Algebric signs of ΔH°, ΔS°, and ΔG° of the combustion of gasoline.?

Question

a) ΔH° - , ΔS° + , and ΔG° +
b) ΔH° +, ΔS° - , and ΔG° +
c) ΔH° + , ΔS° + , and ΔG° -
d) ΔH° - , ΔS° + , and ΔG° -

i think that its D

Answer 1

Actually, none of the answers are correct. The correct answer is:

ΔH° - , ΔS° - , and ΔG° -

which is not one of the choices.

The little superscript circles are normally used to signify "standard state" changes, i.e., changes when both the reactants and products are at a pressure of 1 bar and 298.15 K, the conditions that are by convention normally taken as the standard state.

The standard state enthalpy change (ΔH°) is obviously negative (exothermic) because the combustion of gasoline releases heat under standard conditions.

The standard state free energy change (ΔG°) must also be negative, because the combustion of gasoline is a spontaneous process (once initiated) under standard conditions.

The standard state entropy change, however, cannot be determined simply by inspection. If one makes the simplifying assumption that gasoline is made up of pure octane, one can write the combustion reaction as:

C8H18 + 12.5*O2 -> 8*CO2 + 9*H2O

Under standard conditions, 1 mole of liquid (C8H18) plus 12.5 moles of gas (O2) are converted to 9 moles of liquid (water) plus 8 moles of gas (CO2). Now, if there were an net increase in the amount of gas due to the reaction, we would be safe in assuming that the reaction has a positive entropy change because the gaseous state is the most disordered state and production of a mole of gas is generates significant additional molecular disorder. A net increase in the amount of gas would therefore result in a net increase in disorder (and hence entropy). Changes in the number of moles of condensed phases (i.e., liquids and solids) have a much smaller effect on the entropy change of the reaction, and one cannot, in general predict the sign of the change.

One can look up the standard entropies of the reactants and products of the octane combustion reaction on the NIST Chemistry WebBook (see source). The values are:

S° of C8H18(liquid) = 361.2 J/mol/K
S° of O2(g) = 205.1 J/mol/K
S° of CO2(g) = 213.8 J/mol/K
S° of H2O(liquid) = 69.95 J/mol/K

Using these values to calculate the entropy change for the reaction yields:

ΔS°= 8*213.8 + 9*69.95 - 361.2 - 12.5*205.1 = -585 J/mol octane/K, a negative value.

Answer 2

It is the Greek letter capital delta. Delta is used to signify a difference as in subtraction (or change as in temperature change).

Answer 3

ΔH° is read as "delta H" and refers to enthalpy.
ΔS° is read as "delta S" and refers to entropy
ΔG° is read as "delta G" and refers to Gibbs free energy

Answer 4

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