When heated, lanthanum(III) oxalate decomposes as follows: La2(C2O4)3(s) <===> La2O3(s) + 3 CO2(g) + 3 CO(g) Starting with just the oxalate in a 10.0 L flask, at equilibrium the TOTAL pressure observed is 0.200 atm. What is the value of KP for the equilibrium? (Dalton’s Law of Partial Pressure!)

Partial pressure of [tex]CO_2[/tex] = [tex]\frac{\text{Moles of }CO_2}{\text{Total moles}}\times \text{Total pressure}=\frac{1}{2}\times 0.2atm=0.1atm[/tex]

and,

Partial pressure of [tex]CO[/tex] = [tex]\frac{\text{Moles of }CO}{\text{Total moles}}\times \text{Total pressure}=\frac{1}{2}\times 0.2atm=0.1atm[/tex]

Now we have to calculate the value of equilibrium constant.

The expression of equilibrium constant [tex]K_p[/tex] for the reaction will be:

[tex]K_p=(p_{CO_2})^3\times (p_{CO})^3[/tex]

Now put all the values in this expression, we get :

[tex]K_p=(0.1)^3\times (0.1)^3[/tex]

[tex]K_p=1.0\times 10^{-6}[/tex]

Therefore, the value of [tex]K_p[/tex] for the equilibrium is [tex]1.0\times 10^{-6}[/tex]

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-03-23T07:46:30+01:00

1 × 10⁻⁶ is the value of Kp for the equilibrium of the given equation.

How we calculate the equilibrium constant?

Equilibrium constant for the given reaction in terms of partial pressure will be calculated as Kp = (pCO₂)³ × (pCO)³.

Equilibrium constant will depend only these two quantities because other quantities are present in the pure solid form and there value is 1.

Given pressure for the reaction = 0.200atm

Given chemical reaction is:

La₂(C₂O₄)₃(s) ⇄ La₂O₃(s) + 3CO₂(g) + 3CO(g)

For this first we calculate the partial pressure of CO₂ & CO.

pCO₂ = (moles of CO₂/total moles) × total pressure

pCO₂ = 3/6 × 0.200 = 0.1 atm

pCO = 3/6 × 0.200 = 0.1 atm

Now we put these values on the above equation of equilibrium constant, we get

Kp = (0.1)³ × (0.1)³

Kp = 1×10⁻⁶

Hence, value of Kp is 1×10⁻⁶.

To know more about equilibrium constant, visit the below link:

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