2-14$
P2–10)(e)$
The$amount$of$catalyst$necessary$to$achieve$40$%$conversion$in$a$single$PBR$can$be$found$from$
calculating$the$area$of$the$shaded$region$in$the$graph$below.$
$
P2–10)(f)$
$
)
$
3-1$
Solutions)for)Chapter)3)–)Rate)Laws)
)
P3–1)(a))$
(i)$Individualized$solution$
(ii)$2550$K$
(iii)$Individualized$solution$
)
P3–1)(b)$
(i)$The$equilibrium$concentration$changes,$but$the$equilibrium$conversion$remains$the$same$in$all$the$
three$cases$(50%).$The$time$taken$to$attain$equilibrium$remains$the$same$in$all$the$three$cases.$
$
$
P3–1)(c)$
$
3-2$
P3–1)(c))Continued$
ln 𝑘 =ln 𝐴−
𝐸
𝑅
1
𝑇$
From$the$graph$of$ln$k$vs$1/T$above,$we$get:$
ln 𝐴=27.577,
𝐸
𝑅
=−10889 𝐾$
Thus,$
𝐴=9.474 ∗10!! 𝑠!!, 𝐸=90.531 𝑘𝐽/𝑚𝑜𝑙$
$
P3–1)(d))Example)3-1$
For,$E$=$60kJ/mol$ $ $ $ $$$$$$$$$$$$For,$E$=$240kJ/mol$ $
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $
$
T)(K)$
k)(1/sec)$
1/T$
ln(k)$
$
T)(K)$
k)(1/sec)$
1/T$
ln(k)$
310$
1023100$
0.003226$
13.83918$
$
310$
4.78E–25$
0.003226$
–56.0003$
315$
1480488$
0.003175$
14.2087$
$
315$
2.1E–24$
0.003175$
–54.5222$
320$
2117757$
0.003125$
14.56667$
$
320$
8.77E–24$
0.003125$
–53.0903$
325$
2996152$
0.003077$
14.91363$
$
325$
3.51E–23$
0.003077$
–51.7025$
330$
4194548$
0.00303$
15.25008$
$
330$
1.35E–22$
0.00303$
–50.3567$
335$
5813595$
0.002985$
15.57648$
$
335$
4.98E–22$
0.002985$
–49.0511$
$
$
$
3-3$
P3–1)(e))No$solution$will$be$given$
P3–1)(f)$
A+1
2
B→1
2
C
$
Rate$law:$ $$and$
$
$
$ $ $
𝟐
$
)
P3–2)(a)$
Refer$to$Fig$3-4$
The$fraction$of$molecular$collisions$having$energies$less$than$or$equal$to$35$Kcal$is$given$by$the$area$
$
P3–2(b))$
The$fraction$of$molecular$collisions$having$energies$between$10$and$20$Kcal$is$given$by$the$area$under$
$
P3–2)(c)$
The$fraction$of$molecular$collisions$having$energies$greater$than$the$activation$energy$EA=$25$Kcal$is$
$
)
P3–3)(a)$
$
$
(a)$
(b)$
$
)
P3–4)No$solution$will$be$given$
$
3-4$
P3–5)(a)$
The$fraction$of$collisions$having$energy$between$E$=$3$and$E$=$5$is$the$area$under$the$graph$between$
P3–5)(b)$
$
P3–5)(c)$
$
P3–5)(d)$
Since$f(E,T)$=$0$for$E$>$8$kcal,$the$fraction$with$energies$greater$than$8$kcal$=$0$
)
P3–6)(a)$$
Note:$This$problem$can$have$many$solutions$as$data$fitting$can$be$done$in$many$ways.$
Using$Arrhenius$Equation$
For$Fire$flies:$
T(in$K)$
Flashes/min$
ln(flashes/min)$
294$
9$
2.197$
298$
12.16$
2.498$
303$
16.2$
2.785$
$
Plotting$ln$(flashes/min)$vs.$1/T,$$
We$get$a$straight$line.$
$
For$Crickets:$$
T(in$K)$
1/T$x103$
chirps/min$
ln(chirps/min)$
287.2$
3.482$
80$
4.382$
293.3$
3.409$
126$
4.836$
300$
3.333$
200$
5.298$
$
Plotting$ln$(chirps/min)$Vs$1/T,$$
We$get$a$straight$line.$
➔Both,$Fireflies$and$Crickets$data$$
)
)
)
)
3-5$
P3–6)(b)$$
For$Honeybee:$$
T(in$K)$
1/T$x103$
V(cm/s)$
ln(V)$
298$
3.356$
0.7$
–0.357$
303$
3.300$
1.8$
0.588$
308$
3.247$
3$
1.098$
$
Plotting$ln$(V)$vs.$1/T,$almost$straight$line.$
$
P3–6)(c)$$
For$Ants:$$
T(in$K)$
1/T$x103$
V(cm/s)$
ln(V)$
283$
3.53$
0.5$
–0.69$
293$
3.41$
2$
0.69$
303$
3.30$
3.4$
1.22$
311$
3.21$
6.5$
1.87$
$
Plotting$ln$(V)$$vs.$$1/T,$$
We$get$almost$a$straight$line.$$
$
So$activity$of$bees,$ants,$crickets$and$fireflies$follow$Arrhenius$model.$So$activity$increases$with$an$
increase$in$temperature.$Activation$energies$for$fireflies$and$crickets$are$almost$the$same.$
$
Insect$
Activation$Energy$
Cricket$
52150$
Firefly$
54800$
Ant$
95570$
Honeybee$
141800$
$
P3–6)(d)$$
There$is$a$limit$to$temperature$for$which$data$for$any$one$of$the$insect$can$be$extrapolate.$Data$which$
would$be$helpful$is$the$maximum$and$the$minimum$temperature$that$these$insects$can$endure$before$
$
P3–6)(e)$
1)$The$rate$at$which$the$beetle$can$push$a$ball$of$dung$is$directly$proportional$to$its$rate$constant,$
therefore$
-rA$=c*k,$where$c$is$a$constant$related$to$the$mass$of$the$beetle$and$the$dung$and$k$is$the$rate$constant$
$
3-6$
P3–6)(e))Continued$
From$the$data$given$
$$$$$$$$$-rA$
$$$$$$T(K)$
$
1/T$
ln$k$
6.5$
300$
$
0.003333$
1.871802$
13$
310$
$
0.003226$
2.564949$
18$
313$
$
0.003195$
2.890372$
$
Refer$to$P3–8$(similar$procedure)$
Therefore,$A$=$1.299X1011$$$
$
P3–6)(e))$
2))Individualized$solution$
$
)
P3-7$
There$are$two$competing$effects$that$bring$about$the$maximum$in$the$corrosion$rate:$Temperature$and$
HCN-H2SO4$concentration.$The$corrosion$rate$increases$with$increasing$temperature$and$increasing$
$
)
y$=$-7120.4x$+$25.593$
R²$=$0.9893$
0)
0.5)
1)
1.5)
2)
2.5)
3)
3.5)
0.00318) 0.00321) 0.00324) 0.00327) 0.0033) 0.00333) 0.00336)
ln(k))
1/T))(K-1)$
P3–8)Antidote$did$not$dissolve$from$glass$at$low$temperatures.$
$
)
P3–9)(a)$
If$a$reaction$rate$doubles$for$an$increase$in$10°C,$at$T$=$T1$let$k$=$k1$and$at$T$=$T2$=$T1+10,$let$k$=$k2$=$2k1.$$
Then$with$k$=$Ae–E/RT$in$general,$ $and$ $,$or$
$$$
$$$or$$$$
$
$
Therefore:$
$
which$can$be$approximated$by$ .$Consequently,$for$this$doubling$rate$fule$of$thumb$to$
be$valid,$the$temperature$at$which$the$doubling$will$take$place$must$be$related$to$the$activation$energy$
$
P3–9)(b))Individualized$solution$
$
)
P3–10)$
From$the$$given$data$
-rA(dm3/mol.s)$
T(K)$
k=$-rA/(4*1.5)$
1/T$(𝐾!!)$
ln$(k)$
0.002$
300$
0.00033333$
0.003333$
–8.00637$
0.046$
320$
0.00766667$
0.003125$
–4.87087$
0.72$
340$
0.12$
0.002941$
–2.12026$
8.33$
360$
1.38833333$
0.002778$
0.328104$
Plotting$ln(k)$vs$(1/T),$we$have$a$straight$line:$$
$
3-8$
P3–10)(a)$
Activation)energy)(E),))
$$
P3–10)(b)$
Frequency)Factor)(A),)ln)A)=)41.99)))))$
⇒$𝐴=1.72 ∗10!” !”!
!”#!.!
!)$
P3–10)(c)$
$𝑘=1.72 ∗10!” exp (!!”###
!)$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$(1)$
$
)
P3–11)(a)$$–ra=kCACB$
P3–11)(b)$$–ra=k$
$
)
P3–12)(a)$$C2H6$$$→$$C2H4$$+$$H2$ $ $ Rate$law:$$-rA$$=$ $,$k$=$[$s-1$]$
P3–12)(b)$$C2H4$$+$$1/2O2$$→$$C2H4O$ $ Rate$law:$$-rA$$=$ $,$k$=$[$dm1.5/s.mol0.5$]$
3-9$
P3–12)(g))(CH3CO)2O$+$H2O$→$2CH3COOH$ $
)
P3–13)(a))$
(1) –rA$=$kACACB
2$
(2) –rA(=$kACB$
$
)
P3–14)(a)$ $
$
We$need$to$assume$a$form$of$the$rate$law$for$the$reverse$reaction$that$satisfies$the$equilibrium$
condition.$If$we$assume$the$rate$law$for$the$reverse$reaction$(B–>A)$is$ =$ $$
then:$
$$
$$$$
$
From$Appendix$C$we$know$that$for$a$reaction$at$equilibrium:$$KC$
At$equilibrium,$rnet$0,$so:$$
$
P3–14)(b))
$
3-10$
P3–14)(b))Continued$
then:$
$
$
From$Appendix$C$we$know$that$for$a$reaction$at$equilibrium:$$KC$
At$equilibrium,$rnet$0,$so:$$
$
If$we$assume$the$rate$law$for$the$reverse$reaction$ $is$$ $
$
From$Appendix$C$we$know$that$for$a$reaction$at$equilibrium:$$KC$
At$equilibrium,$rnet$0,$so:$$
$
3-11$
P3–15$
Assuming$the$reactions$to$be$elementary:$
2$Anthracene$$–>)$Dimer$
⇒ −rA=
β
k+CAnthracene
2−CDimer
Kc
#
$
%
%
&
‘
(
(“where,”Kc=k+k−
$
Similarly$for$the$second$reaction:$
Norbornadiene$$→$Quadricyclane$
$