/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Due to the environmental concern... [FREE SOLUTION] | 91Ó°ÊÓ

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Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The heat required is 13.399 kJ.

Step by step solution

01

Calculate Molar Mass of Isobutane

Isobutane's chemical formula is \( C_4H_{10} \). Its molar mass can be calculated by adding the atomic masses of its constituent atoms: - Carbon (C) has an atomic mass of 12.01 g/mol. - Hydrogen (H) has an atomic mass of 1.008 g/mol. The molar mass is thus \( (4 \times 12.01) + (10 \times 1.008) = 58.14 \text{ g/mol} \).
02

Convert Mass to Moles

To find the number of moles of isobutane, use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] Substitute the values: \[ \text{moles} = \frac{25.0 \text{ g}}{58.14 \text{ g/mol}} \approx 0.430 \text{ moles} \]
03

Calculate Heat for Temperature Increase in Liquid Phase

Use specific heat capacity to calculate the heat needed to raise the temperature from \(-50^{\circ}C\) (liquid) to \(-12^{\circ}C\) (boiling point). The formula is: \[ q_1 = m \cdot C_{\text{liquid}} \cdot \Delta T \] Where: - \( m = 0.430 \text{ moles} \), - \( C_{\text{liquid}} = 129.7 \text{ J/mol-K} \), - \( \Delta T = (-12 - (-50)) \text{C} = 38 \text{K} \). Calculate \( q_1 \): \[ q_1 = 0.430 \cdot 129.7 \cdot 38 = 2115.03 \text{ J} \approx 2.115 \text{ kJ} \]
04

Calculate Heat for Vaporization

Calculate the heat required to vaporize the liquid at the boiling point (phase change). Use the formula: \[ q_2 = n \cdot \Delta H_{\text{vap}} \] Where: - \( n = 0.430 \text{ moles} \), - \( \Delta H_{\text{vap}} = 21.3 \text{ kJ/mol} \). Calculate \( q_2 \): \[ q_2 = 0.430 \cdot 21.3 = 9.159 \text{ kJ} \]
05

Calculate Heat for Temperature Increase in Gas Phase

Use specific heat capacity to calculate the heat needed to raise the temperature from \(-12^{\circ}C\) to \(40^{\circ}C\) in the gas phase. The formula is: \[ q_3 = m \cdot C_{\text{gas}} \cdot \Delta T \] Where: - \( m = 0.430 \text{ moles} \), - \( C_{\text{gas}} = 95.2 \text{ J/mol-K} \), - \( \Delta T = (40 - (-12)) \text{C} = 52 \text{K} \). Calculate \( q_3 \): \[ q_3 = 0.430 \cdot 95.2 \cdot 52 = 2125.456 \text{ J} \approx 2.125 \text{ kJ} \]
06

Calculate Total Heat Required

Add up all the individual heats to find the total heat required: \[ q_{\text{total}} = q_1 + q_2 + q_3 \] Substitute the values: \[ q_{\text{total}} = 2.115 + 9.159 + 2.125 = 13.399 \text{ kJ} \]

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity is a fundamental property of matter. It tells us the amount of heat required to raise the temperature of a given quantity of substance by one degree Celsius (or one Kelvin). Every material has its unique specific heat capacity, so this property plays a critical role in understanding how different substances absorb and lose heat.
For isobutane, the problem specifies the specific heat capacities for both its liquid and gaseous states:
  • Liquid phase: 129.7 J/mol-K
  • Gas phase: 95.2 J/mol-K
To calculate the heat required for temperature changes, we use the formulas:
  • For the liquid phase: \( q = m \cdot C_{\text{liquid}} \cdot \Delta T \)
  • For the gas phase: \( q = m \cdot C_{\text{gas}} \cdot \Delta T \)
Where \( q \) represents the heat added, \( m \) is the number of moles, \( C \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. By knowing these values, we can precisely calculate the energy required to change the temperature of isobutane as it transitions between phases.
Phase Change
Phase changes are transformations between different states of matter, such as from liquid to gas or solid to liquid. These transformations occur without any change in temperature, despite the exchange of heat.
For isobutane in the original problem, the critical phase change is vaporization—the transition from liquid to gas. This occurs at its boiling point, -12°C, and requires a specific amount of heat known as the heat of vaporization.
The heat of vaporization for isobutane is specified as 21.3 kJ/mol. This means that each mole of isobutane requires 21.3 kJ of energy to change from liquid to gas at the boiling point. The formula used for the heat required for vaporization is:
  • \( q = n \cdot \Delta H_{\text{vap}} \)
Where \( q \) is the heat added, \( n \) is the number of moles, and \( \Delta H_{\text{vap}} \) is the heat of vaporization. This calculation is essential for accurately determining the total energy needs for phase changes, playing a critical role in thermodynamic calculations.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's a crucial property used to convert between the mass of a substance and the number of moles, which is essential for stoichiometric calculations in chemistry.
For isobutane, the chemical formula is \( C_4H_{10} \). To find its molar mass, sum the atomic masses of all the atoms contained in the molecule:
  • Carbon (C): 12.01 g/mol, with four atoms contributing to a total of \( 4 \times 12.01 \) g/mol.
  • Hydrogen (H): 1.008 g/mol, with ten atoms contributing to a total of \( 10 \times 1.008 \) g/mol.
This results in a molar mass of 58.14 g/mol for isobutane. Having the molar mass allows us to convert a given mass of isobutane into moles using the formula:
  • \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)
This is a critical step in the calculation, ensuring accurate conversion from mass to moles, which is necessary for calculating the required heat for phase changes and temperature changes.
Boiling Point
The boiling point is the temperature at which a liquid changes to a gas, under a given pressure (usually 1 atmosphere for standard boiling points). At this temperature, the vapor pressure of the liquid equals the external pressure surrounding the liquid.
For isobutane, the boiling point is given as -12°C. This is the temperature at which isobutane will begin to vaporize. Understanding the boiling point is crucial for thermodynamic calculations, as it indicates when a phase change from liquid to gas will occur, and lets us apply the heat of vaporization at the correct moment in these calculations.
When a substance reaches its boiling point, any heat input will be used solely to transform the liquid into a gas, rather than increasing the temperature. This temperature plateau during a phase change is a significant characteristic of thermodynamics. The boiling point helps define the conditions under which a substance will change its phase, which is vital for any process involving heating or cooling of materials.

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Most popular questions from this chapter

It often happens that a substance possessing a smectic liquid crystalline phase just above the melting point passes into a nematic liquid crystalline phase at a higher temperature. Account for this type of behavior.

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is \(\mathrm{a}-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melt- ing points of the salts that form between the 1 -alkyl3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p},=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(\left(\mathrm{m} \cdot \mathrm{p},=40^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} \cdot=10^{\circ} \mathrm{C}\right),\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (m.p. \(\left.=-61^{\circ} \mathrm{C}\right) .\) Why does the melting point decrease as the length of alkyl group increases?

At room temperature, \(\mathrm{CO}_{2}\) is a gas, \(\mathrm{CCl}_{4}\) is a liquid, and \(\mathrm{C}_{60}\) (fullerene) is a solid. List these substances in order of (a) increasing intermolecular energy of attraction and (b) increasing boiling point.

The table below shows some physical properties of compounds containing O-H groups. \begin{tabular}{lccc} \hline Liquid & Molecular Weight & Experimental Dipole Moment & Boiling Point \\\ \hline \(\mathrm{CH}_{3} \mathrm{OH}\) & 32.04 & 1.7 & \(64.7^{\circ} \mathrm{C}\) \\\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 74.12 & 1.66 & \(117.7^{\circ} \mathrm{C}\) \\ \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 62.07 & 1.5 & \(197.3^{\circ} \mathrm{C}\) \\ \hline \end{tabular} Which of the following statements best explains these data? (a) The larger the dipole moment, the stronger the intermolecular forces, and therefore the boiling point is lowest for the molecule with the largest dipole moment. (b) The dispersion forces increase from \(\mathrm{CH}_{3} \mathrm{OH} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\); since the boiling point also increases in this order, the dispersion forces must be the major contributing factor for the boiling point trend; \((\mathbf{c}) \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) has two groups capable of hydrogen bonding per molecule, whereas \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) have only one; therefore, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) has the highest boiling point.

Suppose you have two colorless molecular liquids \(A\) and \(B\) whose boiling points are \(78^{\circ} \mathrm{C}\) and \(112^{\circ} \mathrm{C}\) respectively and both are at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) Both A and B are liquids with identical vapor pressure at room temperature of \(25^{\circ} \mathrm{C} .(\mathbf{b})\) Liquid \(\mathrm{A}\) must consist of nonpolar molecules with lower molecular weight than B. \((\mathbf{c})\) Both liquids \(A\) and \(B\) have higher total intermolecular forces than water. (d) Liquid \(A\) is more volatile than liquid B because it has a lower boiling point. (e) At \(112^{\circ} \mathrm{C}\) both liquids have a vapor pressure of 1 atm.

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