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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

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is pressurized into liquid and stored in cylinders to be used as a fuel. The normal boiling point of propane is listed as \(-42^{\circ} \mathrm{C}\). (a) When converting propane into liquid at room temperature of \(25^{\circ} \mathrm{C}\), would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid propane is in it? (b) Suppose the fuel tank leaks and a few liters of propane escape rapidly. What do you expect would happen to the temperature of the remaining liquid propane in the tank? Explain. (c) How much heat must be added to vaporize \(20 \mathrm{~g}\) of propane if its heat of vaporization is \(18.8 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this amount of propane occupy at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} ?\)

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

Rubbing alcohol is the common name for 2 -propanol (isopropanol), \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\). (a) Draw the three-dimensional structure of the 2 -propanol molecule and predict the geometry around each carbon atom. (b) Is the 2-propanol molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between the 2 -propanol molecules? (d) The isomer 1 -propanol has the same molecular weight as 2 -propanol, yet 2 -propanol boils at \(82.6^{\circ} \mathrm{C}\) and 1 -propanol boils at \(97.2{ }^{\circ} \mathrm{C}\). Explain the difference.

Hydrazine \(\left(\mathrm{H}_{2} \mathrm{NNH}_{2}\right),\) hydrogen peroxide \((\mathrm{HOOH}),\) and water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) all have exceptionally high surface tensions compared with other substances of comparable molecular weights. (a) Draw the Lewis structures for these three compounds. (b) What structural property do these substances have in common, and how might that account for the high surface tensions?

Rationalize the difference in boiling points in each pair: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\left(-23^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \quad\left(78^{\circ} \mathrm{C}\right)\), (b) \(\mathrm{CO}_{2}\left(-78.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CS}_{2}\left(46.2^{\circ} \mathrm{C}\right),(\mathbf{c}) \mathrm{CH}_{3} \mathrm{COCH}_{3}\left(50.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{COOH}\left(101^{\circ} \mathrm{C}\right)\).

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