The equilibrium constant in terms of pressures for the reduction of tungsten(IV) oxide to tungsten at...

The equilibrium constant in terms of pressures for the reduction of tungsten(IV) oxide to tungsten at 25 °C is K_p = 3.82×10^-4, corresponding to the reaction

WO₂(s) + 2CO(g) = (double arrow) W(s) + 2CO₂(g)

If the total pressure of an equilibrium system at 25 °C is 2.67 atm, calculate the partial pressures of CO(g) and CO₂(g).

P_CO = atm

P_CO₂ = atm

Expert Solution

Few things we have to remember in this type of problems. The total pressure given in the problem is the sum of the partial pressure of the gas components. So here also the sum of the partial pressure of CO₂ and CO is the total pressure. Another thing is that the partial pressure applying by solid components is measured as 1. So, we can reduce the equilibrium constant in terms of pressure (K_p) by eliminating these terms. The partial pressure of both the gases are calculated here. We can do a back calculation also to check the answer if it is right or wrong.

queen_honey_blossom answered 11 months ago

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