Calculate the standard Gibbs energy of the reaction CO(g) + CH3CH2OH(l) → CH3CH2COOH(l) at 298 K, using the values of standard entropies and enthalpies of formation. The data is given below. CO(g) : ΔfH0 = −110.53 kJ mol-1, Sm0=197.67 J K-1 mol−1 at 298 K. CH3CH2OH(l) : ΔfH0 = −277.69 kJ mol-1, Sm0=160.7 J K-1 mol−1 at 298 K. CH3CH2COOH(l) : ΔfH0 = −510 kJ mol-1, Sm0 = 191 J K-1 mol−1 at 298 K.

-7.2 * 10^4 kJ mol-1

Explanation:

First we obtain the change in enthalpy for the reaction;

ΔHrxn= ΔHproducts - ΔHreactants

ΔHrxn=[( −510 ) - (−110.53) + (−277.69)]

ΔHrxn= -121.78 * 3 J mol-1

The we obtain the entropy change of the reaction

ΔSrxn= ΔSproducts - ΔSreactants

ΔSrxn= [(191) - (197.67) + (160.7)]

ΔSrxn= -167.37  J K-1 mol−1

Then we calculate ΔG at 298 K

ΔG = ΔH - TΔS

ΔG = ( -121.78 * 3) - (298) (-167.37)

ΔG = -7.2 * 10^4 kJ mol-1

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank