Problem 16 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The \(\Delta_{f} H^{\circ}\) 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: \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is (a) \(524.1\) (b) \(41.2\) (c) \(-262.5\) (d) \(-41.2\)

Step-by-Step Solution

Answer

The standard enthalpy change for the reaction is \(41.2 \textrm{ kJ/mol}\).

Step 1: Determine the Standard Enthalpies of Formation

Identify the standard enthalpies of formation (\(\triangle_{f}H^{\theta}\) for the substances involved in the reaction. For \(\mathrm{CO}_{2}(\mathrm{g}),\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2}O(\mathrm{g})\) these are \(\-393.5 \textrm{kJ mol}^{-1},\ -110.5 \ \textrm{kJ mol}^{-1}\) and \(\-241.8 \ \textrm{kJ mol}^{-1}\) respectively.

Step 2: Write the Balanced Chemical Equation

Write down the balanced chemical equation for the reaction: \(\mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g}) + \mathrm{H}_{2}O(\mathrm{g})\).

Step 3: Apply Hess's Law

Use Hess's Law to calculate the standard enthalpy change for the reaction (\(\triangle_{r}H^{\theta}\)). \(\triangle_{r}H^{\theta} = \sum \triangle_{f}H^{\theta} \text{(products)} - \sum \triangle_{f}H^{\theta} \text{(reactants)}\).

Step 4: Calculate the Standard Enthalpy Change for the Products

Add the standard enthalpies of formation of the products: \(\triangle_{f}H^{\theta} (\mathrm{CO}) + \triangle_{f}H^{\theta} (\mathrm{H}_{2}O) = \-110.5 \textrm{ kJ/mol} + \-241.8 \textrm{ kJ/mol}\).

Step 5: Calculate the Standard Enthalpy Change for the Reactants

Add the standard enthalpies of formation of the reactants: \(\triangle_{f}H^{\theta} (\mathrm{CO}_{2}) + \triangle_{f}H^{\theta} (\mathrm{H}_{2}) = \-393.5 \textrm{ kJ/mol} + 0 \textrm{ kJ/mol}\) since the standard enthalpy of formation for any element in its most stable form (such as \(\mathrm{H}_{2}\)) is zero.

Step 6: Subtract the Sum of Reactants' Enthalpies from the Sum of the Products' Enthalpies

Subtract the sum of the reactants' standard enthalpies of formation from the sum of the products' standard enthalpies of formation to get the standard enthalpy change of the reaction: \( (\-110.5 \textrm{ kJ/mol} + \-241.8 \textrm{ kJ/mol}) - (\-393.5 \textrm{ kJ/mol} + 0 \textrm{ kJ/mol}) = -352.3 \textrm{ kJ/mol} + 393.5 \textrm{ kJ/mol} = +41.2 \textrm{ kJ/mol}\).

Key Concepts

Enthalpy of Formation

The concept of enthalpy of formation, denoted as \( \Delta_{f}H^{\circ} \), refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states under standard conditions (1 atm pressure and 298.15 K). In essence, it's the heat change that occurs during the creation of a substance from the pure elements. For example, in the equation provided, substances like \( \mathrm{CO}_{2}(g) \) have specific enthalpy values associated with their formation.

Understanding the enthalpy of formation is crucial because it allows us to measure the amount of energy required or released during the synthesis of compounds. This value is fundamental in thermochemical calculations where we predict the energy changes associated with chemical reactions.

Hess's Law

A pivotal principle in thermochemistry is \( \text{Hess's Law} \). It states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This is because enthalpy is a state function, which means its value depends only on the current state of the system, not the path or steps taken to reach that state.

By employing Hess's Law, we can add or subtract known enthalpies of formation to find the enthalpy change of a reaction without directly measuring it. This is particularly useful when direct measurement is difficult or impossible. As we saw in this exercise, Hess's Law simplifies the calculation of the standard enthalpy change of the reaction from the standard enthalpies of formation of reactants and products.

Chemical Thermodynamics

Chemical thermodynamics is the study of energy and work in chemical reactions. It encompasses concepts such as enthalpy, entropy, and Gibbs free energy, and it allows scientists to understand the energetics and feasibility of reactions. At the heart of chemical thermodynamics is the first law of thermodynamics, which asserts the conservation of energy, indicating that the total energy of an isolated system remains constant.

This branch of thermodynamics provides critical insight into how reactions occur and what may influence them—factors like temperature, pressure, and concentration play significant roles. The principles of thermodynamics enable us to predict whether a change in conditions will make a reaction more favorable or less so, which has vast implications in both natural and industrial processes.

Balanced Chemical Equation

A balanced chemical equation is the foundation of any stoichiometric calculation in chemistry. It ensures the law of conservation of mass is obeyed in a chemical reaction by having the same number of atoms for each element on both the reactants and products side.

Equations must be balanced to accurately represent the reactants that are consumed and the products that are formed. Moreover, it gives us mole to mole relationships which are necessary for calculating the amount of reactants needed or products formed. In the exercise example, \( \mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightarrow \mathrm{CO}(g) + \mathrm{H}_{2}O(g) \) is a balanced equation. This allows us to apply Hess's Law and the enthalpy changes of formation to determine the overall energy change of the reaction.