Question

Given that \(\Delta \mathrm{H}^{\circ} \mathrm{CO} 2(\mathrm{~g})=-94.0, \Delta \mathrm{H}^{\circ} \mathrm{CO}(\mathrm{g})=-26,4, \Delta \mathrm{H}^{\circ} \mathrm{H} 2 \mathrm{O}(\ell)\) \(=68.4\) and \(\Delta \mathrm{H}^{\circ} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})=-57.8 \mathrm{Kcal} / \mathrm{mole}\), determine the heats of reaction of (1) \(\mathrm{CO}(\mathrm{g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}),(2)\) \(\mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) and (3) \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short answer

The heats of reaction for the specified reactions are as follows: 1. For \(CO(g) + \frac{1}{2}O_{2}(g) \rightarrow CO_{2}(g)\), the heat of reaction (\(∆H°_{reaction}\)) is -67.6 Kcal/mole. 2. For \(H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)\), the heat of reaction (\(∆H°_{reaction}\)) is 68.4 Kcal/mole. 3. For the phase change from liquid to gas \(H_{2}O(l) \rightarrow H_{2}O(g)\), the heat of reaction (\(∆H°_{reaction}\)) is -126.2 Kcal/mole.

Step-by-step solution

\(∆H°_{reaction} = ∆H°_{CO_2(g)} - ∆H°_{CO(g)} - \frac{1}{2} ∆H°_{O_2(g)}\) Since the standard enthalpy of formation of an element in its standard state is zero, ∆H° for O2(g) is zero. Now, we can use the given ∆H° values for CO2(g) and CO(g):

\(∆H°_{reaction} = (-94.0) - (-26.4) = -67.6 \, \mathrm{Kcal/mole}\) #Step 2: Calculate the heat of reaction for H2(g) + (1/2)O2(g) → H2O(l)# Similarly, we will use Hess's law to find the heat of this reaction:

\(∆H°_{reaction} = ∆H°_{H_2O(l)} - ∆H°_{H_2(g)} - \frac{1}{2} ∆H°_{O_2(g)}\) Since H2(g) is an element in its standard state, its ∆H° value is zero, and we can use the given ∆H° value for H2O(l):

\(∆H°_{reaction} = (68.4) = 68.4 \, \mathrm{Kcal/mole}\) #Step 3: Calculate the heat of reaction for H2O(l) → H2O(g)# This third reaction is actually a phase-change reaction where liquid water becomes water vapor. We can use Hess's law once again to determine the heat of this reaction:

\(∆H_{reaction}° = ∆H°_{H_2O(g)} - ∆H°_{H_2O(l)}\) Now, we can use the given ∆H° values for H2O(g) and H2O(l):

\(∆H°_{reaction} = (-57.8) - (68.4) = -126.2 \, \mathrm{Kcal/mole}\) In conclusion, the heats of reaction for the given reactions are: 1. CO(g) + (1/2)O2(g) → CO2(g): ∆H°reaction = -67.6 Kcal/mole 2. H2(g) + (1/2)O2(g) → H2O(l): ∆H°reaction = 68.4 Kcal/mole 3. H2O(l) → H2O(g): ∆H°reaction = -126.2 Kcal/mole

Key concepts

Hess's Law

Hess's Law is a fundamental concept in thermochemistry, stating that the total enthalpy change for a chemical reaction is the same, no matter how it is carried out, provided the initial and final conditions are the same. This principle is especially useful when the enthalpy change of a reaction cannot be measured directly. Instead, one can calculate it by combining other reactions for which the enthalpy changes are known.

The exercise described utilizes Hess's Law for determining heats of reaction by taking into account known standard enthalpy of formation values. By rearranging known reactions and adding their enthalpy changes, one can find the desired enthalpy change for a new reaction. For example, in the provided problem, the heat of reaction for the formation of CO2(g) from CO(g) and O2(g) is found using known values of individual substances, underlining the practical application of Hess's Law in solving thermochemical equations.

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \( \Delta H^\circ_f \), is a measure of the energy change when one mole of a substance is formed from its elements in their standard states. The standard state refers to the most stable form of an element or compound at 1 atmosphere of pressure and a specified temperature, typically 25°C (298 K). It is a crucial piece of information in thermochemistry as it helps us understand how much energy is involved in the formation of substances.

For instance, in the exercise given, the standard enthalpy of formation is used to calculate the heat of three different reactions. Elements in their standard states, like O2(g) and H2(g), have a \( \Delta H^\circ_f \) of zero. This concept simplifies the calculations of reaction enthalpies by allowing only the \( \Delta H^\circ_f \) values of compounds to be considered in the Hess's Law equation.

Thermochemistry

Thermochemistry is the study of the heat and energy associated with chemical reactions and physical transformations. It helps scientists and engineers understand how energy is absorbed or released during these processes, enabling informed decisions in chemical manufacturing, energy production, and even environmental policy.

The exercise given exemplifies thermochemistry, as it involves calculating the heat of reactions—vital for knowing whether a reaction will release heat to its surroundings (exothermic) or absorb heat (endothermic). Understanding the enthalpy changes during reactions and how temperature, pressure, and composition affect such changes is central to mastering thermochemistry.

Phase-Change Reaction

A phase-change reaction involves a substance transitioning from one phase of matter (solid, liquid, or gas) to another. These reactions are accompanied by energy changes, known as latent heat. For instance, the conversion of water from liquid to vapor requires energy absorption, while the reverse process releases energy.

In the exercise, the phase change from liquid water (H2O(l)) to water vapor (H2O(g)) is determined using the given enthalpy values. The enthalpy change (\( \Delta H^\circ \) value) sign is negative indicating the process is endothermic, meaning that energy is absorbed as heat. This particular case illustrates how phase-change reactions are an integral part of thermochemistry and highlight the energy requirements or releases during these transformations.