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ollegr [7]
3 years ago
8

You subjected the secondary alkyl halides to both SN2 and SN1 reaction conditions. By which mechanism did each alkyl halide reac

t faster, SN2 or SN1
Chemistry
2 answers:
Andre45 [30]3 years ago
7 0

Answer: SN2

Explanation:

While secondary akyl halides react both by SN1 and SN2 mechanism depending on the conditions of the reaction, they react faster under SN2 conditions if the alkyl halide substrate is not hindered by bulky groups. This is because the nucleophile accesses the carbon atom more easily from the backside making SN2 mechanism faster than SN1

meriva3 years ago
5 0

Answer:

The stability of the carbocation can the type SN1 be favored and by decreasing the steric hindrance can the type of reaction SN2 be favored.

Explanation:

The type of SN reaction works through the formation of the carbocation and its rate depends on the stability. The type of SN2 reaction leads to the formation of the transition and its rate is dependent for the secondary alkyl halide. for a secondary alkyl halide, both type SN1 and type SN2 favor this reaction in the same way, but only by increasing the stability of the carbocation can the type SN1 be favored and by decreasing the steric hindrance can the type of reaction SN2 be favored.

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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).

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\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

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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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3 years ago
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LUCKY_DIMON [66]

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