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Ch 342 S02 Homework 9 Solution 15.Apr.
restart: readmylib(pchemconstants):
Problem 1
Formulation and Solutions
eqnRateRatio := r2/r1 = (C2/C1)^p: alpha := solve(subs({r2=2.05,C2=2,r1=1.00,C1=1},eqnRateRatio),p); α :=1. beta := solve(subs({r2=5.62,C2=2,r1=2.05,C1=1},eqnRateRatio),p); β :=1. g := solve(subs({r2=0.21,C2=.05,r1=1,C1=0.01},eqnRateRatio),p); g :=-. delta := solve(subs({r2=1.01,C2=.05,r1=1,C1=0.01},eqnRateRatio),p); δ :=.
reaction orders: alpha = 1 beta = 1.5 gamma = -1 delta = 0
Problem 2
part a
Formulation and Solution
eqn1stOrder := CA = CAoexp(-kt): ta := solve(subs({CA=0.75,CAo=1,k=4.5e-5},eqn1stOrder),t); ta :=6392.
part b
Formulation and Solution
eqn1stOrderHL := th = ln(2)/k: tb := evalf[3](subs(k=4.5e-5,rhs(eqn1stOrderHL))); tb :=15400.
Answers
a: 6400 s or 107 min
b: 15400 s or 257 min
Extra Credit 1
td := solve(eqn1stOrder,t); kh := solve(eqn1stOrderHL,k);
td :=−
ln CA CAo k
kh :=
ln 2( ) th
tc := subs({CA=f*CAo,k=kh},td);
tc :=−
ln( ) f th ln 2 ( )
QED
Extra Credit 2
skAGE := evalf3; skAGE :=7520.
skeleton is at least 7500 years old
Problem 3
part a
Formulation and Solution
eqn2ndOrder := CA = CAo/(1 + ktCAo): ta := solve(subs({CA=0.75,CAo=1,k=4.5e-5},eqn2ndOrder),t); ta :=7407.
part b
Formulation and Solution
eqn2ndOrderHL := th = 1/(k*CAo): tb := evalf[3](subs({k=4.5e-5,CAo=1},rhs(eqn2ndOrderHL))); tb :=22200.
Answers
a: 7410 s or 123 min
b: 22200 s or 370 min
The activation energy and prefactor are
Eact := m*(-Rg)/1000; ko := exp(b);
Eact :=119. ko :=.2652521131 10 -
activation energy: 120 kJ/mol prefactor: 0.27 E-5 /s
The data were created with an activation energy of 125 kJ/mol and a prefactor of 1.5 E-5 /s!
Extra Credit
Two additional lines representing maximum and minimum slopes that fit the data are drawn through the following sets of points:
pmax1 := table([(iT) = 2.48e-3,(lnr) = -48]): pmax2 := table([(iT) = 3.34e-3,(lnr) = -61.2]):
pmin1 := table([(iT) = 2.42e-3,(lnr) = -48]): pmin2 := table([(iT) = 3.34e-3,(lnr) = -60.6]):
The slopes and intercepts of the maximum and minimum fit lines are
mx := (pmax2[lnr] - pmax1[lnr])/(pmax2[iT] - pmax1[iT]); mn := (pmin2[lnr] - pmin1[lnr])/(pmin2[iT] - pmin1[iT]); mx :=-15348. mn :=-13695.
bx := pmax2[lnr] - mxpmax2[iT]; bn := pmin2[lnr] - mnpmin2[iT];
bx :=-9. bn :=-14.
The average value of slope and its uncertainty are
mave := (mx+mn)/2; delm := (mx-mn)/2;
mave :=-14522. delm :=-826.
The average activation energy and its uncertainty are therefore
Eactave := mave(-Rg)/1000; delEact := delm(-Rg);
Eactave :=120. delEact :=6872.
The maximum and minimum prefactors are
komax := exp(bx); komin := exp(bn); komax :=. komin :=.3530974520 10 - The average prefactor and its uncertainty are therefore koave := (komax + komin)/2; delko := (komax - komin)/2; koave :=. delko :=.
The final results from an analysis of the data by hand should be reported as
activation energy: 121 ± 7 kJ/mol
prefactor: 2 ± 2 E-5 / s
Notice the significantly large uncertainty in the prefactor! It is as large as the value itself. This is because a small change in slope of the line leads to a large deviation in the position of the intercept (1/T = 0) point. To determine precise prefactors from experimental kinetic data, the data must have not have significant scatter about a straight line fit.
Problem 6
Formulation
restart: Eact := -R*Diff(ln(k),iT); # note the mistake in the homework assignment
Eact :=− R
iT ln( k )
Given Expressions and Solution
k := k1 + k2: k1 := ko1exp(-E1iT/R): k2 := ko2exp(-E2iT/R): Eact;
− R
iT
ln ko1 e +
^
− (^) E1 iT R ko2 e
^
− (^) E2 iT R
Differentiate k and simplify expression
Eactoverall := simplify(-R*diff(ln(k),iT));
Eactoverall :=
ko1 E1 e +
^
− (^) E1 iT R ko2 E2 e
^
− (^) E2 iT R
ko1 e +
^
− (^) E1 iT R ko2 e
^
− (^) E2 iT R
The numerator and denominator match the required expression.
QED
