Three equations for the oxidation of a metal are known: a) 2M + O2 (g) = 2MO (s)
ID: 475250 • Letter: T
Question
Three equations for the oxidation of a metal are known:
a) 2M + O2 (g) = 2MO (s) : delta G = -290,400 + 46.1T cal.
b) 2M + O2 (g) = 2MO (s) : delta G = -358,754 + 102.6T cal.
c) 2M + O2 (g) = 2MO (s) : delta G = = 298,400 + 55.4T cal.
One of these equations is for the oxidation of solid M, one is for the oxidation of liquid M and one is for the oxidation of gaseous M.
(1) Determine which equation is for which oxidation (justify your choices)
(2) Calculate the melting and normal boiling points of M.
Explanation / Answer
Delta G = -290400 + 46.1 T
It means that entahlpy change = -290400 cal while entropy change = 46.1 Cal
delta G = -358,754 + 102.6T cal,
It means the enthalpy change = -358754 and entropy change is 102.6 cal
delta G = = 298,400 + 55.4T cal.
It means the enthalpy change = 298,400 (endothermic)
Entropy change = 55.4
So both are positive hence Delta G for this process is positive hence the reaction will not be sponateous
The entropy change will be more if there will be change from more ordered phase to less ordered phase
the maximum entropy change is in second equation
2M + O2 (g) = 2MO (s) : delta G = -358,754 + 102.6T cal.
So this is possible that this is the oxidation of solid to solid
the next is
2M + O2 (g) = 2MO (s) : delta G = = 298,400 + 55.4T cal.
So this is possible that this is the oxidation of liquid to solid
and hence 2M + O2 (g) = 2MO (s) : delta G = -290,400 + 46.1T cal.
The oxidation of gas to solid.