Thermochemistry - Energy Changes and Rates of Reactions

Thermochemistry The study of energy transformations that occur with chemical reactions.  Exothermic Reactions  Transfer heat from the chemical system (the reaction), to chemical the surroundings. productsset system cractant changing...

Thermochemistry The study of energy transformations that occur with chemical reactions.  Exothermic Reactions  Transfer heat from the chemical system (the reaction), to chemical the surroundings. productsset system cractant changing t  This transfer increases the thermal energy of the particles in the surroundings, which makes them move faster, thus increasing the temperature of the surroundings. "exothermic leaving system "endothermic" : Inside system Endothermic Reactions  Cause a transfer of heat from the surroundings to the chemical system. stored energy. think  It is changed into potential energy, so the temperature of the chemical system does NOT increase.  However, the temperature of the surroundings will experience a decrease. ↳ test tube feels cold Types of Systems ube  OPEN systems allow matter and energy in and out. (e.g. a car engine) (e.g. a closed container) -  CLOSED systems prevent the flow of matter but not energy.  ISOLATED systems prevent the flow of both energy and matter. (e.g. a closed and insulated container) These are rare. - Calculating Energy Changes in Systems  A formula can be used to calculate energy changes occuring - in systems: - Q = mc∆t ⑯ heat absorbed (endothermin released Lexothermic where: ⑦ heat Q = heat of the system (if it’s a positive value, - heat has been absorbed, and if it’s a negative - value, heat has been released). Measured in joules - (J). m = mass. Measured in grams (g).  c = substance’s specific heat capacity, which is the amount of energy needed to increase the temperature of a gram of substance by one degree Celcius. Measured in J/g∙°C. ∆t = change in temperature. Tfinal – Tinitial. Measured in °C. *Now read the sample problem on p.281. · M ~ / t 8 · ~ · r - - - 7 Y · N - - E 8 N - : · o ( ↑ : i i You can use this equation to solve for any of the variables. E.g: 6 0) : a = 250 C=. m 2 10. 09 ° C ti = 20. 0 =? = 910)(g R: t ) : 2. A : G   The ENTHALPY of a system is the total energy of a system, plus the pressure times the - volume.  Scientists measure the enthalpy change, ∆H, of a system, to study thermochemical changes. This value compares the initial and final states of the chemical system. -  Since there are different enthalpies that can be calculated, subscripts can be used on the enthalpy symbol: # Specific e.g. ∆Hsol = enthalpy change associated with a XH (enthalpies) solute dissolving (solution) e.g. ∆Hcomb = enthalpy change associated with a substance undergoing a combustion L combustion  Govern energy changes  First Law: Energy can be converted from one form to another, but can’t be created or opposites = destroyed. Therefore, Esystem = - Esurroundings  Second Law: When two objects are in thermal contact, heat is always transferred from the object at a higher temperature to the object at a lower temperature until the two objects are the same temperature. At this point, we say they are in ‘thermal equilibrium’.  *  Energy is used to overcome or allow intermolecular forces to act.  Particles remain unchanged at the molecular level. ↳ physicales  Typical enthalpy changes are in the range ∆H = 100 – 102 kJ/mol. -  Examples include changes of state, and dissolving substances.  -  Energy changes overcome the electronic - structure and chemical bonds within the - reactions particles. - chemical  New substances with new chemical bonding (and therefore properties), are formed.  Typical enthalpy changes are in the range - ∆H = 102 – 104 kJ/mol.  Examples include combustion reactions, and other chemical reactions (e.g. single - - displacement, synthesis, etc.)  -  Energy changes overcome the forces between - protons and neutrons in nuclei. breaking atums -  New atoms with different numbers of protons or neutrons are formed.  Typical enthalpy changes are in the range ∆H = 1010 – 1012 kJ/mol.  Examples include nuclear decay, and fission and - fusion reactions. - Enthalpy and Moles  Another formula that can be used to calculate enthalpy is: & where: L ΔH = n ΔH specific associated with the change accompanying ΔH = the overall enthalpy of the reaction (kJ) - A mal of - n = the number of moles of substance (mol) substance h - - ΔHspecific = the molar enthalpy of the substance (kJ/mol) - - - (This value will have a specific short form associated with it, e.g. ΔHcomb, ΔHvap, ΔHfus, etc.) Note: sometimes, a question will give you a mass. If this is the case, F - convert the mass to moles using the molar mass of the substance, (which you will also have to calculate). E.g. Determine the thermal energy released by the Notcoms combustion of 56.78 g of hexane, C6H14(l). Its molar enthalpy of combustion is -4163.2 kJ/mol.  First calculate the molar mass of hexane: MC6H14(l) = 6(12.01g/mol) + 14(1.01g/mol) = 86.2g/mol  Then use the molar mass and mass to calculate the number of moles in this particular mass: > n = m/M - = (56.78g)/(86.2g/mol) = 0.659 mol (approximately)  Wur u  Next substitute into the formula. (Remember to include the formula and units throughout to receive full marks!) ΔH = n ΔHcomb = (0.659mol)(-4163.2kJ/mol) because * no negative = -2743.5488 kJ say released you or -2.74 x 103 kJ ↳ exothermic ↑ L  Final sentence: Therefore approximately 2.74 x 103 kJ was released in - the combustion of 56.78g of hexane. - Representing Enthalpy Changes t  Enthalpy changes are represented from the perspective of the chemical system. Enthalpy is measured in kilojoules. There are 4 different ways you might see this: 1) Writing the energy in the reaction. A positive value is used. The energy is a reactant in endothermic reactions and a product in exothermic reactions.  e.g. H2O2(l) + 285.8 kJ → H2(g) + ½ O2(g) Endothermic (energy on left) CH = 285 8k).  e.g. Mg(s) + ½ O2(g) → MgO(s) + 601.6 kJ Exothermic (energy on right) LH = 601 6K) -. 2) Write the enthalpy change as ΔH after the equation.  e.g. H2O2(l) → H2(g) + ½ O2(g) ΔH = + 285.8 kJ Endothermic (energy is positive)  e.g. Mg(s) + ½ O2(g) → MgO(s) ΔH = - 601.6 kJ Exothermic (energy is negative) 3) The use of a potential energy diagram (enthalpy diagram)  Displays exothermic reactions drawn with the energy of the products lower than the energy o f the reactants (because energy was released). ·  In endothermic reactions, the products are drawn at an energy level higher than the reactants, as energy was absorbed by the chemical system. -  Enthalpy Diagram Examples Exothermic Reactions: · al potent E.g. ↓ ↳ how long it takes for theeaction to proceed Enthalpy Diagram Examples Endothermic Reactions: E.g. 4) Write a molar enthalpy of reaction with the appropriate subscript on the ΔH.  Subscripts on the ΔH indicate which particular enthalpy we are dealing with.  For example, in ΔHr ,“r” stands for the “reaction”.  There are more specific ones. E.g. ΔHcomb represents combustion, while ΔHfre represents freezing.  If at SATP (standard atmospheric temperature and pressure), then the standard molar enthalpy of reaction is used as shown by the degree sign ΔHo.  e.g. CO(g) + 2 H2(g) → CH3OH(l) would be * -auditions written as: meanssat ΔHfo = - 128.6 kJ/mol of CH3OH - C (to indicate a formation reaction.). molar entalpies always have units KJ/md Notes about molar enthalpies:  You have to make sure to properly balance the equation first.  Use the coefficients in the chemical equation you are working with: you must either multiply or divide the enthalpy value accordingly.  E.g if the compound you’re looking at has a coefficient of three in the balanced equation, you must triple (multiply) the molar enthalpy value to obtain the overall enthalpy of the reaction.  If instead you are determining a molar enthalpy using a reaction, you must divide the overall enthalpy by the coefficient in the chemical reaction of the item you are determining the molar enthalpy of.  See the following examples… Examples on Using Molar Enthalpies - - + + Noi caq) AgNO3(s) > Ag (as) - (H = n@H(sol = (Imp) (22 64) Imp. , = 22. 6k] - - plugin Eg.2 D He n4Hccomb) = & mal 72871kJImol) = = 3742k) + 130219 > 8CO2) + 10H2UD + 3742 2CyHo(q) - Calorimetry is the process of measuring energy changes during a physical or chemical change.  Calorimetry is accomplished by using a calorimeter. This is an insulated reaction chamber with holes for a thermometer and stirring device.  The calorimeter holds water, so any change in the thermal energy of the SYSTEM is detected as a temperature change of the water. This Photo by Unknown Author is licensed under CC BY-SA-NC In calorimetry, the total amount of thermal energy absorbed or released by a chemical system is given by Q. Q = mc∆t  Water is the surroundings. We use Q for the surroundings to predict if the change will be exothermic or endothermic.  An increase in the temperature of water indicates an exothermic reaction, and a decrease indicates an endothermic reaction.  4H = n4Hsp - The system’s heat is represented by ∆H. The surrounding’s heat is represented by Q. Therefore, ↳ ∆H = - Q. surroundings the reflect in the change Example: 10 g of urea, NH2CONH3(s), is dissolved insystem 150 - - mL of water in a simple calorimeter. A temperature - I change from 20.4°C to 16.7°C is measured. Calculate the molar enthalpy of solution for the fertilizer. # Density When have you volumes given of water in these questions , Iglm) mL is = g GOOD FOR WATER AND ALL AQUEOUS SOLUTIONS 1) Write down everything the question gives you, separating items that belong to Q from ones that belong to ∆H. XH nLH = &= mclt Q: water (surroundings) 00 ∆H: urea (system) v = 150 mL → 150 g m = 10.0 g M = 60 g/mol (molarmass a o T1 = 20.4°C T2 = 16.7°C n = 10.0 g x 1 mol c = 4.19 J/g∙°C 60 g = 0.167 mol o not add masses # 2) Substitute your values into the appropriate equation. First start with the surroundings. - Q = mc∆t = (150 g)(4.19 J/g∙°C)(16.7°C - 20.4°C) = - 2325.45 J or – 2.33 kJ 3) Relate Q to ∆H. ∆H = - Q = - (- 2.33 kJ) = 2.33 kJ Remember: this makes sense because it represents an endothermic reaction. If you look back at the temperature change of water, it decreased, indicating the system absorbed heat from the surroundings. 4) Now find the molar enthalpy. ∆H = n∆Hsol ∆Hsol = ∆H n = 2.33 kJ 0.167 mol = 13.95 kJ/mol Therefore the molar enthalpy of solution for urea is 13.95 kJ/mol. Additional notes: -  You may do the substitution into ∆H = - Q all at once. like:↳ n∆Hsol = -(mc∆t). 1. Th For example, for the previous question this would look  If you have two items that you are reacting together, use whichever item is the limiting factor to determine the number of moles reacted. This may require you to use the formula c = n/v if using liquids, or n = m/M if using solids. (If both items have the same number of moles, just use that number.) Example of mixing two solutions: p.303 Note some of the differences:  Add the volume of both solutions to get the overall volume (this is “m” in the Q=mc∆t equation)  Determine the amount in moles of each solution to determine which one is the limiting factor (n=cV). Whichever value is LESS, becomes your “n” value for ∆H = n∆Hsol ↑ · # i  Hess’ Law (a.k.a. the law of additivity of reaction enthalpies), states that if a reaction is the sum of a series of reactions then the change in enthalpy is also the sum of the changes of enthalpies of the various steps.  ∆Htarget = ∆H1 + ∆H2 + ∆H3 + … known enthalpies > T or ∆Htarget = ∑ ∆Hknown - the looking for enthalpy of  To use Hess’ Law, we write the target equations as the sum of a series of known reactions and we add their - ∆H values to obtain the overall ∆H value. -  The reactions may need to be reversed which also flips - the sign for ∆H, and/or multiplied by a factor to yield - the correct number of moles of each component.  Also note that if you add the known equations to get the target equation, terms on each side of the equations will cancel, leaving you with just the target equation. (This is a great place to check your work.)  LH - ? ① correct side - - - Harget) & ② correct mols - - - - When doing these questions, try to match one item from each known equation with an item from the target equation, that is only common to both of them. Then check to see if: 1) it’s on the right side or if you need to flip it, and 2) if you have to modify the number of moles.  A writehis ↓  In the first known equation, propanol matches with the target. We need two moles of it and it has to be on ① the product side, so we will multiply the equation and the enthalpy by -2. ·2 multiply -  In the second known equation, carbon matches with the target. It is on the right side, but we need 6 moles 6 of it, so we will multiply the equation and enthalpy by - Q 6.  In the third known equation, hydrogen matches with the target. It is on the right side, but we need 8 moles Q of it, so we will multiply the equation and enthalpy by 8. Notice that when the equations are added, they give you the target. - also with enthalpy ( j &- f - - - I 2 + + - = - y -/ I - rid of common exactants comme - Examples  First use the standard molar enthalpies to calculate ∆H. use char -743 pg - - & and. chem. ear element has H value of zero  Now calculate the molar enthalpy. Termites  Efficiency: the ratio of useful energy output to the amount of energy used, expressed as a percent. Formula: 𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡𝑝𝑢𝑡 Efficiency = x 100% 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑝𝑢𝑡 (You studied a simple version of this calculation in grade 9 electricity.)  Example problem – p.326 Energy input in this question involves figuring out a molar enthalpy (based on enthalpies of formation), and using that to determine an overall enthalpy of the reaction using that amount of propane. Energy output involves calculating energy that is absorbed by the water and pot using Q=mc∆t. Then the formula for efficiency is used.  Most reactions are not efficient. I.e. a lot of energy is lost/transferred to the surroundings. (Remember when we said most systems are open?) But, there are ways that can be used to improve the efficiency of reactions… To improve efficiency - More input energy is required to be converted to useful output energy. - If every energy transformation is less than 100%, the more number of transformations required, the lower the efficiency. So REDUCE the number of transformations. E.g. gas heating vs. gas burning electricity plant - You can also REDUCE waste energy by using it. (E.g. condensing gas furnace) Energy Sources and the Environment - to assess the impact consider the following: 1. Are any waste products or by-products of energy production harmful to living things or the environment? 2. Is obtaining or harnessing the fuel harmful to living things or the environment? 3. Will using the energy resource permanently remove the fuel from the environment? -nonrenewable –takes millions of years to replace itself -renewable –replenished in a short period of time. E.g. human life time absolute valu I reactant : - += + consumption production · volume a mass · temperate  > - new substances new formed with properties Is proper orientation ↳ enough force * You canrelate rates of items chemical in a ratios equation to each other using their mole e. g rate of B = rate of AX - · activation energy diana - ~furne cate complex activation Earev) is the reverse energy for the & from the p 365 reaction runs. products to activation complex  sometime wil i - sa Willstep a be the with the nis reactant #males , - sometimes listed nically cure(grio e overed when highest energet in  activated complex creactants) < products) E Additional Notes on Collision Theory and Rate Law  Threshold Energy  The threshold energy is the minimum energy that a collision must have in order to change kinetic energy into the activation energy. (tip of curve)  The chemical nature of the reactants is what determines the threshold energy. More reactive compounds have a lower threshold energy.  Factors Affecting Reactions  Increasing the surface area or the concentration will increase the number of collisions (because in each case more particles are made available to collide), but not the percentage of effective collisions.  Increasing the temperature will increase the number of collisions and the percentage of effective collisions. (Recall: the kinetic energy of the particles is being increased.)  Factors Affecting Reactions (cont’d)  Catalysts provide an alternative pathway that has a lower activation energy and thus the percentage of end Ge effective collisions increases dramatically.  Catalysts may be homogeneous catalysts (which means they are in the same physical state as the reactants), or they may be heterogeneous catalysts, (which means they are in a different physical state than the reactants).  Rate Law Exponents Zero Order Reactions  Have an exponent of 0  Adjusting the concentration of the reactant has NO effect on the rate  Rate Law Exponents (cont’d) First Order Reactions  Have an exponent of 1  The rate is proportional to the concentration of the reactant (e.g. if the concentration of the reactant is doubled, the rate will double)  Linear relationship (linear graph) I  Rate Law Exponents (cont’d) Second Order Reactions  Have an exponent of 2  If the concentration of a reactant is doubled, the rate will be quadrupled (i.e. rate is raised to the exponent 2)  The graph is curved (e.g. parabolic shape)  ΔH and Activation Energy  You can determine ΔH from a potential energy diagram by looking at the Ea(fwd) and the Ea(rev).  The following formula is used: ΔH = Ea(fwd) - Ea(rev)  If the reaction is exothermic, ΔH is negative and Ea(fwd) < Ea(rev)  If the reaction is endothermic, ΔH is positive and Ea(fwd) > Ea(rev)