Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
I think Kinetic energy forms <em>Motion energy </em>and Potential energy forms <em>Gravitational Potential energy.</em>
Sodium Hydroxide (NaOH) is also known as lye which is a base (very high ph; Alkaline)
Now, in chemistry, equilibrium is what affects the reaction rate of a reaction. If they are in equilibrium, the concentrations of them will not change (both reactants and products).
Now, lets say that to synthesize a certain chemical, we need it to be in an acidic environment with HCL or some other acid as the catalyst for the reaction.
Well, if we were to add Sodium Hydroxide to this which is very alkaline, the ph would change greatly which affects the reaction rate. If we do not have enough energy to overcome the activation barrier, the reaction will not occur (atleast for a very long time).
However, a common mistake is thinking that a catalyst will affect the equilibrium. This is not true. The reaction will still take place but it will have a very slow reaction rate.
TLDR; Adding a catalyst (like NaOH or Sodium Hydroxide) will not change the equilibrium but instead change the reaction rate. The reaction can still occur, although it can take a very, very long time (like diamonds turning into graphite)
Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,
