/*! 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 12 The enthalpy of vaporization of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The enthalpy of vaporization of liquid mercury is \(59.11 \mathrm{kJ} / \mathrm{mol} .\) What quantity of energy as heat is required to vaporize \(0.500 \mathrm{mL}\) of mercury at \(357^{\circ} \mathrm{C},\) its normal boiling point? The density of mercury is \(13.6 \mathrm{g} / \mathrm{mL}.\)

Short Answer

Expert verified
Approximately 2.002 kJ of energy is required to vaporize the mercury.

Step by step solution

01

Calculate the mass of mercury

First, we need to determine the mass of mercury. Use the formula for mass: \( \text{mass} = \text{volume} \times \text{density} \). Therefore, the mass of mercury is: \[ 0.500 \text{ mL} \times 13.6 \text{ g/mL} = 6.8 \text{ g} \].
02

Convert mass to moles

Next, convert the mass of mercury to moles using the molar mass of mercury, which is approximately 200.59 g/mol. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Therefore, the moles of mercury are: \[ \frac{6.8 \text{ g}}{200.59 \text{ g/mol}} \approx 0.0339 \text{ mol} \].
03

Calculate energy required for vaporization

Use the enthalpy of vaporization to calculate the energy required to vaporize the mercury. The formula is: \( Q = n \times \Delta H_{vap} \), where \( n \) is the moles of mercury and \( \Delta H_{vap} \) is the enthalpy of vaporization. Thus, \( Q = 0.0339 \text{ mol} \times 59.11 \text{ kJ/mol} \approx 2.002 \text{ kJ} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mercury
Mercury is a unique element often recognized by its liquid state at room temperature. Unlike most metals, mercury is found in its liquid form, making it quite distinct.
It's also known for its shiny, silver color and its ability to flow, much like water. In this exercise, we explore the process involved in turning liquid mercury into vapor by applying heat. This process is known as vaporization, and it's essential to understand how much energy is needed to bring about this change. If you ever wondered why it takes energy to change the state of a substance, it's because you need to overcome the forces that hold the particles together in the liquid state. For mercury, those forces are relatively weak compared to solids but still significant enough to require attention when calculating energy requirements.
Energy Calculation
When it comes to determining the amount of energy needed to vaporize mercury, we rely on the concept of enthalpy of vaporization. This is expressed in terms of energy per mole, usually in kilojoules per mole (kJ/mol). In our case, mercury requires 59.11 kJ/mol to transition from liquid to gas at its boiling point.
To compute the required energy:
  • First, determine the mass of mercury, using its volume and its known density.
  • Then, convert this mass into moles, using mercury's molar mass (200.59 g/mol).
  • Finally, apply the formula for energy calculation: \[ Q = n \times \Delta H_{vap}, \] where \( n \) represents the moles, and \( \Delta H_{vap} \) is the enthalpy of vaporization.
This simple calculation shows how enthalpy, a fundamental thermodynamic quantity, helps us understand and predict energy requirements for phase changes, offering insight into how substances behave under the influence of heat.
Density
Density is a crucial concept when it comes to calculating energy for substances like mercury. It expresses the mass of a substance per unit volume, often in units such as grams per milliliter (g/mL).
In our scenario, mercury has a density of 13.6 g/mL. This means every milliliter of mercury weighs 13.6 grams. Why is this important? The density allows us to find the mass of the mercury we have, which is a necessary step before we can calculate the moles and subsequently the energy required for vaporization. Here's how it works:
  • Take the given volume of mercury, multiply by its density to get its mass.
  • This mass calculation is fundamental because all further calculations depend on accurately knowing how much mercury you are working with.
By understanding density, we can seamlessly move to subsequent steps in energy calculation, ensuring our answers are both accurate and meaningful.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Water \((10.0 \mathrm{g})\) is placed in a thick walled glass tube whose internal volume is \(50.0 \mathrm{cm}^{3} .\) Then all the air is removed, the tube is sealed, and then the tube and contents are heated to \(100^{\circ} \mathrm{C}\) (a) Describe the appearance of the system at \(100^{\circ} \mathrm{C}\) (b) What is the pressure inside the tube? (c) At this temperature, liquid water has a density of \(0.958 \mathrm{g} / \mathrm{cm}^{3} .\) Calculate the volume of liquid water in the tube. (d) Some of the water is in the vapor state. Determine the mass of water in the gaseous state.

Compare the boiling points of the various isomeric hydrocarbons shown in the table below. Notice the relationship between boiling point and structure; branched-chain hydrocarbons have lower boiling points than the unbranched isomer. Speculate on possible reasons for this trend. Why might the intermolecular forces be slightly different in these compounds? $$\begin{array}{lc} \text { Compound } & \text { Boiling point }\left(^{\circ} \mathrm{C}\right) \\ \hline \text { Hexane } & 68.9 \\ \text { 3-Methylpentane } & 63.2 \\ \text { 2-Methylpentane } & 60.3 \\ \text { 2,3-Dimethylbutane } & 58.0 \\ \text { 2,2-Dimethylbutane } & 49.7 \\ \hline \end{array}$$

Equilibrium vapor pressures of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) at various temperatures are given in the table. $$\begin{array}{cc} \text { Temperature }\left(^{\circ} \mathrm{C}\right) & \text { Vapor Pressure }(\mathrm{mm} \mathrm{Hg}) \\ \hline 7.6 & 40 . \\ 26.1 & 100 \\ 60.6 & 400 \\\ 80.1 & 760 \\ \hline \end{array}$$ (a) What is the normal boiling point of benzene? (b) Plot these data so that you have a graph resembling the one in Figure \(12.17 .\) At what temperature does the liquid have an equilibrium vapor pressure of \(250 \mathrm{mm}\) Hg? At what temperature is the vapor pressure \(650 \mathrm{mm}\) Hg? (c) Calculate the molar enthalpy of vaporization for benzene using the the Clausius-Clapeyron equation.

What factors affect the viscosity of a substance? Which of the following substances, water \(\left(\mathrm{H}_{2} \mathrm{O}\right),\) ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right),\) and glycerol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\right),\) is expected to have the highest viscosity? Should viscosity of a substance be affected by temperature? Explain your answers.

What type of intermolecular forces must be overcome in converting each of the following from a liquid to a gas? (a) \(\mathrm{CO}_{2}\) (c) \(\mathrm{CHCl}_{3}\) (b) \(\mathrm{NH}_{3}\) (d) \(\mathrm{CCl}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.