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Chapter 6: Problem 60

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

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

Step by step solution

01

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: \[ 2 \text{KBr}(s) \rightarrow 2 \text{K}(s) + \text{Br}_2(g) \]
02

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: \[ 2 \text{KBr}(s) + 788 \text{ kJ} \rightarrow 2 \text{K}(s) + \text{Br}_2(g) \]
03

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: \[ \text{Moles of KBr} = \frac{10,000 \text{ g}}{119.0 \text{ g/mol}} = 84.03 \text{ mol} \]
04

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}, \[ \Delta H = 84.03 \text{ mol} \times 394 \text{ kJ/mol} = 33,104 \text{ kJ} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

If you feel warm after exercising, have you increased the internal energy of your body? Explain.

When \(25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added to \(25.0 \mathrm{~mL}\) of 1.00 \(M\) KOH in a coffee-cup calorimeter at \(23.50^{\circ} \mathrm{C}\), the temperature rises to \(30.17^{\circ} \mathrm{C}\). Calculate \(\Delta H\) in \(\mathrm{kJ}\) per mole of \(\mathrm{H}_{2} \mathrm{O}\) formed. (Assume that the total volume is the sum of the volumes and that the density and specific heat capacity of the solution are the same as for water.)

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is \(\mathrm{C}_{12} \mathrm{H}_{26}\) (a) Write a balanced equation, using the simplest whole-number coefficients, for the complete combustion of kerosene to gases. (b) If \(\Delta H_{\mathrm{rn}}^{\circ}=-1.50 \times 10^{4} \mathrm{~kJ}\) for the combustion equation as written in part (a), determine \(\Delta H_{\mathrm{f}}^{\circ}\) of kerosene. (c) Calculate the heat released by combustion of 0.50 gal of kerosene \((d\) of kerosene \(=0.749 \mathrm{~g} / \mathrm{mL})\) (d) How many gallons of kerosene must be burned for a kerosene furnace to produce \(1250 .\) Btu \((1 \mathrm{Btu}=1.055 \mathrm{~kJ}) ?\)

6.54 Does a negative \(\Delta H\) mean that the heat should be treated as a reactant or as a product?

Make any changes needed in each of the following equations so that the enthalpy change is equal to \(\Delta H_{i}^{\circ}\) for the compound: (a) \(\mathrm{Cl}(g)+\mathrm{Na}(s) \longrightarrow \mathrm{NaCl}(s)\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{H}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\)
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