at 300掳C

In this case, we are comparing the same substance at two different temperatures. The higher the temperature, the more kinetic energy and, consequently, enthalpy is possessed by the particles. Therefore, 1 mol of N鈧�(g) at 300掳C has a higher enthalpy than 1 mol of N鈧�(g) at 100掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

CO2 Phase Changes
When discussing the phase changes of carbon dioxide (\[ ext{CO}_2 \]), it's important to understand that this compound can exist in solid, liquid, and gaseous states under different conditions. However, we're focusing on the solid (\[ ext{CO}_2(s) \]) and gaseous (\[ ext{CO}_2(g) \]) phases. At the same temperature, the gaseous state of \[ ext{CO}_2 \] has a higher enthalpy than the solid state.

To explain this, we need to dive into the nature of molecules in each phase. In the gaseous state, CO鈧� molecules move more freely due to higher kinetic energy. This freedom of motion is attributed to the higher energy level, meaning that more heat (enthalpy) is present in the gas form. In contrast, in the solid state, molecules are tightly packed and vibrate in fixed positions, resulting in lower kinetic energy and, consequently, lower enthalpy.

  • Gases usually have higher enthalpies compared to solids.
  • Kinetic energy, which translates to movement, greatly influences enthalpy.
Enthalpy of Formation
The concept of enthalpy of formation is crucial when considering chemical reactions, particularly when discussing the stability and energy content of compounds. It refers to the heat change associated with the formation of one mole of a compound from its elements in their standard states.

For example, when we look at the formation of water (\[ ext{H}_2 ext{O} \]) from hydrogen (\[ ext{H}_2 \]) and oxygen (\[ ext{O}_2 \]):\[ ext{H}_2(g) + 0.5 ext{O}_2(g) ightarrow ext{H}_2 ext{O}(g)\]The enthalpy of formation of water (\[ ext{H}_2 ext{O}(g) \]) is negative. This means that energy is released when water forms, making it more stable and lower in energy compared to its gaseous reactants. Thus, reactants like \[ ext{H}_2(g) \] and \[ ext{O}_2(g) \] have a higher enthalpy than the product, \[ ext{H}_2 ext{O}(g) \].

  • Enthalpy of formation helps determine the energy absorbed or released during formation.
  • Negative enthalpy of formation implies stability and energy release.
Kinetic Energy and Temperature
Kinetic energy and temperature are deeply interconnected concepts in thermodynamics. Essentially, kinetic energy refers to the energy a body possesses due to its motion, which contributes significantly to a system's temperature.

In the context of gases, temperature is a direct measure of the average kinetic energy of the particles. As temperature increases, so does the kinetic energy, making the gas particles move faster and collide more frequently. This increase in motion not only raises the temperature but also the enthalpy, since enthalpy includes energy related to temperature.

For example, consider the comparison of nitrogen gas (\[ ext{N}_2 \]) at two different temperatures:
  • At 100掳C, \[ ext{N}_2(g) \] has less kinetic energy compared to \[ ext{300掳C} \].
  • At 300掳C, \[ ext{N}_2(g) \] particles move more vigorously, leading to higher kinetic energy and enthalpy.
This illustrates that higher temperatures correlate with an increase in both kinetic energy and enthalpy.

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Most popular questions from this chapter

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

(a) When a 0.235 -g sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19\()\) , the temperature rises \(1.642^{\circ} \mathrm{C} .\) When a \(0.265-\mathrm{g}\) sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2}\) is burned, the temperature rises \(1.525^{\circ} \mathrm{C} .\) Using the value 26.38 \(\mathrm{kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{g},\) what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

During a normal breath, our lungs expand about 0.50 L against an external pressure of 1.0 atm. How much work is involved in this process (in J)?

A sodium ion, \(\mathrm{Na}^{+},\) with a charge of \(1.6 \times 10^{-19} \mathrm{Cand}\) a chloride ion, \(\mathrm{Cl}^{-},\) with charge of \(-1.6 \times 10^{-19} \mathrm{C}\) , are separated by a distance of 0.50 \(\mathrm{nm}\) . How much work would be required to increase the separation of the two ions to an infinite distance?

From the enthalpies of reaction $$\begin{aligned} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) & \Delta H=-537 \mathrm{kJ} \\ \mathrm{C}(s)+2 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) & \Delta H=-680 \mathrm{kJ} \\ 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) & \Delta H=+52.3 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2} :\) $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)$$
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