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 \(-393.5\), \(-110.5\) and \(-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy change (in \(\mathrm{kJ}\) ) for the reaction: \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \mathrm{is}\) (a) \(524.1\) (b) \(41.2\) (c) \(-262.5\) (d) \(-41.2\)

Short answer

The standard enthalpy change is \(41.2 \, \mathrm{kJ}\).

Step-by-step solution

Write the Reaction Formula

The given reaction is \( \mathrm{CO}_{2}(\mathrm{~g}) + \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \). We need to calculate the standard enthalpy change for this reaction.

Recall the Formula for Enthalpy Change

The standard enthalpy change for a reaction \( \Delta H_{\text{reaction}}^{\circ} \) is calculated using the formula: \[ \Delta H_{\text{reaction}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants}) \] where \( \Delta H_{f}^{\circ} \) is the standard enthalpy of formation.

Input Values for Products and Reactants

For the products, \( \Delta H_{f}^{\circ}(\mathrm{CO}(\mathrm{g})) = -110.5 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \) and \( \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(\mathrm{g})) = -241.8 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \). For the reactants, \( \Delta H_{f}^{\circ}(\mathrm{CO}_{2}(\mathrm{~g})) = -393.5 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \) and \( \Delta H_{f}^{\circ}(\mathrm{H}_{2}(\mathrm{~g})) = 0 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \) since it is an element in its standard state.

Calculate the Enthalpy Change for the Reaction

Substitute the values into the formula: \[ \Delta H_{\text{reaction}}^{\circ} = [(-110.5) + (-241.8)] - [(-393.5) + 0] \] This simplifies to: \[ \Delta H_{\text{reaction}}^{\circ} = (-352.3) - (-393.5) = 41.2 \, \mathrm{kJ} \]

Key concepts

Enthalpy Change

Enthalpy change, represented by \( \Delta H \), is a crucial concept in thermodynamics. It refers to the heat content variation between reactants and products during a chemical reaction under constant pressure conditions. Understanding enthalpy change helps to determine if a reaction is exothermic (releases heat) or endothermic (absorbs heat). In essence, \( \Delta H \) is calculated using the formula:

• \( \Delta H_{\text{reaction}}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants}) \)

This method involves subtracting the total enthalpy of the reactants from the total enthalpy of the products. If the result is negative, the reaction releases heat, and if positive, it absorbs heat.
In our exercise, we found \( \Delta H_{\text{reaction}}^{\circ} = 41.2 \, \mathrm{kJ} \), indicating the process absorbs heat.

Enthalpy of Formation

The enthalpy of formation, denoted by \( \Delta H_{f}^{\circ} \), is a specific type of enthalpy change. It refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This value helps assess the energy efficiency of producing compounds. Values for the enthalpies of formation are usually given in \( \mathrm{kJ} \, \mathrm{mol}^{-1} \).

• For our chemical reaction calculation:
  • \( \Delta H_{f}^{\circ}(\mathrm{CO}_{2}(\mathrm{g})) = -393.5 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)
  • \( \Delta H_{f}^{\circ}(\mathrm{CO}(\mathrm{g})) = -110.5 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)
  • \( \Delta H_{f}^{\circ}(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})) = -241.8 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)
  • \( \Delta H_{f}^{\circ}(\mathrm{H}_{2}(\mathrm{g})) = 0 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)

These described values are critical for calculating the standard enthalpy change of reactions, as they provide the necessary basis for determining the energy absorbed or released.

Chemical Reaction

Chemical reactions are the basis of chemistry, involving the transformation of reactants into products. This transformation is often coupled with an energy exchange, which can be measured and analyzed through thermodynamic principles.
The specific reaction in our example is:

• \( \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \)

This reaction highlights the conversion of carbon dioxide and hydrogen gas into carbon monoxide and water vapor. Expressing this process helps us understand which bonds are broken and formed, providing insights into the enthalpy changes involved.
Understanding the dynamics of chemical reactions, alongside enthalpy changes, allows scientists and students to predict how energy-efficient and feasible reactions are.