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The combustion of \(2.650 \mathrm{~g}\) of indium raised the temperature of a calorimeter \(1.055 \mathrm{C}\). The energy equivalent of the calorimeter was \(2412.81 \mathrm{cal} /{ }^{\circ} \mathrm{C}\). Calculate the molar enthalpy combustion of indium (Molar mass of indium \(114.8\) ) (a) \(2412.81 \mathrm{cal} / \mathrm{mole}\) (b) \(119683.8 \mathrm{cal} / \mathrm{mole}\) (c) \(110,273.6 \mathrm{cal} / \mathrm{mole}\) (d) None

Short Answer

Expert verified
The answer is (c) 110,273.6 cal/mol.

Step by step solution

01

Understand the Problem

We need to calculate the molar enthalpy of combustion for indium using the given data: mass of indium burned, temperature change in the calorimeter, and the energy equivalent of the calorimeter. Then, we need to find which of the provided options closest matches the calculated value.
02

Calculate Heat Released (q)

The heat released by the combustion can be calculated using the formula: \( q = c \cdot \Delta T \), where \( c \) is the heat capacity of the calorimeter and \( \Delta T \) is the temperature change. Substituting the values, we get: \[ q = 2412.81 \mathrm{~cal/^{ extdegree} C} \times 1.055 \mathrm{~C} = 2545.51555 \mathrm{~cal} \].
03

Calculate Moles of Indium

Using the formula \( \, n = \frac{m}{M_{\mathrm{In}}} \,\), where \( m \) is the mass of the indium and \( M_{\mathrm{In}} \) is the molar mass of indium. Substitute the given values: \( \, n = \frac{2.650 \mathrm{~g}}{114.8 \mathrm{~g/mol}} \, = 0.023080384 \text{ mol} \).
04

Calculate Molar Enthalpy of Combustion

Molar enthalpy of combustion (\( \Delta H_{comb} \)) is calculated as \( \Delta H_{comb} = \frac{q}{n} \). Using the values from the previous steps, we have: \[ \Delta H_{comb} = \frac{2545.51555 \mathrm{~cal}}{0.023080384 \mathrm{~mol}} = 110273.6 \mathrm{~cal/mol} \].
05

Match with Given Options

The calculated value of molar enthalpy of combustion is \( 110,273.6 \mathrm{cal/mol} \), which corresponds to option (c).

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Key Concepts

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

Calorimetry
Calorimetry is a fundamental concept in thermochemistry that measures the amount of heat involved in a chemical reaction or physical change. In the context of this exercise, we use a calorimeter, which is a device capable of measuring the heat of reaction during the combustion of substances. The calorimeter absorbs the heat released during the indium combustion, which raises its temperature.
Understanding how a calorimeter works is essential as it allows us to determine the amount of heat absorbed by the calorimeter. This is calculated using the formula:
  • \( q = c \times \Delta T \), where \( q \) is the heat absorbed, \( c \) is the heat capacity, and \( \Delta T \) is the change in temperature.
The given energy equivalent of the calorimeter is expressed in calories per degree Celsius, which helps in calculating how much heat was released by the indium combustion in our exercise.
Molar Mass
Molar mass is an important concept in chemistry, representing the mass of one mole of a substance, expressed in grams per mole (g/mol). It acts as a conversion factor between the mass of a sample and the amount of substance (in moles).
In our exercise, the molar mass of indium is given as 114.8 g/mol. This value is crucial for calculating the number of moles of indium used in the reaction through the formula:
  • \( n = \frac{m}{M_{\text{In}}} \), where \( n \) is the number of moles, \( m \) is the mass of indium, and \( M_{\text{In}} \) is the molar mass of indium.
Knowing this, we can convert the mass of indium given in the exercise to moles, which is necessary for further calculations of molar enthalpy.
Thermochemistry
Thermochemistry is the branch of chemistry that deals with the heat involved in chemical reactions. It provides insights into energy changes in the form of heat and is vital for understanding reaction energetics.
The key concept in thermochemistry related to our exercise is the molar enthalpy of combustion, which represents the heat released when one mole of a substance is entirely burnt with oxygen under standard conditions. In this exercise, we focus on calculating the molar enthalpy of combustion for indium.
We use the formula:
  • \( \Delta H_{\text{comb}} = \frac{q}{n} \), where \( \Delta H_{\text{comb}} \) is the molar enthalpy of combustion, \( q \) is the heat released, and \( n \) is the number of moles of indium involved.
This calculation helps students understand how much energy, in terms of calories per mole, is involved in the combustion of indium.
Indium Combustion Problem
The indium combustion problem demonstrates the practical application of calorimetry and thermochemistry. It illustrates the steps involved in determining the enthalpy change for a metal combustion reaction. By burning a known mass of indium in a calorimeter and measuring the resultant temperature increase, the problem lets students calculate the molar enthalpy of combustion.
This is done by first measuring the total heat released and then dividing by the amount of indium, in moles, used in the reaction. The problem highlights the importance of understanding each interconnected concept, from calculating heat using a calorimeter to applying the formula for molar enthalpy. This integrated approach reinforces comprehension of theoretical concepts through hands-on measurement and calculation. Learning this helps students see the linkage between theoretical and practical applications in chemistry.

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Most popular questions from this chapter

The enthalpy change involved in the oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Twenty five percent of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one kilometer, what is the maximum distance that a person will be able to walk eating \(120 \mathrm{~g}\) of glucose? (a) \(7.9 \mathrm{~km}\) (b) \(9.7 \mathrm{~km}\) (c) \(4.8 \mathrm{~km}\) (d) \(8.4 \mathrm{~km}\)

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Anhydrous \(\mathrm{AlCl}_{3}\) is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution (ionisation energy of \(\mathrm{Al}=5137 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{H}_{\text {indration }}\) for \(\mathrm{Al}^{+3}=-4665 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}, \mathrm{AH}_{\text {hydation }}\) for \(\left.\mathrm{C}^{-}=-381 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) Ionic (b) Covalent (c) Both (d) None

When a certain amount of ethylene was combusted \(6226 \mathrm{~kJ}\) heat was evolved. If heat of combustion of ethylene is \(1411 \mathrm{~kJ}\). The volume of \(\mathrm{O}_{2}\) (at NTP) that entered into the reaction is (a) \(296.5 \mathrm{ml}\) (b) \(296.5 \mathrm{~L}\) (c) \(6226 \times 22.4 \mathrm{~L}\) (d) \(22.4 \mathrm{~L}\).

Carnot's cycle operates between \(\mathrm{T}\) and \(300 \mathrm{~K}\) (T > 300K). It is \(25 \%\) efficient when temperature of die source (T) is (a) \(350 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(375 \mathrm{~K}\)

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