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Ede4ka [16]
2 years ago
13

3. Identify the reagents you would use to convert 1-bromopentane into each of the following compounds: (a) Pentanoic acid (b) He

xanoic acid (c) Pentanoyl chloride (d) Hexanamide (e) Pentanamide (f) Ethyl hexanoate
Chemistry
1 answer:
zhuklara [117]2 years ago
3 0

Answer:

Explanation:

a )

CH₃CH₂CH₂CH₂CH₂Br + KOH   ⇒ CH₃CH₂CH₂CH₂CH₂OH

CH₃CH₂CH₂CH₂CH₂OH  + acidic potassium dichromate ⇒  CH₃CH₂CH₂CH₂COOH

b )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .

c )

CH₃CH₂CH₂CH₂CH₂Br + KOH   ⇒ CH₃CH₂CH₂CH₂CH₂OH

CH₃CH₂CH₂CH₂CH₂OH  + acidic potassium dichromate ⇒  CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl  chloride ) ⇒ CH₃CH₂CH₂CH₂COCl

d )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂

e )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒  

CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )

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Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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Explanation :

For non-electrolyte solution, the formula used for lowering in freezing point is,

\Delta T_f=k_f\times m

where,

\Delta T_f = lowering in freezing point

k_f = molal depression constant

m = molality

As per question, the molality is same for all the non-electrolyte solution. So, the lowering in freezing point is depend on the k_f only.

That means the higher the value of k_f, the higher will be the freezing point lowering.

From the given non-electrolyte solutions, the value of k_f of carbon tetrachloride is higher than the other solutions.

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