Question

When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount...

When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced.

MO2(s) <-> M(s) + O2(g) Deltra G = 287.3 kJ/mol

When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous.

What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s).

What is the thermodynamic equilibrium constant for the coupled reaction?

K =

Homework Answers

Answer #1

Solution:-

(1) (A) MO2(s) <=> M(s) + O2(g), Delta G1 = 287.3 kJ/mol

(B) C(s) + O2(g) <=> CO2(g), Delta G2 = -394.4 kJ/mol

Now adding equation (A) + equation (B):

MO2(s) + C(s) + O2(g) <=> M(s) + O2(g) + CO2(g)

Cancelling the common terms to get coupled reaction:

MO2(s) + C(s) <=> M(s) + CO2(g)

Delta Go = Delta G1 + Delta G2

= 287.3 + (-394.4)

= -107.1 kJ/mol = -107100 J/mol

(2) Temperature T = 25 deg C = 298.15 K

Molar gas constant R = 8.314 J/mol.K

Equilibrium constant K = exp(-Delta Go/RT)

= exp(-107100/(8.314 x 298.15))

= 5.81 x 10^(18)

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