Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
1 answer:
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
b) To calculate the constant we have the following expression:
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
c) The equilibrium pressure of O₂ over M is:
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Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.
(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of will increase then.