Consider the following reaction: 2SO3(g) 2SO2(g) + O2(g) If 0.164 moles of SO3, 0.253 moles of SO2, and 0.346 moles of O2 are at equilibrium in a 12.5 L container at 1.20×103 K, the value of the equilibrium constant, Kp, is __.


Step 1 of 2

Given data: 0.164 moles of SO3, 0.253 moles of SO2, and 0.346 moles of O2 are at equilibrium in a 12.5 L container at 1.20×103.

Find: we need to find the value of the equilibrium constant, Kp.

Explanation:

we need to use the given concentrations of the reactants and products at equilibrium.

Step 2 of 2

To find the value of the equilibrium constant, Kp, we need to use the given concentrations of the reactants and products at equilibrium.

The balanced equation for the reaction is:

2SO3(g)⟶2SO2(g)+O2(g)

The equilibrium constant expression, Kp, can be written as:

Kp=(P(SO2)2⋅P(O2)) / (P(SO3)2)

We are given the following concentrations at equilibrium:

[SO3] = 0.164 moles / 12.5 L = 0.01312 M

[SO2] = 0.253 moles / 12.5 L = 0.02024 M

[O2] = 0.346 moles / 12.5 L = 0.02768 M

Now, let’s substitute these values into the Kp expression:

Kp=(0.02024)2×0.02768(0.01312)2Kp=6.537

Explanation:

Therefore, the value of the equilibrium constant, Kp, is 6.537.

Final solution

By, using the given concentrations of the reactants and products at equilibrium, the value of the equilibrium constant, Kp, is 6.537.

 

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