Question

When \(1 \mathrm{~mol}\) of \(\mathrm{KBr}(s)\) decomposes to its elements, \(394 \mathrm{~kJ}\) of heat is absorbed. (a) Write a balanced thermochemical equation. (b) What is \(\Delta H\) when \(10.0 \mathrm{~kg}\) of \(\mathrm{KBr}\) forms from its elements?

Short answer

\(\Delta H = 33,104 \text{ kJ}\) when 10.0 kg of KBr forms from its elements.

Step-by-step solution

Write the decomposition reaction

The decomposition of potassium bromide (\text{KBr}) to its elemental forms (\text{K} and \text{Br}_2) can be written as: \[ 2 \text{KBr}(s) \rightarrow 2 \text{K}(s) + \text{Br}_2(g) \]

Include the heat absorbed in the equation

Given that 394 kJ of heat is absorbed when 1 mol of \text{KBr} decomposes, the thermochemical equation for 2 moles of \text{KBr} would be: \[ 2 \text{KBr}(s) + 788 \text{ kJ} \rightarrow 2 \text{K}(s) + \text{Br}_2(g) \]

Convert 10.0 kg of KBr to moles

First, find the molar mass of \text{KBr}. \[ \text{Molar mass of K} = 39.1 \text{ g/mol} \] \[ \text{Molar mass of Br} = 79.9 \text{ g/mol} \] \[ \text{Molar mass of KBr} = 39.1 + 79.9 = 119.0 \text{ g/mol} \] Convert 10.0 kg to grams: \[ 10.0 \text{ kg} = 10,000 \text{ g} \] Calculate moles: \[ \text{Moles of KBr} = \frac{10,000 \text{ g}}{119.0 \text{ g/mol}} = 84.03 \text{ mol} \]

Calculate \Delta H for the reaction

Given that 394 kJ is absorbed for 1 mol of \text{KBr}, for 84.03 mol of \text{KBr}, \[ \Delta H = 84.03 \text{ mol} \times 394 \text{ kJ/mol} = 33,104 \text{ kJ} \]

Key concepts

enthalpy change

Enthalpy change, denoted as \(\text{ΔH}\), is the heat absorbed or released during a chemical reaction at constant pressure. In endothermic reactions, heat is absorbed, resulting in a positive \(\text{ΔH}\). In exothermic reactions, heat is released, leading to a negative \(\text{ΔH}\). Understanding \(\text{ΔH}\) helps us determine the energy changes accompanying reactions.

mole conversions

Mole conversions are crucial to translating between the mass of a substance and the number of moles. This is done using the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]. It's essential for stoichiometric calculations in chemical reactions, helping to predict yields and reagent quantities.

decomposition reactions

Decomposition reactions occur when one compound breaks down into two or more simpler substances. They typically require energy input (heat, light, electricity) to break bonds. An example is the decomposition of potassium bromide (\(\text{KBr}\)), breaking into potassium (\(\text{K}\)) and bromine (\(\text{Br}_2\)). These reactions are pivotal in fields like industrial chemistry for material production.

molar mass calculations

Calculating molar mass is foundational in chemistry, providing the mass of one mole of a given substance. It's the sum of the atomic masses of all atoms in a molecule. For potassium bromide, the molar mass is calculated as: \[ \text{Molar mass of} \text{KBr} = \text{39.1 g/mol (K)} + \text{79.9 g/mol (Br)} = 119.0 \text{ g/mol} \]. Accurate molar mass calculations are essential for correct stoichiometric computations in reactions.