Answer:
Explanation:
N2(g)+O2(g)⇌2NO(g),
N2(g)+2H2(g)⇌N2H4(g),
2H2O(g)⇌2H2(g)+O2(g),
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
