PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel 1.
ANSWER KEY
If the reaction is not elementary, one cannot know the order. If the reaction is nd elementary, the order is 2 in either direction. −[ A ]
=
k 1[A][B]
=
k −1[ R ]
dt d [A ] dt
2.
1
2
1 − d [ NO 2 ] 2
dt
d [ O 2 ]
= −2
=
d [ N 2 O5 ]
dt d [ A ]
dt
3.
The same. ( r A
4.
1760 (s ) 3 6 (mol/m )
5.
No, they are not.
6.
3.0 x10
7.
a)
atm hr
b)
One mole of an ideal gas at 400°K occupie occupiess 32.82 32.82 L ⋅ atm . PV F L ⋅ atm I × 400b K g = 32.82 F L ⋅ atm I = RT = 0.082053 G G J H K ⋅ mol J K H n mol K
=
dt
, regardless of how the stoichiometric equation is written.)
-1
-4
− r A = −
1 3
rB
=
1 2
r R
=−
[A ] dt
=−
1 d [ B] 3 dt
=
1 d [ R ] 2 dt
L mol ⋅ hr -1
-1
F 1 I × 32.82 F L ⋅ atm I = 1.20 × 102 F L I G J G J H atm ⋅ hr J K H H mol K mol ⋅ hr K
3.66 G
8.
If 1 >> k 2[N2O], then the reaction is second order in N2O, and second order overall. If 1 << k 2[N2O], then the reaction is first order in N 2O, and first order overall. If 1 ≈ k 2[N2O], then the order cannot be specified.
PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel
ln
9.
k 1 k 2
k 1
=−
k 2
E F 1
G R H T1
. × 10 = 197
3
−
1 I
J T2 K
2 ANSWER KEY
300 kJ mol−1
=−
8.314 × 10
−3
−1
kJK mol
times faster
2
-1
10.
2.74 x10 kJ mol
11.
E = 45.0 kJ mol
-1
6.0
-1
ln(Running speed) = -5412K + 23.87
5.8
) d 5.6 e e p s g n 5.4 i n n u R ( n l
5.2
5.0
0.00332
0.00336
0.00340
0.00344
0.00348
-1
K
12.
1.5 times
13.
Growth rate = 1/Growing days. -1 28.2 kJ mol
14.
At 60°F (288.6 K), k = 1.333 chirps s -1 At 80°F (299.7 K), k = 2.667 chirps s -1 E = 44.9 kJ mol
15.
Reaction order = ln3/ln2 = 1.585
16.
Order in A = 1/2. Order in B = 2/3.
17.
Order in A = 1. Order in B = 2/3.
-1
0.00352
F 1 − 1 I K 1 = 7586 . 1 G H 923 773J K −
−
PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel 18.
3 ANSWER KEY
First, assume second step is rate limiting. d [ O 2 ] * = k 3 [ NO 3 ] dt * * Apply steady-state approximation to NO3 and NO : d [ NO*3 ] * * * * = 0 = k1[ N 2 O 5 ] − k 2 [ NO 2 ][NO 3 ] − k 3[ NO 3 ] − k 4[ NO ][NO 3 ] dt k 1[ N 2 O5 ] [ NO*3 ] = k2 [ NO 2 ] + k 3 + k 4 [ NO* ] d [ NO* ] dt
[ NO* ] =
=
0 = k3[ NO*3 ] − k 4 [ NO* ][NO*3 ]
k 3 k 4
Therefore, [ NO *3 ] =
k 1[ N 2 O 5 ] k 2 [ NO 2 ] + 2 k 3
Inserting this into the rate equation yields: d [ O 2 ] k1k 3 [ N 2 O 5 ] =
k 2 [ NO 2 ] + 2 k 3
dt
But, this yields first-order kinetics only if k 2[NO2] << k 3, which contradicts the assumption that step 2 is rate-limiting. Therefore, assume step 3 is rate-limiting: d [ O 2 ] k1k 3 [ N 2 O5 ] * * = k 4 [ NO ][NO 3 ] = dt k 2 [ NO 2 ] + 2 k 3 If k 2 << k 3, then this analysis is consistent with first-order kinetics: d [ O 2 ] k 1[ N 2 O5 ] =
2
dt
19.
a) A+A
A* d [ R ] dt
=
k 1 k -1 k 2
k 2 [ A*]
A* + A
R+S
PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel But, K eq b)
=
k 1
4 ANSWER KEY
[A*]
=
[A]
k −1
; therefore
d [ R ]
=
dt
k 2 K eq [ A]
One could attempt to prove the existence of A*, and better yet, measure its steady-state concentration. d [ A*] 2 = 0 = k1[ A ] − k −1[ A*][A ] − k 2 [A*] dt k 1[ A ]2
[A*] =
≅
k −1[A ] + k 2
K eq [ A ]
Or, one could see the effect on rate of a scavenger or trapping agent for the proposed intermediate, A*. 20.
a) b)
Overall first order. k 1
O3
k -1 k 2
O + O3
1 d [ O 3 ]
Rate = − [O⋅] =
2 dt k 1[ O 3 ]
=
k −1[ O 2 ]
1 d [O 2 ] 3
=
dt
=
K eq
O2 + O
fast
2O2
slow
1 d [ O 2 ] 3 dt [O 3 ]
=
k 2 [O⋅][O 3 ]
[O 2 ]
k2 K eq [O 3 ]2[O 2 ]
−1
To test the mechanism, one would wish to prove the participation of O⋅ and demonstrate inhibition using common free radical scavengers. 21. H3PO 2 + H H3PO2* + Ox d [ H 3 PO 3 ] dt
=
k 1 k -1 k 2
k 2 [ H 3 PO *2 ][O x ]
H3PO2* + H H3PO 3
PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel
5 ANSWER KEY
Application of the steady-state approximation to the active intermediate yields, ⊕ k 1[ H 3 PO 2 ][H ] * [ H 3 PO2 ] = k −1[ H ⊕ ] + k 2 [ O x ] and therefore, ⊕ d [ H 3 PO 3 ] k1k 2 [ H ][O x ][H 3 PO 2 ] =
⊕
k −1[ H ] + k 2 [ O x ]
dt
At low [Ox], the right term of the denominator is negligible, and, d [ H 3 PO 3 ] k1 k 2 [O x ][H 3 PO 2 ] =
dt
k −1
At high [Ox], the left term is negligible, and, d [ H 3 PO 3 ] ⊕ = k 1[ H ][H 3 PO 2 ] dt 22.
With the first step rate limiting, the following mechanism is consistent with the experimental rate law: k 1 A+B AB A + AB
23.
K =
a)
k 1 k 2
=
k 2
[X] [A][E]
A2B
and
[ E o ] = [ E ] + [ X]
Combining the two equations above yields, [ X] =
The rate expression is then given by, k1k 3 [ A][E o ] k 3 [A][E o ] − r A = k 3 [ X] = = k 2 k 2 + k 1[ A ] + [A] k 1
k 1[ A][E o ] k2
+ k 1[ A ]
PSC 480/740 Problem Set 2 Questions Chapter 2, Levenspiel b)
6 ANSWER KEY
In this case, apply steady-state approximation. d [ X ] = k1[ A][ E ] − k 2 [ X ] − k 3 [ X ] = 0 dt Substituting [ E ]o = [ E ] + [ X ] and solving for [X] yields, k 1[A][E o ] [ X] = b k2 + k3 g + k 1[A] The rate expression is then given by, k1k 3[ A][E o ] k 3[A][E o ] − r A = k 3[ X] = = b k 2 + k 3 g + k 1[A ] F k 2 + k 3 I + [A] GH k J K 1
Note that the only difference is that the fast equilibrium approach imposes the restriction that k 2 >> k 3; whereas the steady-state approach does not.