In what type of reaction or process does heat flow into the system?
1 answer:
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Answer:
See the image 1
Explanation:
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. (see image 2)
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. (see image 3)
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product.
Explanation:
To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.
Based on this premise, the number of moles of species on the reactant and product side must be the same;
Li + Br₂ → LiBr
Put a,b and c as the coefficient of each species
aLi + bBr₂ → cLiBr
balancing Li;
a = c
balancing Br;
2b = c
let a = 1;
c = 1
b =
or a = 2, b = 1 , c = 2
2Li + Br₂ → 2LiBr
P + Cl₂ → PCl₃
Using the same method;
aP + bCl₂ → cPCl₃
balancing P;
a = c
balancing Cl;
2b = 3c
let a = 1;
c = 1
b =
or
a = 2, b = 3, c = 2
2P + 3Cl₂ → 2PCl₃
iii,
H₂ + SO₂ → H₂S + H₂O
use coefficients a,b,c and d;
aH₂ + bSO₂ → cH₂S + dH₂O
balancing H;
2a = 2c + 2d
balancing S;
b = c
balancing O
2b = d
let b = 1,
c = 1
d = 2
a = 3
3H₂ + SO₂ → H₂S + 2H₂O