General Chemistry 2 Midterm Module 3 PDF

Senior High School General Chemistry 2 Quarter 3 - Module 3 Colligative Properties and Thermodynamics Department of Education Republic of the Philippines  Lesson Colligative Properties and Determination 3.1 of Molar Mass of Solutions What I Need to Know This module discu...

Senior High School General Chemistry 2 Quarter 3 - Module 3 Colligative Properties and Thermodynamics Department of Education Republic of the Philippines  Lesson Colligative Properties and Determination 3.1 of Molar Mass of Solutions What I Need to Know This module discusses about the colligative properties of solutions and determination of molar mass from colligative property data. After going through this module, you are expected to 1. Calculate boiling point elevation and freezing point depression from the concentration of a solute in a solution (STEM_GC11PPIIId-f-117); and 2. Calculate molar mass from colligative property data (STEM_GC11PPIIId-f-118). What’s New Activity 3.1.1. Find me! Locate the words associated with colligative properties in the grid. The words can be running in horizontal, vertical and diagonal directions. 1. ____________________ 2. ____________________ 3. ____________________ 4. ____________________ 5. ____________________ 6. ____________________ 7. ____________________ 8. ____________________ 9. ____________________ 10. ____________________ What Is It In this lesson, we are going to discuss about colligative property and how to determine molar mass of a solution using these colligative properties. Colligative property refers to the property of the solution that depends on the number of solute particles present. Colligative properties of solutions include freezing point depression and boiling point elevation. There are more colligative properties like osmotic pressure and vapor pressure lowering. In this lesson, we’ll focus only on the first two property mentioned above. Colligative Properties of Nonelectrolyte and Electrolyte Solutions Boiling Point Elevation of Nonelectrolyte Solution Previously, you have learned that a liquid boil when its vapor pressure is equal to its atmospheric pressure (1.01 x 105 Pa). The normal boiling point of a liquid is the temperature at which its vapor pressure is 1.01 x 105 Pa. If the vapor pressure decreases in a nonelectrolyte solution, the boiling point of the solution tends to increase. Therefore, to reach the 1.01 x 10 5 Pa vapor pressure, the solution must be boiled at a temperature higher than the normal boiling point temperature of the pure solvent. The increase in the boiling point temperature is called boiling point elevation (∆T b). To determine the boiling point elevation of a solution, we will have to use equation 3.1.1, 48  ∆Tb = Kbm Equation 3.1.1 where Kb is the molal boiling point elevation constant and m is the molality of the solute in the solution. Molality is used in this equation since it relates to mole fraction and, thus, to particles of solute and temperature does not affect it. On the other hand, the boiling point of a solution can be determined using equation 3.1.2, Tb(solution) = Tb(solvent) + ∆Tb Equation 3.1.2 where Tb(solution) is the boiling point of solution, Tb(solvent) is the boiling point of pure solvent and ∆Tb is the boiling point elevation. Let’s do an example! Example 3.1A: Sucrose (C22O11H22, 342 g/mol), like many sugars, is highly soluble in water; almost 1500 g will dissolve in 1000 g of water, giving rise to what amounts to pancake syrup. Using Kb of 0.514 K/mol*Kg, estimate the boiling point elevation of such a sugar solution. We can solve this problem in a straightforward manner. Since all the data needed is already given, we can just directly substitute the given values in equation 3.1.1. But take note, in equation 3.1.1, the boiling point elevation can be solved if we have the values of molality and the molal boiling point elevation constant and since molality is not given, then we have to calculate for it first. Given: Molar Mass(C22O11H22) = 342 g/mol Mass(C22O11H22) = 1500 g Volumewater = 1000 g = 1 Kg Kb = 0.514 K/mol*Kg Unknown: Boiling point elevation (∆T b) Equation: ∆Tb = Kbm molality = Equation 3.1.3 Now that we already listed all the given data, unknown and equation we will use, let’s proceed to the solution part. Solution: Since, we are given the molar mass and mass of the sucrose, let’s get the number of moles of sucrose. Number of moles = Equation 3.1.4 Number of moles of sucrose = Number of moles of sucrose = 4.39 mol That’s it! Now that we have the number of moles of sucrose, we can then solve for the molality of the solution using equation 3.1.3. molality (m) = molality (m) = molality (m) = 4.39 mol sucrose/kg water Finally, we have the values of molality and the K b , we can now go to the last part of the calculation, which is to calculate for the boiling point elevation. ∆Tb = Kbm ∆Tb = (0.514 K/mol*Kg)( 4.39 mol sucrose/kg water) ∆Tb = 2.25 K Therefore, the boiling point elevation of the sugar solution is 2.25 K. 49 Freezing Point Depression Recall that only solvent vaporizes from solution, so solute molecules are left behind. Similarly, only solvent freezes, again leaving solute molecules behind. The freezing point of a solution is the temperature at which its vapor pressure equals that of the pure solvent, that is, when solid solvent and liquid solution are in equilibrium. The freezing point depression (∆Tf) occurs because the vapor pressure of the solution is always lower than that of the solvent, so the solution freezes at a lower temperature; that is, only at a lower temperature will solvent particles leave and enter the solid at the same rate. The freezing-point depression (∆Tf) is defined as the freezing point of the pure solvent (Tf°) minus the freezing point of the solution (Tf): ∆Tf = Tf° + Tf Equation 3.1.5 Because Tf° > Tf, ∆Tf is a positive quantity. Again, ∆Tf is proportional to the concentration of the solution: ∆Tf α m ∆Tf = Kfm Equation 3.1.6 where ∆Tf is the freezing point elevation, m is the concentration of the solute in molality units, and Kf is the molal freezing-point depression constant. The Kf of water is 1.86 °C*Kg/mol. Let’s practice solving for freezing point depression! Example 3.1B: Calculate the mass of ethylene glycol (formula mass = 62.1 g/mol), an antifreeze component, that must be added to 5.0 L of water to produce a solution that freezes at -18.3°C. Assume that the density of the water is exactly 1 g/mL. Given: Kf = 1.86 °C*Kg/mol Tf = -18.3 °C Molar massethylene glycol = 62.1 g/mol Densitywater = 1 g/mL Volumewater = 5.0 L Unknown: Mass of ethylene glycol Solution: The approach in answering this kind of problem is the same with the previous problem, it just differs slightly since this time, it’s the mass that is being asked. ∆Tf = Tf° + Tf ∆Tf = 0 °C – (-18.3 °C) ∆Tf = 18.3 °C Now, we have the value for the boiling point elevation. However, our task is not yet done since it’s the mass that we’re looking for and not the boiling point elevation. Previously, we learned that ∆Tf can also be solved using molality and Kf, as shown is equation 3.1.6. To solve for the molality, we can use this equation and transpose it. ∆Tf = Kf m m= m= m = 9.84 mol/Kg With this, we can now solve for the mass of the ethylene glycol by substitution. Number of moles of ethylene glycol = molality (Volume of water) (density of water)( Number of moles of ethylene glycol = 9.84 mol/Kg (5000 mL) (1 Number of moles of ethylene glycol = 49.2 mol 50  We know that to get the mass of a substance by simply multiplying the number of moles of the substance and its molar mass. Massethylene glycol = 49.2 mol (62.1 g/mol) Massethylene glycol = 3.05 x 103 g = 3.05 Kg Therefore, the mass of ethylene glycol, an antifreeze component, is 3.05 Kg. Colligative Properties of Electrolyte Solution The study of colligative properties of electrolytes requires a slightly different approach than the one used for the colligative properties of nonelectrolytes. The reason is that electrolytes dissociate into ions in solution, and so one unit of an electrolyte compound separates into two or more particles when it dissolves. For example, electrolyte solutions such as sodium chloride dissociates into two ions – Na+ and Cl-. A van’t Hoff factor (i) must be put into the previous equations involving the colligative properties of nonelectrolytes to account this effect. The van’t Hoff factor establishes the relationship between the number of moles of solute dissolved and the moles of particles in a solution. This factor is given by, i= Equation 3.1.7 where the i value for an electrolyte with an AB empirical formula is 2, whereas for AB 2, i is 3. Thus, i should be 1 for all nonelectrolytes. For strong electrolytes such as NaCl and KNO 3, i should be 2, and for strong electrolytes such as Na2SO4 and CaCl2, i should be 3. Table 2.1 shows the experimentally measured values of i and those calculated assuming complete dissociation. As you can see, the agreement is close but not perfect, indicating that the extent of ion-pair formation in these solutions at that concentration is appreciable. Table 3.1.1. The van’t Hoff Factor of 0.0500 M Electrolyte Solutions at 25 degrees Celsius In calculating the colligative properties of electrolytes, such as boiling point elevation and freezing point depression, i is integrated. The general equation now becomes, ∆Tb = iKbm Equation 2.1.8 ∆Tf = iKf m Equation 2.1.9 Let’s practice solving! Example 3.1C: Determine the freezing and boiling point of a solution prepared by dissolving 85.40 g of calcium chloride (CaCl2) in 200 g of water (H2O). Just follow the process we did from the previous examples. Start with listing all the known and unknown data before going to the solution part. Given: Masscalcium chloride = 85.40 g Masswater = 200 g Molar MassCaCl2 =110.98 g/mol Kf = -1.86 Kb = 0.512 Take note that CaCl2 dissociates into 3 ions, 2 ions for Chlorine and 1 ion for Calcium as given in the reaction below. CaCl2(aq) → Ca2+(aq) + 2Cl−(aq) 51  Unknown: Tf and Tb First thing we have to do, is to calculate for the moles of CaCl2. We can do this by dividing the mass of the CaCl2 to its molar mass. Number of moles = Number of moles = Number of moles = 0.770 mol CaCl2 Now, we can solve for the molality of the solution using Equation 2.1.3. molality (m) = molality (m) = molality (m) = 3.85 mol/Kg Using equations 2.1.8 and 2.1.9, we can consecutively solve for ∆Tb and ∆Tf. ∆Tb = iKbm ∆Tb = 3 (0.512 (3.85 mol/Kg) ∆Tb = 5.91 ∆Tf = iKfm ∆Tf = 3 (-1.86 (3.85 mol/Kg) ∆Tf = -21.48 Since the normal boiling point of water is 100 using equation 3.1.2 for boiling point and 3.1.5 for freezing point, we must add the calculated ∆T b to 100 to get the new boiling point of the solution. Therefore, the new boiling point is 105.91. On the other hand, since the normal freezing point of water is 0 , the new freezing point is still -21.48 Using Colligative Properties to Find Solute Molar Mass The colligative properties of nonelectrolyte solutions provide a means of determining the molar mass of a solute. We learned that each colligative property is proportional to solute concentrations. Thus, by measuring the property we can determine the amount (mol) of solute particles and with a given mass, the molar mass can be calculated. Let’s have an example! Example 3.1D: The addition of 250 mg of naphthalene in benzene elevated the boiling point of 100 g of benzene by 0.05°C at 1.01 x 105 Pa. Calculate the molar mass of naphthalene molecules. Benzene has a boiling point of 80.1°C and the K b for benzene is 2.53 °C*Kg/mol. Given this problem, first thing you have to do is to breathe. Relax, we can solve this problem together! Now, you have to list down all the data given and the unknown. You may follow the format below to make your calculations neat and easier to track. Given: MassNaphthalene = 250 mg = 0.250 g MassBenzene = 100 g Pressure = 1.01 x 105 Pa Kb = 2.53 °C*Kg/mol Tb = 80.1°C ∆ Tb = 0.05°C Unknown: Molar Mass of naphthalene molecules All givens in? Already know what we have to calculate? Good! Let’s proceed to the solution part. Solution: With all the given data, next we have to think for an equation where we can use those data given to us. Do you think we can use equation 3.1.1 to solve for the molality of the naphthalene molecules? Absolutely! But we can’t use the equation on its original form since we’re not looking for the ∆T b, we’re looking for the molality. So, using the knowledge you acquire from the subject algebra, we have to transpose the equation and get the value for molality. 52  ∆Tb = Kbm m= m= m = 0.020 We already calculated the molality of the solution. But, we are not done yet since our final goal is to determine the molar mass of the naphthalene molecules. We know that the solution was prepared using 0.100 kg of benzene. Now, we can find the number of moles of naphthalene in the solution. Molality (m) = Number of moles of naphthalene = molality (mass of solvent) Number of moles of naphthalene = (0.020 mol/kg) (0.100 kg) Number of moles of naphthalene = 1.98 x 10-3 mol Previously, we learned that to get the molar mass we can use equation 3.1.4. Molar mass = Molar mass = Molar mass = 128 g/mol The molecular formula of naphthalene is C10H8 and its molar mass is 128 g/mol. What’s More Activity 3.1.2. Solve Me! Calculate the problems below. Write your solution and box your final answer on the answer sheets provided. Follow the flow of solving like the examples given above. You may use the Kb and Kf of water in the examples. 1. Calculate the freezing and boiling points of a solution prepared by dissolving 15.5 g of Al (NO3)3 in 200.0 g of water. (Molar mass of Al (NO3)3 is 212.996 g/mol). 2. A solution is prepared by dissolving 120 grams of NaCl in 450 grams of water. Find the freezing and boiling points of this solution. (Molar mass of NaCl is 58.44 g/mol) What I Have Learned Activity 3.1.3. Imagine Me! Congratulations! You won the lotto and bought a new car. Since, it’s very hot in the Philippines you are required to buy an antifreeze to keep the water in your car’s radiator from boiling. You add 1.00 Kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in your car’s radiator. What are the boiling and freezing points of the solution? What I Can Do Activity 3.1.4. Know Me! Look for a book or search the web for additional information regarding the colligative properties of solution. Refer to the guide questions below. 1. State the effect that dissolving a solute has on each of the following physical properties of a solvent. a. Boiling point b. Freezing point 2. Explain why the colligative properties of a solvent are affected more by the dissolving of an electrolyte compared to an equal amount of a nonelectrolyte. 3. Are the values of Kf and Kb dependent on the identity of the solvent, the solute, or both? Explain. 53  Lesson Laboratory Procedures in Determination 3.2 of Solution’s Concentration What’s In Previously, we discussed about the colligative properties of the solutions. Now, let’s learn how to prepare and determine the concentrations of those solutions in laboratory. What I Need to Know This module discusses about laboratory procedure needed to be followed in preparing and determining the concentration of the solutions. After going through this module, you are expected to describe laboratory procedures in determining concentration of solutions (STEM_GC11PPIIId-f-119); What’s New Activity 3.2.1. Twist me! Rearrange the word to get the correct word. Clue: These words are related to solution! 1. NCTOENCITORA 2. TYOMILAR 3. NILUSOTIO 4. VESLONT 5. AUQEOSU What Is It In performing chemical experiments across the world, the most important procedure is preparing a solution. With this, the concentration of the solution must be determined first. Recall, that a solution is made up of a substance dissolved in liquid. The substance that is being dissolved is called solute while the substance that is dissolving the solute is known as solvent. We can then call the result of this mixture as solution. Most of the time, solution is defined by their solute concentration, which refers to to the measure of how much solute is present per unit of solution. Making solution is easy and requires basic skill only, however, poor technique in doing this may result to a failed experiment. That is why, it is a must that we should know how to properly prepare a solution in a laboratory experiment. Are you familiar with the concept ―Safety First‖? Well, that also applies in performing a laboratory experiment. The first consideration when making a solution, is the safety of those who will perform the experiment. That is why beforehand, your personal protective equipment (PPE) should be complete and equipped to you appropriately. Examples of this PPE are safety goggles, face mask, gloves and laboratory coat. When you’re already equipped with your PPE, the next thing to do is to determine the moles of the solute you will need in order to achieve the desired concentration of your solution. Then convert this value to grams using the molar mass of the chemical. The chemicals can be weighed out using a digital and analytical balance. In an aqueous solution, the solvent to be used is water. In measuring the volume of the water, you can use the graduated cylinder. Most of the time, it is roughly ¾ of the final volume of the solution. It is also highly recommended to use purified water rather than tap water to avoid contamination and compromising the quality of the solution. After measuring the volume of the solvent to be used, you can then transfer it to a beaker which contains a stir bar. The weighed solute can then be added to the purified water. Heating and stirring the solution is also recommended upon mixing to make the process of dissolving faster to finish. 54  Now, these are the most common scenarios in performing a laboratory experiment. But, you might wonder how did we came up with the values of the volume needed right? Usually, a given volume and molarity are required before starting the experiment. Let’s do the calculations! Starting with a Solid Solute Procedure 1. Determine the mass in grams of one mole of solute, the molar mass (MM). 2. Decide volume of solution required, in liters, V. 3. Decide molarity of solution required, M. 4. Calculate mass (M) in grams of solute (g) required using equation 3.2.1, Mass = MM x M x V Equation 3.2.1 Example: Prepare 800 mL of 2 M sodium chloride. MMNaCl = 58.45 g/mol MNaCl = 58.45 g/mol (2 mol/L) (0.8 L) MNaCl = 93.52 g NaCl Therefore, you have to dissolve 93.52 g of NaCl in about 400 mL of distilled water, then add more water until final volume is 800 mL. Basically, in making solution, the following are the most commonly used procedure. Fill Add more volumetric Transfer the Stir until Label the flask Weigh deionized flask with solid to the the solid with the the or distilled deionized volumetric is corresponding solid water until or distilled flask dissolved solution's name the mark water What’s More Activity 3.2.2. Let’s do this! Mikaela is a STEM student, she is currently having her General Chemistry 2 subject and today, she is asked by her teacher to prepare CuSO 4 solution. Help her do her task by following the procedure and answering the question below. 1. Obtain a sample of CuSO4 from the front bench. 2. Using your scale weigh out 18 g of CuSO4. 3. Dissolve the 18 g of CuSO4 in 100 mL of distilled water. 4. Stir until you have reached a homogeneous solution with all the solute dissolved in the solvent. Calculate the concentration of your solution (M in mol/ L of CuSO 4). What I Have Learned Activity 3.2.3. Explain me! Grace needs to prepare a sulfuric acid solution using sulfuric acid and distilled water, however, there is no distilled water left on the laboratory. Marie suggested to use the running water from the faucet but Grace declined. Grace insisted that tap water should not be used in preparing solutions. Explain the basis of Grace’s claims. 55 What I Can Do Activity 3.2.4. Prepare me! Objective: Prepare a sodium chloride (table salt) solution and sugar solution. Equipment: Cup, table salt, sugar, timer/stopwatch, table spoon Procedure: 1. Measure amount of table salt using table spoon. 2. Add the table salt in a ½ cup of water. 3. Stir the solution using the table spoon. 4. Record the time it takes for the salt to be totally dissolved. 5. Picture/draw your set-up in preparing the solution. 6. Do step 1 to 5 for sugar solution. Guide Question: 1. In preparing the solutions, which between the two solutes was dissolved faster? 56 Lesson First Law of Thermodynamics 3.3 What’s In Previously, we discussed about how to prepare and determine the concentrations of those solutions in laboratory. Now, let us proceed to learning what is thermodynamics all about. Let’s start with the first law of thermodynamics. What I Need to Know This module discusses about first law of thermodynamics. After going through this module, you are expected to explain the first law of thermodynamics (STEM_GC11PPIIId-i-124). What’s New Activity 3.3.1. Word twist! Rearrange the word to get the correct word. 1. HIRTONSADMEMCY 2. NEGEYR 3. TEHA 4. KORW 5. CELCINAHAM What Is It Thermochemistry is part of a broader subject called thermodynamics, which is the scientific study of the interconversion of heat and other kinds of energy. The laws of thermodynamics provide useful guidelines for understanding the energetics and directions of processes. In this section we will concentrate on the first law of thermodynamics, which is particularly relevant to the study of thermochemistry. The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed. It states that the change in internal energy of a system, ∆U, equals the net heat transfer into the system, Q, plus the net work done on the system, W. This is shown in equation 3.3.1. ∆U = W + Q Equation 3.3.1 Here, ΔU is the change in internal energy of the system. Q is the net heat transferred into the system—that is, Q is the sum of all heat transfer into and out of the system. W is the net work done on the system. So positive heat, Q, adds energy to the system and positive work, W, adds energy to the system. This is why the first law takes the form it does in equation 3.3.1. It simply says that you can add to the internal energy by heating a system, or doing work on the system. We can think of the first law of thermodynamics as an energy balance sheet, much like a money balance sheet kept in a bank that does currency exchange. You can withdraw or deposit money in either of two different currencies (like energy change due to heat exchange and work done). However, the value of your bank account depends only on the net amount of money left in it after these transactions, not on which currency you used. Work We have seen that work can be defined as force F multiplied by distance d: W=Fxd Equation 3.3.2 Where, F is the distance and d is the distance. In thermodynamics, work can be defined in many ways such as mechanical work, electrical work and surface work. An example of a mechanical work is the compression and expansion of a gas. In this way, work can be defined using equation 3.3.3. 57  W = - P(∆V) Equation 3.3.3 Where, ∆V, the change in volume, is given by Vf – Vi and P is the external atmospheric pressure. Let’s have an example! Example 3.3.1. A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm. The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. What is the conversion factor between L*atm and J? (a) Because the external pressure is zero, no work is done in the expansion. W = - P(∆V) W = - 0 ( 6.0 L – 2.0 L) W = 0 atm*L (b) The external, opposing pressure is 1.2 atm, so W = - P(∆V) W = - 1.2 atm (6.0 L – 2.0 L) W = - 4.8 atm*L To convert the answers to Joules, J. W = - 4.8 atm*L ( ) 2 W = - 4.9 x 10 J Because this is gas expansion (work is done by the system on the surroundings), the work done has a negative sign. Heat The other component of internal energy is heat, Q. For example, it takes 4184 J of energy to raise the temperature of 100 g of water from 20°C to 30°C. This energy can be gained (a) directly as heat energy from a Bunsen burner, without doing any work on the water; (b) by doing work on the water without adding heat energy (for example, by stirring the water with a magnetic stir bar); or (c) by some combination of the procedures described in (a) and (b). This simple illustration shows that heat associated with a given process, like work, depends on how the process is carried out. It is important to note that regardless of which procedure is taken, the change in internal energy of the system, ∆U, depends on the sum of (Q + W). If changing the path from the initial state to the final state increases the value of Q, then it will decrease the value of W by the same amount and vice versa, so that ∆U remains unchanged. What’s More Activity 3.3.2. Solve me! 1. Manuel has a container that contains a sample of nitrogen gas and a tightly fitting movable piston that does not allow any of the gas to escape. During a thermodynamics process, 200 Joules of heat enter the gas, and the gas does 300 Joules of work in the process. What was the change in internal energy of the gas during the process described above? 2. The work done when a gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process. What I Have Learned Activity 3.3.3. Explain me! Juliet was studying about thermodynamics. She was doing well in understanding the lesson except for one concept. She was confused between the two terms, heat and temperature. Is heat and temperature the same? Explain the difference between the two terms. 58 Lesson Enthalpy and Hess’ Law 3.4 What’s In Previously, we discussed about first law of thermodynamics. Now, let us proceed to learning what enthalpy is and how to calculate for enthalpy of a reaction using Hess Law. What I Need to Know This module discusses about enthalpy and the calculation of enthalpy of a reaction using Hess Law. After going through this module, you are expected to: 1. Explain enthalpy of a reaction (STEM_GC11TCIIIg-i-125) 2. Calculate the change in enthalpy of a given reaction using Hess Law (STEM_GC11TCIIIg-i-127) What’s New Activity 3.4.1. Find me! Locate the words associated with enthalpy in the grid. The words can be running in horizontal, vertical and diagonal directions. 1. ____________________ 2. ____________________ 3. ____________________ 4. ____________________ 5. ____________________ 6. ____________________ 7. ____________________ 8. ____________________ 9. ____________________ 10. ____________________ What Is It Our next step is to see how the first law of thermodynamics can be applied to processes carried out under different conditions. Specifically, we will consider two situations most commonly encountered in the laboratory; one in which the volume of the system is kept constant and one in which the pressure applied on the system is kept constant. Enthalpy is a thermodynamic property of a system. It is the sum of the internal energy added to the product of the pressure and volume of the system. It reflects the capacity to do non-mechanical work and the capacity to release heat. Enthalpy is denoted as H; specific enthalpy denoted as h. Enthalpy in a throttling process is constant. We can define enthalpy using equation 3.4.1. H = U + PV Equation 3.4.1 Where, H is the enthalpy of a reaction, U is the internal energy of the system, P is the pressure and V is the volume of the system. Because U and PV have energy units, enthalpy also has energy units. Common units used to express enthalpy are the joule, calorie, or BTU (British Thermal Unit.). 59  Change in enthalpy is calculated rather than enthalpy, in part because total enthalpy of a system cannot be measured since it is impossible to know the zero point. However, it is possible to measure the difference in enthalpy between one state and another. Enthalpy change may be calculated under conditions of constant pressure. One representation of this is of a firefighter who is on a ladder, but the smoke has covered his view of the ground. He cannot see how many steps are below him to the ground but can see there are three steps to the window where a person needs to be rescued. In the same way, total enthalpy cannot be measured, but the change in enthalpy (three ladder steps) can. This change in enthalpy is defined as ∆H. Because most reactions are constant-pressure processes, we can equate the heat change in these cases, to the change in enthalpy. For any reaction of the type reactants products we define the change in enthalpy, called the enthalpy of reaction, ∆Hrxn, as the difference between the enthalpies of the products and the enthalpies of the reactants: ∆Hrxn = H(Products) – H(Reactants) Equation 3.4.2 Remember that the enthalpy of reaction can be positive or negative, depending on the process. For an endothermic process (heat absorbed by the system from the surroundings), ∆H is positive (that is, ∆H > 0). For an exothermic process (heat released by the system to the surroundings), ∆H is negative (that is, ∆H < 0). Another way of calculating the change of enthalpy is by using Hess’ law. Hess’ law states that, the enthalpy change accompanying a chemical change is independent of the route by which the chemical change occurs. In other words, Hess's Law is saying that if you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps. Let’s have an example for better understanding! Example 3.4.1. Find the enthalpy change for the reaction. CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) when: C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol This is an example of a trial and error problem. To solve this, it is better to start with a reaction with only one mole of reactant or product in the reaction. Our reaction needs one CO2 in the product and the first reaction also has one CO2 product. C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ/mol This reaction gives us the CO2 needed on the product side and one of the O2 needed on the reactant side. The other two O 2 can be found in the second reaction. S(s) + O2(g) → SO2(g) ΔHf = -296.8 kJ/mol Since only one O2 is in the reaction, multiply the reaction by two to get the second O2. This doubles the ΔHf value. 2 S(s) + 2 O2(g) → 2 SO2(g) ΔHf = -593.6 kJ/mol Combining these equations gives 2 S(s) + C(s) + 3 O2(g) → CO2(g) + SO2(g) The enthalpy change is the sum of the two reactions: ΔHf = -393.5 kJ/mol + -593.6 kJ/mol = -987.1 kJ/mol This equation has the product side needed in the problem but contains an extra two S and one C atom on the reactant side. Fortunately, the third equation has the same atoms. If the reaction is reversed, these atoms are on the product side. When the reaction is reversed, the sign of the change in enthalpy is reversed. 60  CS2(l) → C(s) + 2 S(s); ΔHf = -87.9 kJ/mol Add these two reactions together and the extra S and C atoms cancel out. The remaining reaction is the reaction needed in the question. Since the reactions were added together, their ΔHf values are added together. ΔHf = -987.1 kJ/mol + -87.9 kJ/mol ΔHf = -1075 kJ/mol Therefore, the change in enthalpy for the reaction is -1075 kJ/mol. What’s More Activity 3.4.2. Calculate me! Given the thermochemical equation 2 SO2(g) + O2(g) 2 SO3(g) ∆H = - 198.2 kJ/mol calculate the heat evolved when 87.9 g of SO 2 (molar mass 64.07 g/mol) is converted to SO3. What I Have Learned Activity 3.4.3. Solve me! 1. Calculate the heat of hydrogenation of ethane, C2H4 given the following thermochemical equations: 2 C (graphite) + 3 H2 (g) → C2H6 (g) ΔHf = - 84.5 kJ/mol 2 C (graphite) + 2 H2 (g) → C2H4 (g) ΔHf = 52.3 kJ/mol 2. Calculate the ΔH for the reaction: CS2 (l) + 2 O2 (g) CO2 (g) + 2 SO2 (g) Given: ΔHf CO2 (g) = - 393.5 kJ/mol; ΔHf SO2 = - 296.8 kJ/mol; ΔHf CS2 (l) = 87.9 kJ/mol What I Can Do Activity 3.4.4. Complete Me! Complete the table by providing the information needed. Importance Application Enthalpy 61 Summary Colligative Properties Colligative property refers to the property of the solution that depends on the number of solute particles present. Colligative properties of solutions include freezing point depression and boiling point elevation. The increase in the boiling point temperature is called boiling point elevation (∆T b). To determine the boiling point elevation of a solution, we will have to use equation, ∆Tb = Kbm where the value of the Kb is 0.514 K/mol*Kg On the other hand, the boiling point of a solution can be determined using equation, Tb(solution) = Tb(solvent) + ∆Tb The freezing point depression (∆Tf) occurs because the vapor pressure of the solution is always lower than that of the solvent, so the solution freezes at a lower temperature; that is, only at a lower temperature will solvent particles leave and enter the solid at the same rate. The equation of freezing point depression is ∆Tf = Kfm where Kf is 1.86 °C*Kg/mol In solving electrolytes solution, a van’t hoff factor must be incorporated to the previous equations. Laboratory Procedures in Concentration’s Determination Fill Add more volumetric Transfer the Stir until Label the flask Weigh deionized flask with solid to the the solid with the the or distilled deionized volumetric is corresponding solid water until or distilled flask dissolved solution's name the mark water First Law of Thermodynamics The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed. It states that the change in internal energy of a system, ∆U, equals the net heat transfer into the system, Q, plus the net work done on the system, W. Enthalpy and Hess’ Law Enthalpy is a thermodynamic property of a system. It is the sum of the internal energy added to the product of the pressure and volume of the system. It reflects the capacity to do non-mechanical work and the capacity to release heat. Hess’ law states that, the enthalpy change accompanying a chemical change is independent of the route by which the chemical change occurs. 62