The U. S. requires automobile fuels to contain a renewable component. The fermentation of glucose from corn produces ethanol, which is added to gasoline to fulfill this requirement: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Calculate ΔH o , ΔS o and ΔG o for the reaction at 25°C. Is the spontaneity of this reaction dependent on temperature?

Explanation:

The chemical equation for the reaction is :

The standard enthalpy of formation  of the above equation is as follows:

= -1274.4 kJ/mol

= -277.7 kJ/mol

= -393.5 kJ/mol

where ;

= stochiometric coefficients of products

stochiometric coefficients of reactants

= formation standard enthalpy of products

= formation standard enthalpy of reactants

For ;

The standard enthalpy of formation of of the reactant and the products are :

The is as follows:

  (to kJ/K)

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

As is negative; the reaction is spontaneous

= negative

= positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction  is dependent on temperature.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j