Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Transcript

☰ Search Explore Log in Create new account Upload × Chapter 6: THERMOCHEMISTRY Thermochemistry- Study of the relationship between chemical reactions and energy changes involving heat. Kinetic Energy- Energy an object has because of its motion. K.E. = (1/2)mv2 v = velocity(m/s) m = mass of moving object in kilograms Potential Energy – Energy an object has due to its position OR energy that is stored for example in a chemical bond. Chemical reaction is an example of potential energy. P.E. = mgh h = height from ground g = gravitational acceleration (9.8 m/s2) m = mass of object in kilograms Electrostatic Energy – Form of potential energy that is derived from the interaction of charged particles or objects E = (KQ1Q2)/d ; Q is charge, d is distance between charges, K is proportionality constant. Units of Energy SI unit of energy is the joule. The metric units of the joule is kg (m2/s2) Ex. 2kg object moving at 1m/s has (1/2)(2kg)(1m/s)2 = 1kg(m2/s2) = 1joule Another unit of energy is calorie; 1 calorie = 4.184 joules, the energy needed to raise the temperature of 1 gram of water 1 degree Celsius. 1000 calories = 1 kilocalorie Calorie in food. Thermochemistry refers to heat energy. Heat energy moves between a system under study and its surroundings. System- What is under study, i.e. contents in a chemical reaction. Consider a piston containing gases. ├────╨────┤ │H2 O2 N2 │ The gases are the system; piston and everything else are the surroundings. Surroundings- Everything other than the system. Closed System- One in which energy is exchanged with the surroundings, but not matter. Force- Push or pull on an object. F = ma, F is force, m is mass in kilograms, a is acceleration in meters per second square. Work (w)- Energy used to cause an object to move against a force. W = F x d, d is distance. Also, W = -PV ; P is pressure, V is change in volume. First Law of Thermodynamics- Energy is conserved. Mathematically, it is expressed this way: U = q + w; U is the internal energy of the system. U is the change in internal energy of system. It is a state function. q- heat energy absorbed (q is positive) or given off (q is negative) by the system. w- work done by the system(w is negative) on its surroundings or done on the system(w is positive) by its surroundings. Ex. A piston full of gases absorbs 70 kJ of heat, causing the gases in the piston to expand and do 50 kJ of work on the surroundings. What is the change in internal energy of the gaseous system? U = q + w = 70 kJ – 50 kJ = 20 kJ. Endothermic Process- The change in the system internal energy is positive (U > 0). Exothermic Process- The change in the system internal energy is negative (U <0). State Function- Value depends only on the initial and final conditions. U is a state function; so is pressure (P) and volume(V). q and w are not state functions. U is associated with a chemical change at constant volume. H(Enthalpy)- heat energy associated with a chemical reaction at constant pressure.(i.e. Activity series reactions and Metathesis reactions done in the lab). H = U + PV, H is also a state function. Can be endothermic H = U + (PV) H = U + (nRT) (positive) or exothermic (negative). Associated with all reactions is an enthalpy of reaction, Hrxn Hrxn = enthalpy for reaction that takes place at standard pressure(1atm) and temperature(for thermochemistry, it is 298 K, for ideal gas law, 273 K). Enthalpy is an extensive property. Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890kJ/mol If one doubles the coefficients of this reaction, then the respective enthalpy change is doubled: 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l) H = -1780kJ Enthalpy change for the reverse reaction is equal in magnitude but opposite in sign(+890kJ). Enthalpy change for a reaction depends on the state of the reactants and products. Example: 2H2O(l) 2H2O(g) H = 88 kJ/mol 2H2O(s) 2H2O(l) H = 12 kJ/mol Calorimetry- The measurement of heat flow in a reaction. Calorimeter- An apparatus that measures heat flow Heat Capacity- The amount of heat energy needed to raise the temperature of a certain amt. of an object by 1 K or Celcius. Specific Heat Capacity (Cp)- is amount of heat capacity for 1g of substance: Cp = quantity of heat transferred (grams sub.)(temp. change) Cp = q/(mT); q = mCpT Example: 209J is needed to raise the temperature of 50grams of water by 1 K. What is the specific heat? Therefore Cp = q/(mT) = 209J)/{(50g)(1 K)} = 4.18J/g-K Heat liberated in a solutionConstant pressure calorimetry: q = (specific heat soln) x (grams soln) x T Ex. 50ml 1M HCl soln + 50ml NaOH soln raises temp. of mixture from 21 to 27.5 C. Assume Cp of soln is 4.18j/g-K; density of soln is 1g/ml. Then T is 27.5 - 21.0 = 6.5 : total mass is 100ml x 1g/ml = 100g q = mCp∆T; HCl + NaOH → NaCl + H2O ∆Hrxn = 2.7kJ/0.05mol = 54kJ/mol q soln. = (4.18J/g-K)(100g)(6.5C) = 2700J (negative because heat was liberated). When two masses at different temperatures come into contact, heat from the warmer mass will transfer to the cooler mass to reach thermal equilibrium (Teq = equilibrium temperature). Therefore: qwarm = qcool Ex. If 20g of warm water (90C) is mixed with 30g of cool water (6.0C), what is Teq? mCpT1 = mCpT2 (20g)(4.184 J/g-K)( 90C-Teq) = (30g)(4.184 J/g-K)(Teq - 6.0C) 7531.2 - 83.68Teq = 125.52Teq - 753.12 209.2Teq = 8284.32 Teq = 39.6C ; Teq is always between warm and cool temperature. Bomb Calorimerty- For this condition, q = Cpbomb x T ; Cpbomb is heat capacity of the bomb calorimeter. Ex. 4.00 grams CH6N2 (Mol Wt. 46 grams) is combusted in a bomb calorimeter; temp of calorimeter is raised 14.5C; Cp of bomb calorimeter is 7.794kJ/C. What is heat capacity per mole of compound? q = 7.794kj/C x 14.5C = 113kJ 4 grams CH6N2 is : (4/46) = 0.087mole , molar heat capacity is (113kJ/0.087) = 1299kJ/mole Hess’s Law- If a reaction is carried out in a series of reaction steps, H for the resultant reaction is equal to the sum of the H’s for the individual step reactions. Ex. Step 1 CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Hrxn1 = -890kJ Step 2 2H2O(l) 2H2O(g) Hrxn2 = 88kJ --------------------------------------------------------------------CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = Hrxn1 + Hrxn2 = -802kJ Standard Enthalpy of Formation (Hf)- For a compound, it is the change in enthalpy for the reaction that forms 1 mole of the compound from its elements in their stable states at standard conditions. For element in their stable states, Hf is zero (i.e. Na atom, O2, H2, S8) Ex. 2Cgraphite + 3H2(g) + (1/2)O2(g) C2H5OH formation of C2H5OH. Hf = -277.7kJ for the Then the enthalpy of a reaction at standard (Hrxn) conditions is: Hrxn = nHf (products) - nHf (reactants) The Hf values are tabulated in textbooks, usually in kilojoules. Given 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) What is Hrxn? Hrxn = nHf(products) - nHf(reactants) Hrxn = [12Hf(CO2) + 6 Hf(H2O)] - [2Hf(C6H6) + 15Hf(O2)] = [12(-393.5 kJ) + 6(-285.8 kJ)] - [2(49.0 kJ + 15(0 kJ)] = -6534.8 kJ for 2 moles of benzene burned OR -3267 kJ/mol CHAPTER 6 – Thermochemistry Problems & Solutions Prof. Onwuachi 1. Calculate the total change in internal energy, U, of a system when 400 J heat is applied to expanding O2 (g) and the gas does 350 J of work on its surroundings. ΔU = q + w = 400J – 350J = 50J 2. Calculate the kinetic energy in J, calories, and kJ of a particle (mass = 9.11x10-28 g) at 6.00x105 m/s. E = (1/2)mv2 = (1/2)(9.11 x 10-31kg)(6.0 x 105 m/s)2 = 1.64 x 10-19J = (1.64 x 10-19J) ÷ 4.184J/cal = 3.93 x 10-20 calories = 1.64 x 10-22kJ 3. Given the following data: a) 1/2 Br2 (l) Br (g) H0 = 111.75 kJ b) 1/2 Cl2 (g) Cl (g) H0 = 121.38 kJ c) 1/2 Br2 (l) + 1/2 Cl2 (g) BrCl (g) H0 = 14.7 kJ determine H0rx for reaction; Br (g) + Cl (g) BrCl (g) Reverse a,=> ΔHº = -111.75 kJ Reverse b, => ΔHº = -121.38 kJ Add to c. => ΔHº = 14.7 kJ ΔH = -218.4 kJ 4. Calculate H0 in kJ for reaction; CuO(s)+ CO (g) Cu(s) + CO2 (g) given the following thermochemical data : (a) 2CO(g) + O2 (g) 2CO2(g) H0 = - 566.1 kJ (b) 2Cu(s) + O2 (g) 2CuO(s) H0 = - 310.5 kJ Multiply (a) by ½ => ΔHº = -566.1 kJ/2 = -283.05 kJ Reverse (b) & multiply by ½ => ΔHº = 310.5 kJ/2 = 155.25 kJ ΔHº = -128 kJ 5. Consider the following reactions (a) & (b) : Calculate Hf0 for H2O2 (l) ; H2 (g) + O2 (g) H2O2 (l) a) H2O2 (l) H2O (l) + 1/2 O2 (g) H0 = -98.3 kJ b) H2 (g) + 1/2 O2 (g) H2O (l) H0 = -285.8 kJ Reverse (a) => ΔHº = 98.3 kJ Add to (b) => ΔHº = -285.8 kJ ΔHº = -187.5 kJ 6. A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion 10.5 kJ/g. Calculate the heat capacity of the calorimeter apparatus in kJ/0C . Equation for bomb calorimeter: q = CbombΔT 10.5 kJ = Cbomb(32.0 ºC – 28.5 ºC) Cbomb = 3.00 kJ/ºC 7. What is the resulting temperature when 35.0 g of water at 75 0 C is mixed with 15.0 g of water at 10 0C? (Heat capacity (Cp) of water = 4.184 J/g. 0C) Equation: m1CpΔT1 = m2CpΔT2 (35g)(Cp)(75-Tf) = (15g)(Cp)(Tf – 10) Tf = 55.5 ºC 8. A 20.0 g piece of a metal with specific heat of 0.080 cal/g.0C at 68 0C dropped into 15.0 ml water in a calorimeter at 23 0C. Calculate the final equilibrium temperature of the mixture. Equation: m1CmetalΔT1 = m2CpΔT2 (20g)(0.08cal/g)(68 -Tf) = (15g)(1.0 cal/gºC)(Tf – 23) Tf = 27.34 ºC 108.8 – 1.6Tf = 15Tf - 345 453.8 = 16.6Tf 453.8/ 16.6 = Tf = 27.34 ºC 9. When 72 g of a metal at 97.00C is added to 100 g of water at 25.0 0C, the final temperature is 45.5 0C. What is the heat capacity per grams of the metal? (Heat capacity of H2O = 4.184 J/g. 0C) Equation: m1CmetalΔT1 = m2CpΔT2 (72g)(Cmetal)(97 – 45.5) = (100g)(4.184J/gºC)(45.5 – 25) Cmetal = 2.31 J/gºC 10. The heat capacity of lead is 0.13 J/g.0C. How many joules of heat would be required to raise the temperature of 15.0 g of lead from 22 0C to 38 0C? Equation : q = mCmetalΔT = (15g)(0.13J/gºC)(38 – 22) = 31.2 J Download 1. Science 2. Physics thermochem-prob-solns.doc chapter6-PEX.doc CHM 1411 Chapter 6.doc 1411-Quiz-6.doc Heat of fusion wax CHAPTER 6 – Thermochemistry (Answers) DR. PAHLAVAN 3.0 Hess's Law Thermochemistry PPT Worksheet 6A on Thermodynamics chapter-5-chem-ii1 Practice Problems ch 9 Steady State Nonisothermal Reactor Design Ch 8. Thermochemistry Chapter 9 Chemical Bonding studylib © 2017 DMCA Report