Lecture 5 Professor Hicks Inorganic Chemistry (CHE152) review - PDF document

Lecture 5 Professor Hicks Inorganic Chemistry (CHE152) review Internal Energy (E) change in internal energy  E = q + w like cash or checks q heat released w work done by system – (negative in sign) - (negative in sign) system surroundings q heat absorbed work done on system + (positive in sign ) + (positive in sign) review examples of q and w • inside a piston fuel burns work releasing heat and releases heat -q expanding pushing the piston up doing work system • a disposable first aid cold pack does not change size absorbs much so work = 0, it absorbs heat +q system heat when it is activated makes surrounding cooler 1

Calculate the change in internal energy,  E, when a review system expands doing 155 J of work while absorbing 206 J of heat.  E = q + w absorbing heat adds energy to the system q = + 206 J doing work (expanding) removes E from system w = – 155 J  E = 206 J -155 J = 51 J q heat absorbed by system + (positive in sign) +206 J w work done by system - (negative in sign) - 155 J system surroundings review Enthalpy (H) • heat released under constant pressure = enthalpy change (  H) • Constant pressure - Conditions for reactions of life - Conditions of most chemical reaction unless they are run in constant volume containers • only reflects part of the internal energy change since it is the heat only not the work • the alternative to constant pressure is constant volume like the aging of wine in a glass bottle review Enthalpy is heat released under constant pressure change in change in enthalpy internal energy  H = q  E = q + w heat work C 6 H 12 O 6 (g) + 6O 2 (g)  6CO 2 (g) + 6H 2 O (g) system surroundings 2

review First law of thermodynamics energy cannot be created or destroyed applied when energy changes forms energy like a liquid Rudolf Clausius heat kinetic energy potential energy review When things go downhill… A 10 kg bowling ball at a height of 100 m has potential energy due to gravity Potential energy = mgh = 10 kg x 9.8 m/s2 x10 m = 9800 J As it falls it speeds up converting potential energy into kinetic energy Right before it hits ground all the potential energy has become kinetic energy = 9800 J Heat 9800 J After it hits the ground it is not moving so kinetic energy is zero- all the energy has become heat = 9800 J review Energy changes forms (but does not get destroyed) 1) potential energy = 9800 J 9800 J 9800 J on 9800 J as ball falls impact PE  KE KE  heat potential kinetic heat energy energy atmospheric P is constant so 2) kinetic energy = 9800 J heat =  H = -9800 J 3) heat 9800 J 3

First law and chemical reactions • If a process is reversed an equal amount of energy must be exchanged in the opposite direction   H and  E have the opposite sign •  H and  E depend upon the amount of material reacted so doubling, tripling reaction doubles, triples etc. the  H and  E • For a multistep process simply add up each steps  H and  E to get overall value (Hess’ Law)  For a cyclic process  H and  E = 0 review First law and chemical reactions • Example: Determine the  H o for 2CO 2 (g)  2C (s) + 2O 2 (g)  H o = 2 x 393 kJ = 786 kJ given that C (s) + O 2 (g)  CO 2 (g)  H o = - 393 kJ Hess’ law 4

Second Law • Heat only flows from hotter to cooler • Explains why processes that release heat tend to not reverse themselves Sadi Carnot Second law • Simplest way to state the second law (and it works for most cases) is “Heat only flows from hotter to cooler” • the second law explains why processes that release heat do not spontaneously reverse themselves q (heat) heat will not flow between objects at COLD HOT same temperature BOTH SAME TEMPERATURE 5

Second Law when the ball hit the ground and the energy becomes heat the ball warms up a little. bowling ball is trapped (bonded) to the earth because it has lost the energy it would need to go back to the top of building heat 9800 J system system surroundings surroundings hotter same temperature cooler Second law and chemical reactions • chemical reactions absorb/release heat • Obeying second law leaves them stuck at lowest energy like the bowling ball CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) + heat CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) + heat system hotter surroundings cooler  CO 2 (g) + 2H 2 O (g) system same surroundings temperature heat is like a limiting reagent for reverse reaction Fuels have chemical potential energy Sometimes some the rest comes Is used to do work out as heat CO 2 + H 2 O new compounds with less chemical potential energy First law says the fuels cannot be recovered without putting energy back in Second law says the energy that has left as heat will not flow back because heat will only flow from hotter to cooler 6

The Standard State • to be able to compare reactions a standard state for every elements is defined • all substances at concentration of 1.0 M gases at a partial pressure of 1.0 atm • Standard Enthalpy Change (  H o ) Standard Formation Reactions • form 1 mole of a substance from elements in most stable state • most elements just element solids • some elements exist as diatomic gases F 2 , Cl 2 , N 2 , O 2 , H 2 • Br 2 and Hg liquid and I 2 is a solid • Standard Enthalpy for this reaction is called standard enthalpy of formation  H o f Standard Formation Reactions • Example: a) Write standard formation reactions for CO 2 (g), H 2 O (l), C 8 H 18 (l),  H o f and CHCl 3 (l). • 8C (s) + 9H 2 (g)  C 8 H 18 (l) • C(s) + 3/2 Cl 2 (g) + ½H 2 (g)  CHCl 3 (l) • C (s) + O 2 (g)  CO 2 (g) • H 2 (g) + ½ O 2 (g)  H 2 O (l) Use the Standard Formation Reaction Data Tables to determine H for these reactions 7

See tables full size appended to the end of the lecture packet Elements in the standard state junction standard state is like a train junction where you transfer for your final destination destination home station elements in standard state +enthalpy formation for B -enthalpy formation for A  H o -  H o f (B) f (A)  H o = ? -  H o f (A) +  H o substance A substance B f (B) Hess’ Law  H o =  n p  H o f -  n r  H o f reaction elements in standard state +  H of (CO 2 ) -  H of (CH 4 ) + + +  H of (H 2 O) x -  H of (O 2 ) x 2 2 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H o = ? 8

See tables full size appended to the end of the lecture packet Hess’ Law  H o =  n p  H of -  n r  H of reaction elements in standard state -  H of (CH 4 )  H of (CO 2 ) -  H of (O 2 ) x 2  H of (H 2 O) x 2 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H o = ? -802.5 kJ  H o = 1 x (-393.5) + 2 x (-241.8) - (1 x - 74.6 + 2 x 0) = products - reactants  n p  H o f -  n r  H o f Form groups of two or three. Every group is assigned one reaction for which to find  H o 6C (g) + 18O (g) + 12H (g)  H o5  H o1  H o6 C 6 H 12 O 6 (s) + 6 O 2 (g) 3C 2 H 2 (g) + 3H 2 O 2 (l) + 6O 2 (g)  H o7  H o2  H o4  H o8  H o3 6CO 2 (g) + 6H 2 O (g) 6CO (g) + 6H 2 O (g) + 3O 2 (g) 9

Entropy • Entropy (S) - imaginary substance - associated with disorder • matter or energy spreads out  entropy increases • Standard Entropy Change (  S o ) Ludwig Boltzmann entropy (disorder) • In a chemical reaction the side that has more moles of gas has more entropy A (solid)  B(gas) Entropy increases  S positive • The side that has a material dissolved in a solvent has more entropy A (solid)  A (aqueous) entropy increases so  S positive Is  S positive or negative ? 2C 8 H 18 (g) +25O 2 (g)  16CO 2 (g) + 18H 2 O (g) 27 moles gas 34 moles gas (less moles gas) (more moles gas) (less entropy) (more entropy) V = nRT Under constant pressure and temperature more P moles gas will have to occupy larger volume And is therefore more spread out V n   Change in entropy  S = S final – S initial more – less + positive + 10

Entropy is created when heat flows from hotter to cooler •  S > q/T • T must be in Kelvins • Processes where entropy is created are said to be Irreversible • Irreversible does not mean they can never be reversed, but rather that energy will have to be added to reverse it • The most precise way to state the second law is “In any spontaneous process the entropy of the universe increases” Example: Estimate the entropy created when 1000 Joules of heat flows from a Bunsen burner flame at 600  C into a boiling water bath at 100  C. Example: When heat is absorbed by boiling water the temperature does not rise. The energy is used to turn liquid water into steam which contains the added energy. If the system is taken to be the liquid and gaseous water, how much entropy is created when 1 mole of liquid water undergoes evaporation? 11

Entropy and Spontaneity • Processes that have positive  S values tend to not reverse themselves • The more matter spreads the lower its concentration/partial pressure • From a kinetic standpoint reverse reaction is slower and slower… H 2 SO 4 (aq) + Na 2 CO 3 (aq) H 2 O (l) + CO 2 (g) Gibbs Free Energy two reasons a process can be spontaneous: -q 1) heat released system (  H negative) 2) matter spreads out +  S (  S positive) system Josiah Willard Gibbs Both should be visualized as products forming that escape preventing reaction from reversing  G  H  S 12