Question

You are given the following data: \(\begin{aligned} \mathrm{H}_{2}(g) & \longrightarrow 2 \mathrm{H}(g) & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{Br}(g) & & \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{HBr}(g) & \Delta H^{\circ} &=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Calculate \(\Delta H^{\circ}\) for the reaction\(\mathrm{H}(g)+\mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g)\).

Short answer

The enthalpy change for the reaction is \( 278.25 \mathrm{~kJ/mol} \).

Step-by-step solution

Write the Given Reactions with Enthalpy Changes

We have three given reactions and their respective enthalpy changes: 1. \( \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \) with \( \Delta H^{\circ} = 436.4 \mathrm{~kJ/mol} \).2. \( \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \) with \( \Delta H^{\circ} = 192.5 \mathrm{~kJ/mol} \).3. \( \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \) with \( \Delta H^{\circ} = -72.4 \mathrm{~kJ/mol} \).

Use Hess's Law to Combine Reactions

According to Hess's Law, we can add or subtract thermochemical equations to find \( \Delta H^{\circ} \) for a reaction. We want the reaction \( \mathrm{H}(g)+\mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g) \). Start by writing both the desired reaction and breaking down given reactions to align with it.

Break Down and Rearrange Given Reactions

We want the formation of HBr from H and Br, i.e., \( \mathrm{H}(g) + \mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g) \), not from \( \mathrm{H}_{2} \) and \( \mathrm{Br}_{2} \), so we need: - From reaction 1: HALF of \( \mathrm{H}_{2} \longrightarrow \mathrm{H}(g) \) i.e., \( \Delta H^{\circ} = \frac{436.4}{2} = 218.2 \mathrm{~kJ/mol} \).- From reaction 2: HALF of \( \mathrm{Br}_{2} \longrightarrow \mathrm{Br}(g) \) i.e., \( \Delta H^{\circ} = \frac{192.5}{2} = 96.25 \mathrm{~kJ/mol} \).- Reverse reaction 3 for half of 2\( \mathrm{HBr}(g) \to \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \), resulting in \( \Delta H^{\circ} = \frac{72.4}{2} = 36.2 \mathrm{~kJ/mol} \).

Calculate Net Enthalpy Change

Add the adjusted enthalpy changes:\[\Delta H^{\circ} = 218.2 + 96.25 + (-36.2)\]Which results in a total of \( \Delta H^{\circ} = 278.25 \mathrm{~kJ/mol} \).

Conclude the Calculation

The enthalpy change for the reaction \( \mathrm{H}(g) + \mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g) \) is \( 278.25 \mathrm{~kJ/mol} \).

Key concepts

Enthalpy Change

Enthalpy change, often represented as \( \Delta H \), is a crucial concept in thermodynamics. It refers to the heat absorbed or evolved during a chemical reaction at constant pressure. When bonds are broken or formed in a reaction, energy changes occur, influencing enthalpy. The sign of \( \Delta H \) indicates the nature of the process:

• A positive \( \Delta H \) shows an endothermic reaction, where energy is absorbed.
• A negative \( \Delta H \) points to an exothermic reaction, where energy is released.

In our example, different reactions had their specific \( \Delta H^{\circ} \) values, which needed to be combined using Hess’s Law to find the total enthalpy change for the target reaction.

Thermochemical Equations

Thermochemical equations are chemical equations that also display the enthalpy change associated with the reaction. They provide the energy dynamics along with the stoichiometric information. These equations include both the reactants and products, along with a \( \Delta H \) value to indicate the enthalpy change. For instance, the equation: \[ \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ/mol} \] illustrates that 436.4 kJ/mol of energy is required to dissociate one mole of \( \mathrm{H}_{2} \) into hydrogen atoms. Using these kinds of equations allows us to perform calculations and combine reactions via Hess’s Law to arrive at the enthalpy for complex reactions.

Reaction Enthalpy

Reaction enthalpy is the change in enthalpy when a reaction occurs between specified amounts of reactants under standard conditions. Standard conditions usually mean a pressure of 1 atm and a temperature of 298 K. When using a reaction enthalpy value from a standard reference condition, like \( \Delta H^{\circ} \), it ensures that calculated results are consistent for comparison across different reactions. In our original problem, Hess's Law helped determine the enthalpy change for deriving \( \mathrm{HBr} \) from \( \mathrm{H} \) and \( \mathrm{Br} \). This demonstrates how reaction enthalpy can be derived for new reactions by cleverly using known values for similar reactions. Such calculations are vital in predicting the energy efficiency and feasibility of chemical processes in industrial applications.