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Using the appropriate tables, determine the change in

Fundamentals of Engineering Thermodynamics | 8th Edition | ISBN: 9781118412930 | Authors: Michael J. Moran ISBN: 9781118412930 139

Solution for problem 6.4 Chapter 6

Fundamentals of Engineering Thermodynamics | 8th Edition

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Fundamentals of Engineering Thermodynamics | 8th Edition | ISBN: 9781118412930 | Authors: Michael J. Moran

Fundamentals of Engineering Thermodynamics | 8th Edition

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Problem 6.4

Using the appropriate tables, determine the change in specific entropy between the specified states, in Btu/lb ? 8R. Show the states on a sketch of the Ts diagram. (a) water, p1 5 10 lbf/in.2 , saturated vapor; p2 5 500 lbf/in.2 , T2 5 7008F. (b) ammonia, p1 5 140 lbf/in.2 , T1 5 1608F; T2 52108F, h2 5 590 Btu/lb. (c) air as an ideal gas, T1 5 808F, p1 5 1 atm; T2 5 3408F, p 5 5 atm. (d) oxygen as an ideal gas, T1 5 T2 5 5208R, p1 5 10 atm, p2 5 5 atm.

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C117 Chapter 13 Notes- Chemical Kinetics and Rates of Reactions 3-4-16  Homogeneous reaction- reactants in same phase, ie. combustions, acid/base  Heterogeneous reaction- reactants in different phases, ie. oxidation of metal o Surface area can strongly influence rates  Reaction rate- measure of how fast reaction occurs +2 +2 o Ex. 2 )6 + →3 2 )5 3 + 2 o −Δ o Reaction rate = Δ = average rate over time change Δ  Negative if dealing with reactant, positive if product  Depends on concentrations of reagents o If we make Δ smaller Δ → 0 , rate becomes instantaneous rate o Here, reaction rate =.0117−0.041= 8.95 ∙ 10−5 555−222 o Units  For solutions- concentrations (molarity r) usually represented in [square brackets]  For gases- can be expressed i , but more common to use partial pressures (atm) and represent with  For ideal gases- Dalton’s Law of Partial press= + +  From ideal gas law = : o = o For given : (partial pressure proportional to moles of gas A over volume) o Stoichiometry and Reaction Rate  Not all reactions have one mole and one mole like above example  Ex. 4 )+ 2 2 ) → 2 )+ 2 2 ()  If we monitor partial pressure o2 , is rate the same or different than we get by monitoring the 4artial pressure o Rate monitoring 2= twice the rate of monitoring 4  In general, for the reaction + → + : −1 Δ ] −1 Δ 1 Δ 1 Δ ] o Rate = Δ = Δ = Δ = Δ  The Rate Law and Reaction Order o In general, rate of reaction can be described by equation like = [] … where is rate constant; this is a rate law o Values of powers must be determined from experiments, except for elementary reactions o One way to get info on powers is to mix known amounts of reagents, then very quickly determine reaction rate (initial rate); this experiment repeats for a number of different initial concentrations − − o Ex. 2 + 3 → + 3 3  Methyl acetate + hydroxide → acetate + methanol  = ] [ − ] Note: there may be other terms in rate law that are not 3 3 in overall equation for reaction  What are and  = 1; double concentration and double rate for methyl acetate  = 1 as well  With equation = [ ] [ [ , exponents give order of reaction; if = 1, reaction is first order with respect to , if = 2 second order, etc.  = [ ] First order w/ respect to methyl acetate and to hydroxide, 3 3 second order overall o Rate law and reaction orders must be determined experimentally, not from balanced chemical equations o Some catalysts or products can appear in rate law o Ex. For the reaction + → , the initial rate is measured for several different initial concentrations of and . What is the rate law  Find experiment where one held constant and other changes to determine changes in rate  Reaction order w/ respect to = ; concentration increases by factor of 4, rate by 2 2  Reaction order w/ respect to = 0; concentration doubles, rate does not change ln o Reaction order = ln( )  Used when cannot solve by inspection  One fixed, one changes  Ex. ln( ) 1ln(4.5)  Order for A = = 3 33.52 = −0.50 ln( ) 1 ln( ) 3 89 o As more A added, reaction slows down; inhibits reaction ln( ) 1ln(4.5)  Order for B = 1 2 210.2 = 2.50 ln 2 ln17)  Rate = ]−0.5[]2.50  Ex. The following information was obtained for the reaction + + → [ , , [ , , 1) 18 10 16 1.60 ∙ 10−2 −2 2) 36 10 16 4.53 ∙ 10 3) 18 10 12 9.00 ∙ 10−3 4) 18 20 12 4.50 ∙ 10−3  For = ] [] [ , what are , , and o For , find 2 experiments where [] changes and [] and [] are constant: = 1 ln( 2 1 ln( 2 1.60∙10 ln4.53∙10)  = 18 ln(36  = 1.50 3 ln( 4 o For , find 2 where [] changes and and [] are constant: = [ ] ln(3) −3 [4 ln(.00∙10)  = 4.50∙10 ln( ) 20  = −1.0 1 ln( 3 o For , find 2 where [] changes and [] and [] are constant: = [1 ln[3) 1.60∙10 ln9.00∙10)  = 16 ln(22  = 2.0 o So = [ ] 1.5[] −1.0 ]2.; = 1.50, = 1.0, = 2.0 o Determining Rate Constant  When we know rate law, value for easily obtained from initial rates  Ex.  Rate = ] 2 −5  = = 4.0∙10 = 0.004 −1−1 []2 [0.1002 2  Units  For zero order: = ] 0 = in (same as rate units) 1 −1 −1  First order: = ] = [] in = −1  Second order: = [ ] 2 = in = −1 −1 ] 2 −1 1− −1  In general: = ] = ] in =  For solutions, use = as above; for gases, use partial pressures in o Zero-Order Reactions  = [ ] [] 0=  Changing concentration has no effect on rate (rare)  Can occur in gas/surface reaction where surface becomes saturated with absorbed reagent; changing pressure of reagent has no effect on coverage or rate  Integrated rate law for zero-order reactions- = − + [ ] 0  Differential equation: [ ]0 o ] = o − = o = − o ∫ = − ∫ []0 0 o − [ ]0 = − o ]= − + [ ]0 o First-Order Reactions  Reaction rate proportional to concentration or partial pressure of reagent  Constant fraction of molecules react in given time period  Ex. If 10% of a reagent reacts per unit of time and we start with 1000 molecules in a first- order reaction, there is exponential deca ( = # unreacted molecules at time ):  Integrated rate law for first-order reactions-= − + ln ]0 OR [] = ]0 −  Differential equation: o = [] ] o − = [] o ]= − [] [ ] o ∫ 0 [] = − ∫0 [ 1 o ∫ ] = − ∫0 0 ] o ln ln [ ]0 = − o ln ] = − + ln [ ] − 0 o ]= [ ]0 o Second-Order Reactions  In first-order, all molecules have same probability of reacting in particular time window; reaction is random event [ ][ ] [ ]2  In second-order, = or ; reaction rate depends on concentration of and or on that of squared  Integrated rate law for second-order reactions: 1 1  For → : = + []0 o Differential equation:  = ]2 ] 2  − = ] ]  2 = − ]  ∫[ ]= − ∫ []0 ] 0 [ 1  []0[]2 = − 0 [  [] []−2 = − ∫0 0 ]  − ] |1 [ ] = − 1 10  − + = − 0  − 1 = − − 1 [0 1 1  = + [0]] 1 [ 0  For + → : = ( − ])ln [] ] 1 1 ] 0 0 0 o = − = − = [ ] o Pseudo-First-Order Kinetics  If one reagent, ie. , present in excess, then things get much simpler; effectively constant, rate law simplifies:  = = where = [] with units −1 ′  Follows first-order integrated law: ln = − + ln ] 0 o Plot of ln against gives straight line with slope −′; if known, value for can be found  [ ≫ can occur naturally; in other cases, we manipulate conditions to make studying reaction easier (pseudo-first easier than second because [] does not need to be followed) o Ex. The following data was obtained for the gas phase decomposition of nitrogen dioxide at 300℃. What is the order of the reaction, and what is the rate constant 1  Balanced equation: 2 )→ ()+ 2 ) 2 1  To distinguish between first and second order, plot ln[2 ] and against time, determine [2] which is linear  If neither linear, try 2 ] against time for zero-order 1  linear, so second-order [2]  To determine , use initial concentration and another concentration at any time : 1 1  []= + []0 1 1  125 = 50 + 100 1 1  50 = 125 − 100 1 1 1  = 50 125 − 100) −5 −1 ≈ 4 ∙ 10  Half-life/- time (usually in ) for concentration of a reactant to fall by factor of 2 o For first-order, independent of concentration: = ln2 1/2 o For second-order, dependent on concentration: = 1 1/2 ]0  Becomes longer as reaction proceeds o Ex. The decomposition of hydrogen peroxide is a first-order reaction with a reaction constant of 1.767 ∙ 10−5 −1. What is the half-life  2 2 + 2 2  = ln2 ≈ 39,227 1/2 1.767∙105−1  Elementary reactions- one-step process whose equation describes which molecules collide and make/break bonds o 2 types:  Unimolecular- individual molecule spontaneously changes structure or dissociates  Isomerization: →  Dissociation: → +  Bimolecular- two molecules collide and associate or rearrange into products  Association: + →  Reaction: + → + o Many complex reactions built up of sequences of elementary ones; some occur in single step o Ex. Overall reaction + ) → ( ) + occurs through 2 elementary 4 3 3 3 3 3 reactions:  )4→ )3 + unimolecular  )3+ 3 3 → 3( 3 3 bimolecular o Ex. The isomerization of cis-2-butene to trans-2-butene is unimolecular:   Exothermic- trans form lower in energy because s most spread out, least steric hindrance  Not all cis molecules automatically convert to trans: − − o Ex. ()+ 3 ()→ 3()+ is bimolecular  Collision geometry important- if iodide approached from different direction, much more difficult to react  Effect of Temperature on Reaction Rate o Usually increase as temperature rises o Ex. + ( )→ + − () 3 3()  Plot shows rate constants as function of temperature: o As temperature increases, KE increases, as well as collision energy; larger fraction of collisions lead to reaction (or faster reaction)   The Relationship Between Rate Constant and Temperature o Information on Arrhenius  Born 1859 near Uppsala in Sweden  One of founders of physical chemistry  Received Nobel prize for electrolytic conductivity work  First to speculate raising concentration in atmosphere would cause greenhouse effect − / o Arrhenius Equation- =  = frequency factor; gives # times the barrier is encountered per second (related to collision frequency ∙ steric factor for bimolecular reactions)  −/ = fraction of collisions with sufficient energy to overcome energy barrier  = energy barrier (activation energy of reaction)  = 8.314 ∙(gas constant)  = temperature in o Take natural logs: ln = ln − ( ) 1  Same as ln = ln − 1  Plot of ln against is linear with slope and intercept ln o Ex. − + → + − () 3 () 3()  Intercept = ln = 23.53  = 23.53 ∙ = 1.66 ∙ 1010 ∙  Slope = − = −9.18 ∙ 10  = − ∙ = 9.18 ∙ 10 ∙ 8.314 ∙ = 76.3  Now can use Arrhenius equation to calculate at any temperature  What is the rate constant at 50℃  = 50℃ + 273.15 = 323.15  = −/ −76,000 = (1.66 ∙ 1010 ) 8.31∙ 323.15 ∙ 10 −28.4 = (1.66 ∙ 10 ∙) −3 = 7.70 ∙ 10 ∙ o Can be given rate constants at 2 temperatures and asked to determine activation energy  = −/1 and = −/2 1 2  Divide to eliminate : 1 −/1 = −/2 2 − / 2= 2 1 −/1 / 1 −/2 = ∙ [1− 1] = 1 2  Take natural logs: ln[ ] = [ 1 − 1 ] 1 1 2  After found, can be found by substituting into expression for either rate constant  Rate Laws for Elementary Reactions o Because ERs occur in single step, chemical equation describes exactly what happens; we can predict rate law from chemical equation o Unimolecular always first order:  → = ]  Ex. -2-butene to -2-butene isomerization driven by heat o Bimolecular always second order:  Can be + → or + → +  + → = [ ]  + → = ] 2  Ex. Stratospheric ozone depletion reaction )+ 3 )→ 2 )+ 2 )occurs in single step o Termolecular reactions- involve simultaneous collision of 3 molecules; almost always not elementary  Ex. 2 ()+ 2 )→ 2 ()occurs through 2-step process involving an intermediate + ⇆ 2 2 2 2+ →22 2  Sequence of elementary reactions like this known as reaction mechanism o Can only predict rate law from equation if known to be elementary; CANNOT from overall chemical equation o To predict rate law for a reaction, must know reaction mechanism  Reaction Mechanisms o Equations that describe sequence of steps in non-elementary reactions o Reaction rate determined by slowest step- rate-determining step o Ex. Oxidation of iodide ion by hydrogen peroxide in acidic solution given by 2 ()+ 2 2 )+ 2 3 ()→ 2 + 4 2 () −  Experimental rate law: = ][ 2 2 ]  Accepted reaction mechanism:  + → + (slow) 2 2 − −  + → + 2 (fast)  2 + 2 3 4 (f2st) − +  2 + 2 22 3 + 4 2 2 Overall  and are reaction intermediates- do not appear in overall equation; only present in small quantities  Reaction rate controlled by slowest step in sequence of elementary reactions; rate law predicted from rate-determining step:  = ][ ] 2 2 o Correctly predicting rate law does not prove reaction mechanism is correct; other ways to confirm mechanism o Ex. Reaction mechanism with fast initial step: 2 + → 2 () 2 ) ()  Accepted mechanism:  + ⇆ (fast) 2 2  +2 → 2 (slow)  2 ()+ 2 ) → 2 () Overall  Predicted rate law: = 2][ ]  Since i2 intermediate, difficult to measure its concentration  Under steady conditions, rate of formation of = rate of destruction of 2 2 (known as Steady State Approximation)  = 1 = 2 ] −1[ 2 2 2][ ]  1= rate constant for formation (reaction 1)  −1 = rate constant for destruction by reversing reaction 1  2= rate constant for destruction by reacting the intermediate  Second step in mechanism is slow; −1[ 2 2[ 2 ] ] o If ≫ , then + ≈ (smaller term can be ignored) o So we can ignore 2 2][ , and 1 = 2[−1 2  Rearranging, get expression for [2]: [ ][ ] o 1 = 2 −1 2 o [ =2] 1 [ 2] −1  Substitute into equation for rate law for second elementary reaction:  = 2 2][ ] = 1 2[ ] [ ] −1 2 = ] [2]  Overall rate constant is combination of those for all elementary steps  Catalysts o Speed up reactions by allowing reaction to occur by a different mechanism (usually with smaller activation energy) o Not consumed in reaction, does not appear in overall equation o Ex. Uncatalyzed reaction for -2-butene → -2-butene occurs with rate law = [-2-butene] 1/2  Can be catalyzed by gaseous molecular iodine; rate law is = [-2-butene]2  1-order usually indicates that half the molecule is involved in rate-determining step 2  Catalyzed reaction believed to occur in 5-step process:  (1) 2→ 2  (2)  (3)  (4)  (5) 2 → 2  Activation energy barrier lowered, causing rate to increase:  Steps in mechanism:  2→ 2  + 3 = 3 → 3 − 3  − → − 3 3 3 3  3 − 3 )→ 3 = 3 )+  2 → 2  3 = 3 )→ 3 = 3 ) o Enzymes- biological catalysts that allow reactions in cells to occur rapidly under mild conditions (ie. at atmospheric pressure and body temperature)  Usually globular proteins held in a particular 3D shape by intermolecular forces  A particular enzyme is usually specific to a particular substrate; results from “lock and key” type of intermolecular attractions occurring between enzyme and substrate  When enzyme and substrate bind, may be conformational changes to both (an induced fit); substrate may be held in a way that helps promote the reaction  Most follow this mechanism:  + ⇆ (fast)  → + (slow)  → Overall  Apply steady state approximation to :]  = [ + with ≫ [ ] 1 −1 2 −1 2  [ = 1 [ ] −1  = 2 = ] 2 1 [ ] −1  The Rate-Limiting Reagent  In general, either catalyst or substrate can be rate-limiting  For enzymes, at low , increasing [] causes reaction rate to increase o At high [], all the converted to and a further increase in [] does not lead to increase in rate o Enzyme Inhibition  Enzymes (and catalysts in general) can be inhibited (or poisoned) by blocking the active site with a molecule that has similar structure to substrate  Many drugs designed to block specific enzymes  Mechanism:  + ⇆ (fast)  + ⇆ (fast)  → + (slow)  If drug has higher affinity for enzyme (ie. sticks more strongly), will block activity of enzyme; no free enzyme to bind to substrate o Temperature Dependence  Enzyme-catalyzed reactions unusually often depend on temperature  Rate increases with temperature; suddenly decreases when enzyme denatured (unfolded due to heat)

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Chapter 6, Problem 6.4 is Solved
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Textbook: Fundamentals of Engineering Thermodynamics
Edition: 8
Author: Michael J. Moran
ISBN: 9781118412930

The answer to “Using the appropriate tables, determine the change in specific entropy between the specified states, in Btu/lb ? 8R. Show the states on a sketch of the Ts diagram. (a) water, p1 5 10 lbf/in.2 , saturated vapor; p2 5 500 lbf/in.2 , T2 5 7008F. (b) ammonia, p1 5 140 lbf/in.2 , T1 5 1608F; T2 52108F, h2 5 590 Btu/lb. (c) air as an ideal gas, T1 5 808F, p1 5 1 atm; T2 5 3408F, p 5 5 atm. (d) oxygen as an ideal gas, T1 5 T2 5 5208R, p1 5 10 atm, p2 5 5 atm.” is broken down into a number of easy to follow steps, and 100 words. Fundamentals of Engineering Thermodynamics was written by and is associated to the ISBN: 9781118412930. Since the solution to 6.4 from 6 chapter was answered, more than 251 students have viewed the full step-by-step answer. This full solution covers the following key subjects: atm, lbf, ideal, States, btu. This expansive textbook survival guide covers 14 chapters, and 1738 solutions. The full step-by-step solution to problem: 6.4 from chapter: 6 was answered by , our top Engineering and Tech solution expert on 11/14/17, 08:39PM. This textbook survival guide was created for the textbook: Fundamentals of Engineering Thermodynamics, edition: 8.

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Using the appropriate tables, determine the change in