Question

If you were to leave calcium metal in contact with oxygen at a partial pressure of 1.00 atm at 298 k, would the calcium oxide form spontaneously? Use free energy to answer this.

Answer 1

The reaction that takes place is

2 Ca (s) + O₂ (g) <====> CaO (s)

The equilibrium constant for the reaction can be written in terms of the partial pressures of the gaseous components. Since O₂ is the only gaseous component involved, we can write

K_P = 1/P_CO2

Given P_CO2 = 1.00 atm, we have,

K_P = (1/1.00) = 1.00

The free energy change of the reaction at 298 K is given by

ΔG⁰ = -RTln K_P = -(8.314 J/mol.K)*(298 K)*ln (1.00) = 0.0

Since the free energy change of the reaction is 0.0, the reaction is at equilibrium at 298 K. Hence, CaO will not form spontaneously at 298 K at a partial pressure of 1.00 atm O₂. At equilibrium, formation and decomposition of CaO takes place such that the rates of formation and decomposition of CaO are equal and the reaction seems to be at “rest”.

Since R and T are both positive here, hence the free energy change of the reaction will be negative and hence the reaction will be spontaneous for partial pressures of O₂ greater than 1.00 atm, i.e., when P_CO2 > 1.00 atm at 298 K, the reaction will be spontaneous.