With the correct choice of acid, acid hydrolysis of ethanamide could produce A) ethanoic acid and ammonium chloride. B) ethanoic
acid and methylamine. C) ethanol and ammonia. D) acetaldehyde and ammonium hydroxide. E) formic acid and ethylamine
1 answer:
Answer:
A) Ethanoic acid and ammonium chloride.
Explanation:
Amides go through acid hydrolysis to produce carboxylic acid and ammonia by heating in aqueous acid. The acid hydrolysis reaction is produced through the <u>nucleophilic addition of water to the protonated amide</u>, then comes the transference of a proton from the oxygen to the nitrogen to make nitrogen a better leaving group, and finally, the subsequent elimination (see attachment).
In this case, the products formed with the reaction of ethanamide and aqueous HCl will be <u>ethanoic acid and ammonium chloride</u>.
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Answer:
1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )
Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
<u>2) Entropy change for Decomposition of mercuric oxide </u>
2HgO (s) → 2Hg(l) + O₂ (g)
Δs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C
hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Answer:
2
Explanation:
Half-life
Concentration
We have the relation
So
Comparing the exponents we get
The order of the reaction is 2.
The rate constant is