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Energy in Chemistry

by: Nick Manning

Energy in Chemistry CHEM - 10060 - 001

Nick Manning
GPA 4.0
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Notes from 2/29 to 3/4, covers Calorimetry, endothermic and exothermic reactions, a bunch of formulas and equations, and hopefully clears up some things. Or maybe just helps review. Anyways, it'll ...
Class Notes
Chemistry, Energy, endothermic, exothermic





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This 9 page Class Notes was uploaded by Nick Manning on Sunday February 28, 2016. The Class Notes belongs to CHEM - 10060 - 001 at Kent State University taught by TBA in Fall 2015. Since its upload, it has received 25 views. For similar materials see GENERAL CHEMISTRY I in Chemistry at Kent State University.


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Date Created: 02/28/16
Week of 2/29- 3/4 ENERGY IN CHEMISTRY Energy is the ability to do work. It is neither created nor destroyed, simply transferred. Kinetic energy and potential energy are present in chemical bonds SOURCES OF KINETIC ENERGY IN CHEMICAL SYSTEM -particles move and rotate through space -bonds vibrate -electrons move around nucleus SOURCES OF POTENTIAL ENERGY IN CHEMICAL SYSTEM -bonded atoms attract and repel -the attraction between nucleus and electrons, also the repulsion between electrons and electrons - Attraction b/w nucleus and shared electrons During chemical reactions, change occurs in BONDING - Changes in the way atoms are bonded involve changes in chemical potential energy - System- the reaction itself - Surroundings- immediate area around reaction - Everything else equals the universe - Heat goes from system  surroundings or surroundings  system - You have to clearly define the system in order to measure energy changes Systems are classified by how it interacts Open system Closed system Isolated Matter & heat exchanged only heat exchanged neither heat nor energy exchanged FIRST LAW OF THERMODYNAMICS ⧍E system = -⧍E surroundings Also⧍Energy system= E fin- Einitial Total energy of the system is a state function State Function- independent of pathway, doesn’t matter how you got there as long you get there; only depends on initial and final states Can either be released as all heat (q) or as heat and work (q and w) ⧍E=q+w Think of a hobo fire think of a car, moving & burning gas is expressed as Joules (J) or kilojoules (kJ). when a system releases heat… qsys< 0 when a system absorbs heat… q sys> 0 If Energy is transferred as Heat only, tqesys =-qsurroundings When the system does work on surroundings, W sys< 0 When work is done on system by surroundings, W sys>0 Energy transferred as only work sys= -Wsurroundings Usually demonstrated with gasses and pistons W= (-p)(⧍V) ⧍E= q+w ⧍E= q- (p⧍V) - To determine ⧍E, we need to we need to know q & w - Instead we use H, means enthalpy H= Heat constant of system @ constant pressure (p) - ⧍H= energy transferred as heat from constant p - ⧍H can be determined from only q, if p kept constant o ⧍E=⧍H=qp When no gasses are involved, ⧍v= 0 (volume) If total moles of gas is unchanged ⧍V=0 CALORIMETRY Exothermic Reaction- heat is released to surroundings; feels hot ⧍Tsurr 0 Endothermic Reaction - absorbs heat energy from surroundings ⧍Tsurr 0 Specific Heat Capacity- amount of heat required to change the substance 1ᵒ o How much heat do you need to make _____ increase by 1 degree? **Temperature wise, 1ᵒC ≠ 1ᵒK, but ⧍1ᵒC does = ⧍1ᵒK ** q=mc⧍T heat energy in Joules (J) Specific Heat Capacity (J/g * K) Mass in G Temp changefininitial) ** You should probably know this formula, Dr. Leslie doesn’t like to give formulas on tests and it is important!!* EXAMPLE 250g of H2O is heated, receiving 752 J of heat. What’s the final temperature if the initial temperature is equal to 22.0ᵒC? (c for H2O= 4.18 J/g * K) Okay, so we know so far that m= 250g , q=752 J, c=4.18, ani T = 22.0ᵒC. 752=(250)(4.18)(x-22.0) 752=1045(x-22.0) .7196=x-22.0 22.7196=x w/ sig figs the final temp = 22.7 ᵒC - For simple heat exchange b/w two objects: o Masses do not combine o q is exchanged until objects reach the same q(system)=-q(sample) q(cal)=-q(sample) endothermic) higher amount of system heat (q sys before reaction, lower amount of system heat after reaction + qsys = - qsys Exothermic) lower q sysefore reaction, higher sysafter - qsys = + qsys For Chemical reactions in a calorimeter -the reaction is the system (A+B⧍ C) - everything else (including H2O & resulting solution) = SURROUNDINGS Stored chemical potential energy Heat energy / surroundings THERMOCHEMICAL EQUATIONS As we know, an exothermic process releases heat to its surroundings. ⧍H (enthalpy) = finalHinitialproductreactants ⧍H < 0 And we know endothermic absorbs heat from surroundings (ice melting for example, the ice takes heat from the surrounding water) ⧍H > 0 Now we have to learn to make the graphs from chemical equations. Best way to do this is with a good ol’ example. CH4 + 2 O2 ⧍ CO2 + 2 H2O; ⧍H = 890 kJ Negative, so exothermic CH4+2O 2 reactants H CO2+2H 2 products The arrow is always drawn from the reactants to the products. It represents the ⧍H, which is negative here so the arrow goes down on the H chart. Now that we got that, we move on to the next thing; calculating certain things from the amount of mols and grams in a thermochemical equation. CH4 (g) + 2O2 (g) ⧍ CO2 (g) +2 H2O (g); ⧍H = 890 kJ If asked “How many kJ of heat is released from 1 mol of4CH ?” how would we do this? Easy, this thermochem equation shows 1 mol of CH4reacting so it would just be 890 kJ. Okay, well how about how much is released from 1 mol of 2 ? Well this shows 2 mols, so we divide the ⧍H by 2 and get 445 kJ. Let’s move up on the difficulty scale and try “What is the amount of heat released from 4.50 g 4H ?” How do we get from grams to moles? Molar mass, homes. 4.50 g CH4* 1 mol CH 4 16.04 g CH4* -890 kJ/ 1 mol C4 = -250. kJ released MANIPULATING THERMOCHEMICAL EQUATIONS We know ⧍H is always given in an equation as written. 2H2O2 (g) + O2 (g) ⧍2 H2O (l); ⧍H =572 kJ If the equation is changed, the ⧍H is also changed. H2O2 (g) + ½ O2 (g) ⧍ H2O (l); ⧍H =286 kJ When you reverse the reaction, the ⧍H stays the same but with the opposite sign. 2 H2O (l) ⧍2H2O2 (g) + O2 (g); ⧍H = 572 kJ When ⧍H is immeasurable due to the reaction being part of a sequence of reactions that all happen at nearly the same time, or it is too dangerous, ori t takes too long to wait for, we use Hess’s law to break the reaction in to parts and add up the ⧍H of the separate part to get the whole. We will be talking about this more on Monday, but I do have an example to check out. S(s) +3/2 O2 (g) ⧍ SO3 (g); what is ⧍H? S(s) + O2 (g) ⧍ SO 2 (g) ⧍H= -296.8 kJ 2 SO2 (g) + O2 (g)2⧍SO3 (g) ⧍H= -198.4 kJ We have to get the elements to equal the equation where we are trying to find ⧍H, so we have to mess around with some equation manipulation. Let’s aim to get the O to 3/2. SO2 (g) +½ O2 (g) ⧍ SO3 (g) ⧍H= -198.4/2 = -99.2 kJ S(s) + O2 (g) ⧍ SO 2 (g) ⧍H= -296.8 kJ We can cross out the 2O because it is on both sides of the equations we are trying to add up. SO2 (g) +½ O2 (g) ⧍ SO3 (g) ⧍H= -198.4/2 = -99.2 kJ S(s) + O2 (g) ⧍ SO 2 (g) ⧍H= -296.8 kJ S(s) +3/2 O2 (g) ⧍ SO3 (g); ⧍H = -99.2 + -296.8 = 396 kJ


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