© 2014 Pearson Education, Inc. Thermochemistry: Chemical Energy.

© 2014 Pearson Education, Inc. Energy and Its Conservation Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form into another. First Law of Thermodynamics Energy: The capacity to supply heat or do work. Kinetic Energy (E K ): The energy of motion. Potential Energy (E P ): Stored energy. ∆E = q + w

© 2014 Pearson Education, Inc. Energy and Its Conservation Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object. Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.

© 2014 Pearson Education, Inc. Some definitions Exothermic: energy is given off (temperature increases) Endothermic: energy is required (temperature decreases) Enthalpy,  H: Internal energy in kJ/mol -  H: exothermic +  H: endothermic

© 2014 Pearson Education, Inc. Coffee-cup calorimetry q = -mc  T q: quantity of heat in joules, J m: mass of liquid, g c: specific heat capacity, 4.18J/g o C  T: change in temperature in o C, T final – T initial q = nΔH or ΔH = q/n

© 2014 Pearson Education, Inc. There are two kinds of coffee- cup calorimetry problems. 1 st kind: Grams of a chemical are added to a given mass or volume of water. What to do: Grams of water will be plugged into m.  T, Last temperature minus first temperature. Solve for q and change joules to kJ. Grams of the other chemical convert to moles. Solve for  H by putting kJ over moles and divide.

© 2014 Pearson Education, Inc. Example #1 In a coffee-cup calorimeter, 5g of sodium are dropped in 300g of water at 20 o C the temperature rises to 35 o C. Calculate the enthalpy change for this reaction.

© 2014 Pearson Education, Inc. Example #2 A 1.5g sample of calcium is dropped into 275mL of water at 20 o C in a coffee-cup calorimeter. The temperature rises to 23 o C. The density of water is 1g/mL. What is the energy change for this reaction in kJ/mol?

© 2014 Pearson Education, Inc. There are two kinds of coffee- cup calorimetry problems. 2 nd kind: Two volumes of a solution are given. What to do: Add them and plug into m (the density of most solutions = 1g/mL, so the mL = the grams)  T, Final temperature minus Initial temperature. Solve for q and change joules to kJ. Pick one of the chemicals and multiply (L)(M) to find moles. Solve for  H by putting kJ over moles and divide.

© 2014 Pearson Education, Inc. Example #3 450mL of 2M HCl are mixed with 450mL of 2M NaOH in a coffee-cup calorimeter. The temperature rises from 23 o C to 25 o C. Calculate the enthalpy change for this reaction.

© 2014 Pearson Education, Inc. Example #4 In a coffee-cup calorimeter, 275mL of 0.5M Ca(OH) 2 react with 275mL of 0.5M H 2 SO 4. The temperature increases from 25 o C to 30 o C. Calculate the energy in kJ/mol.

© 2014 Pearson Education, Inc. Worked Example Calculating the Amount of Heat Released in a Reaction According to the balanced equation, 852 kJ of heat is evolved from the reaction of 2 mol of Al. To find out how much heat is evolved from the reaction of 5.00 g of Al, we have to find out how many moles of aluminum are in 5.00 g. Solution The molar mass of Al is 26.98 g/mol, so 5.00 g of Al equals 0.185 mol: Because 2 mol of Al releases 852 kJ of heat, 0.185 mol of Al releases 78.8 kJ of heat: ✔ Ballpark Check Since the molar mass of Al is about 27 g, 5 g of aluminum is roughly 0.2 mol, and the heat evolved is about (852 kJ/2 mol)(0.2 mol), or approximately 85 kJ. Strategy How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a stoichiometric amount of Fe 2 O 3 ?

© 2014 Pearson Education, Inc. Two kinds of Bomb also: 1 st kind: Enthalpy (kJ/mol) is given What to do: Convert grams to moles Multiply the moles by the enthalpy. This answer plugs in for q Find C or  T (whichever it asks for)

© 2014 Pearson Education, Inc. Example #5 A 5g sample of hydrogen is burned in a bomb calorimeter. The enthalpy of combustion for hydrogen is -242kJ/mol. What is the heat capacity of the calorimeter if the temperature rises 5 o C?

© 2014 Pearson Education, Inc. Example #6 30g of magnesium are burned in a bomb calorimeter that has a heat capacity of 18kJ/ o C. If the enthalpy of combustion for magnesium is -602kJ/mol, how much will the temperature rise?

© 2014 Pearson Education, Inc. Still Bomb 2 nd kind: Enthalpy is asked for (or energy change in kJ/mol) What to do: Convert grams to moles. Plug in C and  T and solve for q. Solve for  H by putting kJ over moles and divide.

© 2014 Pearson Education, Inc. Example #7 If 10g of Ca are burned in a bomb calorimeter that has a heat capacity of 12.8kJ/ o C, the temperature increases by 12.38 o C. What is the enthalpy of combustion of calcium?

© 2014 Pearson Education, Inc. Example #8 An 18g sample of sulfur is burned in a bomb calorimeter. The heat capacity of the calorimeter is 8.9kJ/ o C and the temperature increases 18.74 o C. What is the enthalpy of combustion for sulfur?

What is “Thermodynamics”? The study of energy and its interconversions.

© 2014 Pearson Education, Inc. Some terms: Enthalpy: Internal energy (energy stored in the bonds)  H kJ/mol (+ endothermic, - exothermic) Entropy: disorder of a substance  S J/mol. K gases  aqueous  liquids  solids most disordered to least disordered Gibbs Free Energy: stored energy less the degree of disorder  G kJ/mol

© 2014 Pearson Education, Inc. How to calculate  H,  S &  G  H o =   H f o for products –  H f o for reactants  S o =  S o of products –  S o for reactants  G o =   G f o of products –  G f o for reactants O superscripted means standard states (1atm, 273K, etc)

© 2014 Pearson Education, Inc. Standard Heat of Formation (∆H° f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. ∆H° f = -74.8 kJ CH 4 (g)C(s) + 2 H 2 (g) Standard states 1 mol of 1 substance Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. c C + d Da A + b B ∆H° = ∆H° f (Products) - ∆H° f (Reactants) ReactantsProducts ∆H°= [c ∆H° f (C) + d ∆H° f (D)] - [a ∆H° f (A) + b ∆H° f (B)] Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H° = ? H°= [∆H° f (C 6 H 12 O 6 (s))] - [6 ∆H° f (CO 2 (g)) + 6 ∆H° f (H 2 O(l))] = 2802.5 kJ [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] ∆H°= [(1 mol)(-1273.3 kJ/mol)] - Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. How to calculate  H,  S &  G  S o and  G o can be calculated the same way as  H o  S o =  S o of products –  S o for reactants  G o =   G f o of products –  G f o for reactants O superscripted means standard states (1atm, 273K, etc)

© 2014 Pearson Education, Inc. Predicting Entropy To decide which substance has more entropy, first look at its state (solid, liquid, etc.). The more disordered state will have more disorder. If they have the same state, the larger molecule will have more entropy. Both gaseous, higher temperature, less pressure, more disorder.

© 2014 Pearson Education, Inc. Predicting  S change in a reaction. Look at both sides of the reaction, identify the state of the chemicals, how many moles of chemicals are present (more particles, more disorder)and how big the particles are. These are in order of importance. If the right has more disorder,  S goes up. If the right has less disorder,  S goes down.

© 2014 Pearson Education, Inc. Worked Example Predicting the Sign of Δ S for a Reaction Look at each reaction, and try to decide whether molecular randomness increases or decreases. Reactions that increase the number of gaseous molecules generally have a positive ΔS, while reactions that decrease the number of gaseous molecules have a negative ΔS. Solution a. The amount of molecular randomness in the system decreases when 2 mol of gaseous reactants combine to give 1 mol of liquid product, so the reaction has a negative ΔS°. b. The amount of molecular randomness in the system increases when 9 mol of gaseous reactants give 10 mol of gaseous products, so the reaction has a positive ΔS°. Strategy Predict whether ΔS° is likely to be positive or negative for each of the following reactions: a. b.

© 2014 Pearson Education, Inc. Spontaneity A spontaneous reaction occurs without the need of outside energy. Do not confuse spontaneous with instantaneous. Spontaneous happens on its own, instantaneous happens right away.

© 2014 Pearson Education, Inc. How can you tell?  S + tends to be spontaneous  H – tends to be spontaneous The only way to know for sure is if  G is -, then it is guaranteed to be spontaneous.  G o =  H o - T  S o  G =  H - T  S

© 2014 Pearson Education, Inc. Example #4 Predicting the signs of ΔH, ΔS and ΔG A. When solid ammonium chloride is dissolved in water, it forms an aqueous solution and the temperature drops. B. When hydrogen gas is mixed with oxygen gas and a spark ignited, a fire ball to the ceiling occurs. 2H 2(g) + O 2 (g)  2H 2 O (g)

© 2014 Pearson Education, Inc. Worked Conceptual Example Predicting the Signs of Δ H, Δ S, and Δ G for a Reaction First, decide what kind of process is represented in the drawing. Then decide whether the process increases or decreases the entropy (molecular randomness) of the system and whether it is exothermic or endothermic. Solution The drawing shows ordered particles in a solid subliming to give a gas. Formation of a gas from a solid increases molecular randomness, so ΔS is positive. Furthermore, because we’re told that the process is nonspontaneous, ΔG is also positive. Because the process is favored by ΔS (positive) yet still nonspontaneous, ΔH must be unfavorable (positive). This makes sense, because conversion of a solid to a liquid or gas requires energy and is always endothermic. Strategy What are the signs of ΔH, ΔS, and ΔG for the following nonspontaneous transformation?

© 2014 Pearson Education, Inc. More info about  G If  G is negative the reaction is spontaneous (runs forwards) If  G is positive it is not spontaneous forwards (can be spontaneous backwards) If  G = 0 it is at equilibrium (no net forward or backward movement)

© 2014 Pearson Education, Inc. Temperature and spontaneity You can change the temperature of reactions and they will become spontaneous at their new temperature. Phase changes: The boiling point and freezing (melting) points are temperatures where  G =0.

© 2014 Pearson Education, Inc. Worked Example Using the Free-Energy Equation to Calculate Equilibrium Temperature The spontaneity of the reaction at a given temperature can be found by determining whether ΔG is positive or negative at that temperature. The changeover point between spontaneous and nonspontaneous can be found by setting ΔG = 0 and solving for T. Solution At 25 °C (298 K), we have Because ΔG is positive at this temperature, the reaction is nonspontaneous. The changeover point between spontaneous and nonspontaneous is approximately The reaction becomes spontaneous above approximately 1120 K (847 ° °°C). Strategy Lime (CaO) is produced by heating limestone (CaCO 3 ) to drive off CO 2 gas, a reaction used to make Portland cement. Is the reaction spontaneous under standard conditions at 25 ° °°C? Calculate the temperature at which the reaction becomes spontaneous.

© 2014 Pearson Education, Inc. Haber Process: Multiple-Step Process N2H4(g)N2H4(g)2 H 2 (g) + N 2 (g) Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. 2 NH 3 (g)3 H 2 (g) + N 2 (g) ∆H°= -92.2 kJ ∆H° 1 = ? 2 NH 3 (g)N 2 H 4 (g) + H 2 (g) ∆H° reaction = -92.2 kJ 2 NH 3 (g)3 H 2 (g) + N 2 (g) ∆H° 2 = -187.6 kJ Calculating Enthalpy Changes Using Hess’s Law

© 2014 Pearson Education, Inc. It often takes some trial and error, but the idea is to combine the individual reactions so that their sum is the desired reaction. The important points are that: All the reactants [CH 4 (g) and O 2 (g)] must appear on the left. All the products [CO 2 (g) and H 2 O(l)] must appear on the right. All intermediate products [CH 2 O(g) and H 2 O(g)] must occur on both the left and the right so that they cancel. A reaction written in the reverse of the direction given [ ] must have the sign of its ΔH° reversed (Section 8.7). If a reaction is multiplied by a coefficient [ is multiplied by 2], then ΔH° for the reaction must be multiplied by that same coefficient. Worked Example Using Hess’s Law to Calculate Δ H ° Strategy Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water: Use the following information to calculate ΔH° in kilojoules for the combustion of methane:

© 2014 Pearson Education, Inc. As in Worked Example 8.6, the idea is to find a combination of the individual reactions whose sum is the desired reaction. In this instance, it’s necessary to reverse the second and third steps and to multiply both by 1/2 to make the overall equation balance. In so doing, the signs of the enthalpy changes for those steps must be changed and multiplied by 1/2. Note that CO 2 (g) and O 2 (g) cancel because they appear on both the right and left sides of equations. Worked Example Using Hess’s Law to Calculate Δ H ° Strategy Water gas is the name for the mixture of CO and H 2 prepared by reaction of steam with carbon at 1000 °°C: The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information to calculate ΔH° in kilojoules for the water-gas reaction:

© 2014 Pearson Education, Inc. Example #8 Using the reactions: C (s) + O 2(g)  CO 2(g)  H = -393.5kJ/mol CO (g) + ½ O 2(g)  CO 2(g)  H = -283kJ/mol Calculate the enthalpy for: C (s) + ½ O 2(g)  CO (g)

© 2014 Pearson Education, Inc. Example #9 Calculate  H for the reaction: 2C (s) + H 2(g)  C 2 H 2(g) Given the following: 2C 2 H 2(g) + 5O 2(g)  4CO 2(g) + 2H 2 O (l)  H = -2599.2kJ/mol C (s) + O 2(g)  CO 2(g)  H = -393.5kJ/mol 2H 2(g) + O 2(g)  2H 2 O (l)  H = -571.6kJ/mol

© 2014 Pearson Education, Inc. Example #10 Given the following data: H 2(g) + ½ O 2(g)  H 2 O (l)  H = -286kJ N 2 O 5(g) + H 2 O (l)  2HNO 3(l)  H = -77kJ ½ N 2(g) + 3 / 2 O 2(g) + ½ H 2(g)  HNO 3 (l)  H = -174kJ Calculate the  H for: 2 N 2(g) + 5 O 2(g)  2 N 2 O 5(g)

© 2014 Pearson Education, Inc. Example #11 Calculate  G for the reaction 2CO (g) + O 2(g)  2CO 2(g) From the following: 2CH 4(g) + 3O 2(g)  2CO (g) + 4H 2 O (g)  G = - 1088kJ/mol CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)  G = -801kJ/mol

© 2014 Pearson Education, Inc. Example #12 Calculate  G for the reaction C 2 H 4 (g) + H 2 O (l)  C 2 H 5 OH (l) From the following: 4C (s) + 6H 2(g) + O 2(g)  2C 2 H 5 OH (l)  G = - 700kJ/mol 2H 2(g) + O 2(g)  2H 2 O (l)  G = - 474kJ/mol 2C (s) + 2H 2(g)  C 2 H 4(g)  G = 136kJ/mol

© 2014 Pearson Education, Inc. More Ways to find  G!  G o = - R T lnK  G =  G o + RT lnQ K: equilibrium constants (Kc, Kp, Ka, Kb, Kw, Ksp….) Q: reaction quotient R: 0.008314kJ/mol. K T: temperature in Kelvin

© 2014 Pearson Education, Inc. Example #13 At 127 o C the synthesis of ammonia has the following equilibrium concentrations: [NH 3 ] = 0.031M [N 2 ] = 0.85M [H 2 ] = 0.0031M N 2(g) + 3H 2(g)  2NH 3(g) Calculate  G o for this reaction using the equilibrium concentrations.

© 2014 Pearson Education, Inc. Example #14 Calculate  G for the synthesis of methanol at 25 o C where carbon monoxide is present at 5atm and hydrogen gas is present at 3atm. CO (g) + 2H 2(g)  CH 3 OH (l) Is this reaction spontaneous with these pressures at this temperature?

© 2014 Pearson Education, Inc. Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H° = ? H°= [∆H° f (C 6 H 12 O 6 (s))] - [6 ∆H° f (CO 2 (g)) + 6 ∆H° f (H 2 O(l))] = 2802.5 kJ [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] ∆H°= [(1 mol)(-1273.3 kJ/mol)] - Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. 1.Reverse the “reaction” and reverse the sign on the standard enthalpy change. C(s) + O 2 (g)CO 2 (g) Why does the calculation “work”? ∆H° = 393.5 kJ Becomes CO 2 (g)C(s) + O 2 (g) ∆H° = -393.5 kJ C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2) Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. 1.Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. 6 C(s) + 6 O 2 (g)6 CO 2 (g) Why does the calculation “work”? ∆H°= 6(393.5 kJ) = 2361.0 kJ Becomes C(s) + O 2 (g)CO 2 (g) ∆H°= 393.5 kJ C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2) Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. 2.Reverse the “reaction” and reverse the sign on the standard enthalpy change. C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H° = 2802.5 kJ Why does the calculation “work”? ∆H°= 285.8 kJ Becomes ∆H°= -285.8 kJ (3)(1)(2) H 2 (g) + O 2 (g)H 2 O(l) 2 1 H 2 (g) + O 2 (g) 2 1 Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. 2.Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. 6 H 2 (g) + 3 O 2 (g)6 H 2 O(l) Why does the calculation “work”? ∆H°= 6(285.8 kJ) = 1714.8 kJ Becomes C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H°= 2802.5 kJ (3)(1)(2) ∆H° = 285.8 kJ H 2 (g) + O 2 (g)H 2 O(l) 2 1 Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H°= 1714.8 kJ Why does the calculation “work”? 6 H 2 (g) + 3 O 2 (g)6 H 2 O(l) C 6 H 12 O 6 (s)6 C(s) + 6 H 2 (g) + 3 O 2 (g) 6 C(s) + 6 O 2 (g)6 CO 2 (g) ∆H°= 2802.5 kJ ∆H°= -1273.3 kJ ∆H°= 2361.0 kJ C 6 H 12 O 6 (s) + 6 O 2 (g)6 CO 2 (g) + 6 H 2 O(l) ∆H°= 2802.5 kJ (3)(1)(2) Calculating Enthalpy Changes Using Standard Heats of Formation

© 2014 Pearson Education, Inc. The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state and is thus the amount of energy released when the bond forms. or Standard enthalpy changes for the corresponding bond-breaking reactions. Bond Dissociation Energy: Calculating Enthalpy Changes Using Bond Dissociation Energies

© 2014 Pearson Education, Inc. 2 HCl(g)H 2 (g) + Cl 2 (g) (2mol)(432 kJ/mol) ∆H°= D(Reactant bonds) - D(Product bonds) ∆H°= (D H-H + D Cl-Cl ) - (2 D H-Cl ) ∆H°= [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] - = -185 kJ Calculating Enthalpy Changes Using Bond Dissociation Energies

© 2014 Pearson Education, Inc. Worked Example Using Bond Dissociation Energies to Calculate Δ H ° Identify all the bonds in the reactants and products, and look up the appropriate bond dissociation energies in Table 8.3. Then subtract the sum of the bond dissociation energies in the products from the sum of the bond dissociation energies in the reactants to find the enthalpy change for the reaction. Solution The reactants have four C–H bonds and three Cl–Cl bonds; the products have one C–H bond, three C–Cl bonds, and three H–Cl bonds. The approximate bond dissociation energies from Table 8.3 are: Strategy Use the data in Table 8.3 to find an approximate ΔH° in kilojoules for the industrial synthesis of chloroform by reaction of methane with Cl 2.

© 2014 Pearson Education, Inc. Worked Example Using Bond Dissociation Energies to Calculate Δ H ° Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the approximate enthalpy change for the reaction: The reaction is exothermic by approximately 330 kJ. Continued

© 2014 Pearson Education, Inc. Thermochemistry: Chemical Energy.

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