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larisa [96]
3 years ago
5

On the addition of HI

Chemistry
1 answer:
victus00 [196]3 years ago
8 0

Answer:

CH_3-CH(I)-CH(CH_3)-CH_3

I-CH_2-CH_2-CH(CH_3)-CH_3

Explanation:

Given the compound CH_2=CH-CH(CH_3)-CH_3

The steps of reaction:

HI \longrightarrow H^+ + I^-

CH_2=CH-CH(CH_3)-CH_3 + H^+ \longrightarrow CH_3-CH^+-CH(CH_3)-CH_3

<u>But this intermediate product has a resonance: </u>

CH_3-CH^+-CH(CH_3)-CH_3 \longleftrightarrow C^+H_2-CH2-CH(CH_3)-CH_3

The reaction with I-

CH_3-CH^+-CH(CH_3)-CH_3 + I^- \longrightarrow CH_3-CH(I)-CH(CH_3)-CH_3

C^+H_2-CH2-CH(CH_3)-CH_3 + I^- \longrightarrow I-CH_2-CH_2-CH(CH_3)-CH_3

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2 years ago
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Answer:

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

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Equation of the reaction:

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First:

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For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

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