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.

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.

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