Prices of Response
1
) Use the speculative information to identify the price consistent for the complying with response.
3A + B 
C Test [
A]
[ B
s(b) 0.06 L 2/ mol 2 · s( c )3.6 L 2/ mol 2 · s( d )18 L/mol · s 2
) Use the speculative information to establish the price formula for
the complying with response . 3A +B C Test [A].
[ B] First Rate 1 0.20 0.20 0.145 2 0.20 0.60 0.435 3 0.60 0.20 1.305(a)Rate=
k [A] [B](b)Rate=k [A] 2 [4″>.
B
|
] 2 |
(c)Rate=k [ |
A] |
2 [ |
|||||||||||||||
|
B
|
]
|
(d)Rate=k [A] [B]
|
||||||||||||||||
|
.
|
2 3
|
)Use
|
the speculative information
|
|||||||||||||||
|
to figure out the price formula
for the adhering to response.
|
3A+B
|
C Test [
A ] . [B] First Rate 1 0.20 0.20 0.150 2 0.20 0.60 0.150 3 0.60 0.20 1.350(a)Rate=k [A] [B] 2 (b)Rate=k [. A] 2 [B] 2 ( c ) Rate =k [A] 2 [B]( d)Rate= k [ A] 2 4 )Use the speculative information to establish the price consistent for the adhering to response. 3A +B
|
C Test [
|
. A]
|
[B]
|
Preliminary Rate
|
1
|
0.20
|
0.20
|
0.156
|
2
|
0.40
|
0.20
|
0.312
|
3
|
0.20
|
0.40
|
0.312
|
Price=k
[#000000″>. A] [B](a)0.26 L/mol · s(b)3.9 L/mol · s(c)3.9 L 2/ mol 2 · s( d )0.26 L 2
/ mol 2 · s 5) Use the speculative information to establish the price formula for the complying with response
. 3A+ B C Test [A
] [B] Preliminary Rate 1 0.30 0.30 0.170 2 0.90 0.30 0.170 3 0.30 0.90 1.530(a )Rate=k [
A] [B](b)Rate=k [A] 2 [B]4″>.
(
|
c)Rate=k [ |
. B] 2( |
d)Rate=k [ |
A] |
|
[ B]
. 2 6
|
)Use
|
the speculative information
|
|
|
to identify the price continuous
for the complying with response.
|
3A+B
|
C Test [A] [B]
Preliminary Rate 1 0.30 0.30 0.170 2 0.90 0.30 0.170 3 0.30 0.90 1.530 Price=k [#FF0000″ size=”4″>B] 2 (a) 0.53 L/mol ·s(b)1.9 s -1( c) 1.9 L/mol · s(d)0.53 s -1 House Online Assignments  |
