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Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula: C4H10- Calculate the mole percent of isobutane in an equilibrium mixture at 25° C, given that the standard Gibbs energy of formation of butane is -15.9 kJ/mol and that of isobutane is -18.0 kJ/mol. Does your result support the notion that straight-chain hydrocarbons (that is, hydrocarbons in which the C atoms are joined in a line) are less stable than branched-chain hydrocarbons?

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Answer:

Explanation:

Butane ⇒ Isobutane

ΔG = RT ln K

ΔG = - 18 + 15.9 = - 2.1 kJ / mol

- 2.1 x 10³ = 8.3 x 298 x lnK

lnK = .85

K = 2.33

[ isobutane ] / [ butane ]  = 2.33

isobutane / total mixture = 2.33 / 3.33 = .7

% of isobutane = 70 % .

70% is isobutane and 30 %  butane . Since equilibrium is tilted towards isobutane , it is more stable . Butane is straight chain and isobutane is branched chain compound .

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