Question

For the thermochemical equation \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\ell) \rightarrow 2 \mathrm{HBr}(\mathrm{g}) \Delta H=-72.6 \mathrm{~kJ}\) what mass of HBr will be formed when \(553 \mathrm{~kJ}\) of energy are given off?

Short answer

308.61 g of HBr will be formed.

Step-by-step solution

Understand the Problem

You need to find the mass of \( \text{HBr} \) formed when 553 kJ of energy is released. Given the reaction \( \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\ell) \rightarrow 2 \mathrm{HBr}(\mathrm{g}) \) and \( \Delta H = -72.6 \text{ kJ} \), this means that 72.6 kJ of energy is released per the moles of reactants used in the balanced equation.

Determine Moles of HBr Formed

Use the given \( \Delta H \) to calculate the moles of \( \text{HBr} \) formed with 553 kJ released. The balanced reaction releases 72.6 kJ per 2 moles of \( \text{HBr} \). Calculate the moles using the ratio: \[ \frac{553 \text{ kJ}}{72.6 \text{ kJ per 2 moles}} = 2 \times X \] Hence, \[ X = \frac{553}{72.6 \times 2} \approx 3.81 \text{ moles} \].

Calculate Mass of HBr

To find the mass of \( \text{HBr} \), use the molar mass of \( \text{HBr} \). The molar mass is approximately 81 g/mol (Hydrogen = 1 g/mol, Bromine = 80 g/mol).Multiply the moles of \( \text{HBr} \) by its molar mass:\[ 3.81 \text{ moles} \times 81 \text{ g/mol} \approx 308.61 \text{ g} \].

Key concepts

Enthalpy Change

The concept of enthalpy change is fundamental to understanding thermochemical equations. Enthalpy, often denoted by \( \Delta H \), represents the heat change at constant pressure. For the reaction \( \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\ell) \rightarrow 2 \mathrm{HBr}(\mathrm{g}) \), the enthalpy change is noted as \( -72.6 \) kJ. This means that 72.6 kJ of energy is released as heat when the reaction occurs for the stoichiometric amounts given in the equation.

• A negative \( \Delta H \) value indicates an exothermic reaction, meaning it releases heat into the surroundings.
• The magnitude of \( \Delta H \) is crucial because it dictates the amount of energy change per mole or set of moles for the substances involved in the reaction.

To solve problems like this, relate the given energy change to the stoichiometry of the chemical reaction. Knowing how much energy corresponds to a specific change in number of moles can help determine quantities of reactants or products.

Molar Mass Calculation

Calculating the molar mass of a compound is essential for converting moles of substance into grams. The molar mass is calculated by summing the atomic masses of all atoms in a molecule.

• For Hydrogen Bromide (HBr), calculate it as follows: The atomic mass of hydrogen (H) is approximately 1 g/mol and that of bromine (Br) is about 80 g/mol.
• Add these to get the molar mass of HBr, which is roughly 81 g/mol.

Knowing how to accurately calculate molar masses allows you to transform mole quantities into tangible, measurable masses, facilitating practical laboratory work. This is a core skill needed when dealing with chemical reactions and their accompanying calculations, such as determining the amount of product created in a reaction.

Mole-Mass Conversion

The practical use of chemistry often involves converting between moles and mass, a skill that enables chemists to measure and create desired amounts of substances. In this problem, once you've determined the moles of HBr formed, the next step is to calculate the mass.

• Using the calculated moles from the enthalpy change calculation: \( X \approx 3.81 \) moles of HBr.
• Multiply the moles by the molar mass of HBr (81 g/mol) to get the mass: \[ 3.81 \text{ moles} \times 81 \text{ g/mol} = 308.61 \text{ g} \].

This process exemplifies the conversion between the theoretical, easily calculated moles from a reaction stoichiometry, to the real-world measurable quantity of grams. Understanding this conversion is vital not only in scientific studies but also in practical applications like pharmaceuticals, where dosing requires precise mass measurements.