Question

The \(\Delta \mathrm{H}_{\mathrm{f}^{\circ}}\) for \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), are \(\begin{array}{llll}-393.5,-110.5 & \text { and } & -241.8 & \mathrm{KJ} & \mathrm{mol}^{-1}\end{array}\) respectively. The standard enthalpy change (in \(\mathrm{KJ}\) ) for the reaction \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is (1) \(524.1\) (2) \(41.2\) (3) \(-262.5\) (4) \(-41.2\)

Short answer

The standard enthalpy change is 41.2 KJ, so the correct answer is (2).

Step-by-step solution

- Write down the given formation enthalpies

The given formation enthalpies are: - \(\text{\Delta H}_{\mathrm{f}^{\circ}}\) for \(\text{CO}_{2}(\text{g})\) is \(-393.5 \text{ KJ mol}^{-1}\)- \(\text{\Delta H}_{\mathrm{f}^{\circ}}\) for \(\text{CO}(\text{g})\) is \(-110.5 \text{ KJ mol}^{-1}\) - \(\text{\Delta H}_{\mathrm{f}^{\circ}}\) for \(\text{H}_{2} \text{O}(\text{g})\) is \(-241.8 \text{ KJ mol}^{-1}\)

- Write down the reaction and identify the reactants and products

The reaction is: \(\text{CO}_{2}(\text{g}) + \text{H}_{2} (\text{g}) \rightarrow \text{CO}(\text{g}) + \text{H}_{2} \text{O}(\text{g})\). The reactants are \(\text{CO}_{2}(\text{g})\) and \(\text{H}_{2} (\text{g})\). The products are \(\text{CO}(\text{g})\) and \(\text{H}_{2} \text{O}(\text{g})\).

- Determine the sum of the formation enthalpies for reactants

For the reactants, \(\text{CO}_{2}(\text{g})\) and \(\text{H}_{2} (\text{g})\), add the respective formation enthalpies: \([\text{\Delta H}_{f}^{\circ} (\text{CO}_{2}(\text{g}) )]\)Since \(\text{H}_{2} (\text{g})\) is in its standard state, its formation enthalpy is zero. Therefore, the sum for the reactants is: \(-393.5 \text{ KJ mol}^{-1}\)

- Determine the sum of the formation enthalpies for products

For the products, \(\text{CO}(\text{g})\) and \(\text{H}_{2} \text{O}(\text{g})\), add the respective formation enthalpies: \(\text{\Delta H}_{f}^{\circ} (\text{CO}(\text{g}) ) + \text{\Delta H}_{f}^{\circ} (\text{H}_{2} \text{O}(\text{g}) ) \) is \(-110.5 \text{ KJ mol}^{-1} + (-241.8 \text{ KJ mol}^{-1}) = -352.3 \text{ KJ}\)

- Calculate the enthalpy change of the reaction

Use the equation for the enthalpy change of the reaction: \(\text{\Delta H}_{\text{reaction}} = [ \text{Sum of formation enthalpies of products} ] - [ \text{Sum of formation enthalpies of reactants} ]\). Substitute the values obtained: \(\text{\Delta H}_{\text{reaction}} = -352.3 - (-393.5) = 41.2 \text{ KJ}\)

Key concepts

standard enthalpies of formation

Standard enthalpies of formation, denoted as \( \text{ΔH}_{\text{f}}^{\text{º}} \), are the changes in enthalpy when one mole of a compound is formed from its elements in their standard states. These values are crucial in calculating the enthalpy changes in chemical reactions by using Hess's Law.

Enthalpy of formation values are typically given in units of kilojoules per mole (KJ/mol). For example, the standard enthalpies of formation for \( \text{CO}_{2}(\text{g}) \), \( \text{CO}(\text{g}) \), and \( \text{H}_{2}\text{O}(\text{g}) \) are \( -393.5, -110.5, \text{ and } -241.8 \text{ KJ mol}^{-1} \) respectively.

It is important to remember that the standard enthalpy of formation for any element in its standard state is zero. For example, \( \text{H}_{2} (\text{g}) \) has an enthalpy of formation of zero because it is in its elemental form.

These standard enthalpies of formation allow us to calculate the total change in enthalpy for a reaction by summing the enthalpies of the products and subtracting the enthalpies of the reactants.

thermochemistry

Thermochemistry is the branch of chemistry that studies the energy changes that occur during chemical reactions and changes in state. It focuses on the heat exchange between the system and its surroundings.

In thermochemistry, we often deal with the concept of enthalpy (H), which is the total heat content of a system at constant pressure. The change in enthalpy (ΔH) during a reaction indicates whether the process is exothermic (releases heat) or endothermic (absorbs heat).

To measure these changes, chemists use calorimetry, an experimental technique where the heat evolved or absorbed in a reaction is measured. The units commonly used are joules or kilojoules.

For chemical reactions, the relationship between the enthalpies of reactants and products is given by:
\[ \text{ΔH}_{\text{reaction}} = \text{ΣΔH}_{\text{f}} (\text{products}) - \text{ΣΔH}_{\text{f}} (\text{reactants}) \]

This equation represents the fundamental principle in calculating reaction enthalpies using standard enthalpies of formation.

Hess's law

Hess's Law is a central principle in thermochemistry. It states that the total enthalpy change for a reaction is the same, regardless of the pathway taken, provided the initial and final conditions are the same. This means that we can calculate the enthalpy change of a reaction by using the sum of known enthalpy changes for a series of steps that lead to the overall reaction.

To apply Hess's Law, follow these steps:

• Identify the known enthalpies of formation for all reactants and products involved in the reaction.
• Write down the balanced chemical equation for the reaction.
• Sum the standard enthalpies of formation for the products and subtract the sum of the standard enthalpies of formation for the reactants.

In the given exercise, we used Hess's Law to find the enthalpy change for the reaction: \[ \text{CO}_{2}(\text{g}) + \text{H}_{2}(\text{g}) \rightarrow \text{CO}(\text{g}) + \text{H}_{2}\text{O}(\text{g}) \]

By summing the enthalpies of formation for the products (\( -110.5 + (-241.8) = -352.3 \text{ KJ} \)) and subtracting the enthalpies of formation for the reactants (\( -393.5 \text{ KJ} \)), we calculated the enthalpy change as \( \text{ΔH}_{\text{reaction}} = -352.3 - (-393.5) = 41.2 \text{ KJ} \).

Hess's Law simplifies calculating reaction enthalpies, especially for complex reactions that are difficult to measure directly.