C+1/2O2→CO
2Ag+1/2O2→Ag2O
Mg+1/2O2→MgO
CO+1/2O2→CO2
C(s)+O2(g)→CO2(g): Entropy of solids is negligible. So here one molecule of gas is resulting in one molecule of gas. Hence there is almost no net entropy. So there will be no slope, it is completely horizontal.
C(s)+1/2O2(g)→CO(g): Here one mole of gas is giving you two moles of gas as products. So here the entropy will be positive. And as a result, this curve will go downwards.
Hence, the correct option is C+1/2O2→CO