The answers in exercise 5.23 should show that changing the partial pressure changes the instantaneous Δ rxn G even though the ratio of partial pressures stays the same ( that is, 1 : 1 : 1 for standard pressure conditions is equal to 0.5 : 0.5 : 0.5 for the given conditions). This suggests the interesting possibility that at some equal partial pressure p of all components, the reaction reverses; that is, the instantaneous Δ rxn G becomes negative. Determine p for this equilibrium. (You will have to use the properties of logarithms as mentioned in the chapter to find the answer.) Is your answer of value to those who work with gases at high pressures, or at low pressures? What is your reasoning?
The answers in exercise 5.23 should show that changing the partial pressure changes the instantaneous Δ rxn G even though the ratio of partial pressures stays the same ( that is, 1 : 1 : 1 for standard pressure conditions is equal to 0.5 : 0.5 : 0.5 for the given conditions). This suggests the interesting possibility that at some equal partial pressure p of all components, the reaction reverses; that is, the instantaneous Δ rxn G becomes negative. Determine p for this equilibrium. (You will have to use the properties of logarithms as mentioned in the chapter to find the answer.) Is your answer of value to those who work with gases at high pressures, or at low pressures? What is your reasoning?
Solution Summary: The author explains that the value of p for the given equilibrium is 1.34times 10-3atm.
The answers in exercise 5.23 should show that changing the partial pressure changes the instantaneous
Δ
rxn
G
even though the ratio of partial pressures stays the same ( that is,
1
:
1
:
1
for standard pressure conditions is equal to
0.5
:
0.5
:
0.5
for the given conditions). This suggests the interesting possibility that at some equal partial pressure
p
of all components, the reaction reverses; that is, the instantaneous
Δ
rxn
G
becomes negative. Determine
p
for this equilibrium. (You will have to use the properties of logarithms as mentioned in the chapter to find the answer.) Is your answer of value to those who work with gases at high pressures, or at low pressures? What is your reasoning?
bre
The reaction sequence shown in Scheme 5 demonstrates the synthesis of a
substituted benzene derivative Q.
wolsd works 2
NH2
NaNO2, HCI
(apexe) 13× (1
HNO3, H2SO4
C6H5CIN2
0°C
HOTE
CHINO₂
N
O
*O₂H (
PO
Q
Я
Scheme 5
2 bag abouoqmics to sounde odi WEIC
(i)
Draw the structure of intermediate O.
[2 marks]
to noitsmot od: tot meinedogm, noit so oft listsb ni zaupaib bas wa
(ii) Draw the mechanism for the transformation of aniline N to intermediate O.
Spoilage
(b)
[6 marks]
(iii) Identify the reagent X used to convert compound O to the iodinated compound
[tom E
P.
vueimado oilovonsa ni moitos nolisbnolov ayd toes ai tedw nisiqx
(iv) Identify the possible structures of compound Q.
[2 marks]
[2 marks]
[shom 2]
(v)
bus noires goiribbeolovo xnivollot adj to subora sidab
Draw the mechanism for the transformation of intermediate P to compound Q.
[5 marks]
vi
(vi) Account for the regiochemical outcome observed in the reaction forming
compound Q.
[3 marks]
PROBLEM 4 Solved
Show how 1-butanol can be converted into the following compounds:
a.
PROBLEM 5+
b.
d.
-C=
N
Which alkene is the major product of this dehydration?
OH
H2SO4
heat
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