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

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in   melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.