Click the "draw structure" button to launch the drawing utility. Propane (CH3CH2CH3) has both 1° and 2° hydrogens. Draw the carb
on radical formed by homolysis of a 1° C―H bond in propane. draw structure ... Draw the carbon radical formed by homolysis of a 2° C―H bond in propane.
1 answer:
Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
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Answer:
[CO] = 7.61x10⁻³M
7.61x10⁻³x10³ = 7.61
Explanation:
For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:
We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:
[COCl₂] = 7.73/10.0 = 0.773 M
So, the equilibrium will be:
COCl₂(g) ⇆ CO(g) + Cl₂(g)
0.773 0 0 <em>Initial</em>
-x +x +x <em> Reacts</em>
0.773-x x x <em>Equilibrium</em>
Supposing that x<<0.773, then:
7.5x10⁻⁵ = x²/0.773
x² = 5.7975x10⁻⁵
x = √5.7975x10⁻⁵
x = 7.61x10⁻³ M
The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61