Question

Enter vour answer in t he provided box. Many metal oxides are converted to the free metal by reduction with other elements, such as C or Si. For the following reaction, calculate the enthalpy change and the entropy change. Also, determine whether the process is spontaneous at all temperatures or whether the process is nonspontaneous at all temperatures: MnO2(s)+ Si(s) -» Mn(s) + SiO2(s) kJ ASrx,- This process is spontaneous at all temperatures. This process is nonspontaneous at all temperatures

Answer 1

I am taking the values for standard enthalpies of formation, ?H⁰_f and standard entropies, S⁰ from internet sources.

Element/Compound	?H0f (kJ/mol)	S0 (J/mol.K)
MnO2 (s)	-520.0	53.0
Si (s)	0	18.8
Mn (s)	0	32.0
SiO2 (s)	-910.7	41.8

The standard enthalpy of the reaction is given as

?H_rxn = ?n.?H⁰_f (products) - ?n.?H⁰_f (reactants)

= [(1 mole Mn)*?H⁰_f (Mn, s) + (1 mole SiO₂)*?H⁰_f (SiO₂, s)] – [(1 mole MnO₂)*?H⁰_f (SiO₂, s)] – [(1 mole MnO₂)*?H⁰_f (MnO₂, s) + (1 mole Si)*?H⁰_f (Si, s)]

= [(1 mole)*(0 kJ/mol) + (1 mole SiO₂)*(-910.7 kJ/mol)] – [(1 mole)*(-520.0 kJ/mol) + (1 mole)*(0 kJ/mol)]

= (-910.7 kJ) – (-520.0 kJ)

= -910.7 kJ + 520.0 kJ = -390.7 kJ (ans).

The standard entropy of the reaction is given as

?S_rxn = ?n.S⁰_f (products) - ?n.S⁰_f (reactants)

= [(1 mole Mn)*S⁰_f (Mn, s) + (1 mole SiO₂)*S⁰_f (SiO₂, s)] – [(1 mole MnO₂)*S⁰_f (SiO₂, s)] – [(1 mole MnO₂)*S⁰_f (MnO₂, s) + (1 mole Si)*S⁰_f (Si, s)]

= [(1 mole)*(32.0 J/mol.K) + (1 mole)*(41.8 J/mol.K)] – [(1 mole)*(53.0 J/mol.K) + (1 mole)*(18.8 J/mol.K)]

= [(32.0 J/K) + (41.8 J/K)] – [(53.0 J/K) + (18.8 J/K)]

= (73.8 J/K) – (71.8 J/K) = 2.0 J/K (ans).

The standard free energy change of the reaction is given as

?G_rxn = ?H_rxn – T*?S_rxn where T = room temperature = 25°C = (25 + 273) K = 298 K.

Plug in values and get

?G_rxn = (-390.7 kJ) – (298 K)*(2.0 J/K) = (-390.7 kJ) – (596 J)

= (-390.7 kJ) – (596 J)*(1 kJ/1000 J) = (-390.7 kJ) – (0.596 kJ)

= -391.296 kJ ? -391.3 kJ (ans).

Since ?G_rxn is negative, the reaction is spontaneous at room temperature. However the reaction will become non-spontaneous when the temperature increases. We can determine T at which the reaction becomes non-spontaneous.

We find T when ?G_rxn = 0; therefore,

?H_rxn – T*?S_rxn = 0

====> T = ?H_rxn/?S_exn = (-390.7 kJ)/(2.0 J/K) = (-390.7 kJ)*(1000 J/1 kJ)/(2.0 J/K) = -195350 K.

Since, we cannot have negative temperatures in the Kelvin scale, hence, the reaction is spontaneous at all temperatures (ans).