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Rzqust [24]
3 years ago
11

The alkanes are a homologous series of compounds containing only carbon and hydrogen that have the general formula

ps://tex.z-dn.net/?f=C_nH_%7B2n%2B2%7D" id="TexFormula1" title="C_nH_{2n+2}" alt="C_nH_{2n+2}" align="absmiddle" class="latex-formula">. Members of this series include butane (C₄H₁₀), 3,3-dimethylpentane (C₇H₁₆), hexane (C₆H₁₄), and heptane (C₇H₁₆).
The boiling points for a set of compounds in a homologous series can be qualitatively predicted using intermolecular force strengths. Using their condensed structural formulas, rank the homologous series for a set of alkanes by their boiling point.
Chemistry
1 answer:
AnnZ [28]3 years ago
8 0

Answer:

Heptane > hexane> 3,3-dimethylpentane > butane

Explanation:

All things being equal, it is expected that the boiling point of the members of the alkane homologous series increase as we move down the group.

Hence for example, butane is expected to have a higher boiling point than ethane. Another important factor we would be considering in this answer is the presence of a branched alkane amongst the answers in the question. A straight-chain alkane will have a boiling point which is higher than that of a branched-chain alkane due to the greater surface area in contact, thus the greater van der Waals forces, between adjacent molecules

Now if we are to rank the compounds on the basis of boiling point, the answer is as follows:

Heptane > hexane > 3,3-dimethylpentane> butane

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or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
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Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

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