Hi I am quite new to thermodynamic please help to provide so

Hi I am quite new to thermodynamic please help to provide some explanation. Thank you.

The reduction of FeO by carbon monoxide is represented by the equation below: FeO + CO right arrow Fe + Co_2 Calculate the entropy change at 298 K. The data for the related compounds are given below (Entropy chang = diff in entropy between products and reactants): S degree (FeO, 298) = 53.97 J/mol.K S degree (CO,298) = 197.9 J/mol.K S degree (CO_2,298) = 213.63 J/mol.K S degree (Fe,298) = 27.16 J/mol.K In Qn 2, compute entropy change of the following reaction at 600 K and 1 atm pressure.

Solution

change in entropy = entropy of products - entropy of reactants

= ( entropy of Fe + entropy of Co2) - ( entropy of FeO + entropy of CO)

= (27.16 + 213.63 ) - (53.97 + 197.9)

= 240.79 - 251.87

=-11.08 J/mol.K