Chapter 6.docx

Chapter 6 Exam Preparation: Concepts/Skills – Check List Chapter 6 - First Law of Thermodynamics The first law of thermodynamics (law of conservation of energy) states that the total energy in the universe is conserved. = + = 0 universe system surroundings - Definition of a State Function State function: the internal energy of a system, dependent on the current state (composition, volume, pressure and temperature) and not the path taken to reach the state. - Definition of System and Surroundings System: part of the universe that we are focusing on Surrounding: everything in the universe not including the system - Sign convention from the System view Energy Transferred as Heat Only 1. Heat flowing out of the system: q = negative and = negative 2. Heat flowing into the system: q = positive and = positive Energy Transferred as Work Only 1. Work done by the system: q = negative and = negative 2. Work done on the system: q = positive and = positive q w + + + + + - Depends on size of q and w - + Depends on size of q and w - - - - Use of Hess’ Law of Heat Summation Hess’ Law: the enthalpy change of an overall process is the sum of the enthalpy changes in the individual steps overall 1+ 2+ 3 + … - Be able to calculate heat released given number of moles, mass, or be able convert heat to number of moles generated (all permutations) Sample Problem 6.3 Finding the Quantity of Heat from a Temperature Change Problem A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25°C to 300.°C? The specific heat capacity (c) of Cu is given in Table 6.2. Plan We know the mass (125 g) and c (0.387 J/g·K) of Cu and can find ΔT in °C, which equals ΔT in K. We then use Equation 6.7 to calculate the heat.AT T Tinitial final 300 °C 25°C 275°C 275 K q c x mass (g) x AT 0.387 Jlg.K X 125 g x 275 K 1.33x 104 J water final nitial (28.49°C 25.10PC) 3.39°C 3.39 K AT AT solid final nitial (28.490C 100.00 C) E 71.51°C 71.51 K CHo x massilo xATilo 4.184 J/g K x 50.00 g x 3.39 K 0.450 J/g K solid massed X ATeed 22.05 g X (-71.51 K) AT T Tinitial final 300 °C 25°C 275°C 275 K q c x mass (g) x AT 0.387 Jlg.K X 125 g x 275 K 1.33x 104 J water final nitial (28.49°C 25.10PC) 3.39°C 3.39 K AT AT solid final nitial (28.490C 100.00 C) E 71.51°C 71.51 K CHo x massilo xATilo 4.184 J/g K x 50.00 g x 3.39 K 0.450 J/g K solid massed X ATeed 22.05 g X (-71.51 K)Solution Calculating ΔT and then q: Determining the Specific Heat Capacity of a Solid Problem You heat 22.05 g of a solid in a test tube to 100.00°C and add it to 50.00 g of water in a coffee‐ cup calorimeter. The water temperature changes from 25.108C to 28.49°C. Find the specific heat capacity of the solid. Plan We are given the masses of the solid (22.05 g) and of H O (50.00 g2, and we can find the temperature changes of the water and of the solid by subtracting the given values, always using T final T initialsing Equation 6.7, we set the heat released by the solid (−q solidequal to the heat absorbed by the water (q water). The specific heat of water is known, and we solve for c solid Solution Finding ΔT solidnd ΔT water Solving for c solid Determining the Enthalpy Change of an Aqueous Reaction Problem You place 50.0 mL of 0.500 M NaOH in a coffee‐ cup calorimeter at 25.00°C and add 25.0 mL of 0.500 M HCl,