For the reaction below, Kp = 29.95 at 800 K. Calculate the equilibrium partial pressures of...

For the reaction below, Kp = 29.95 at 800 K. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are P_PCl5 = 0.3900 atm and P_PCl3 = 0.4300 atm. Assume Cl2 is 0 atm. PCL5 (g) > PCl3 (g) + Cl2 (g)

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Expert Solution

Let's write the reaction once again:

PCl₅ ----------------------> PCl₃ + Cl₂

i) 0.39 0.43 0

eq) 0.39 - x 0.43+x x

Kp = pCl₂ pPCl₃ / pPCl₅

Now, let's solve for x:

29.95 = (0.43 + x) (x) / (0.39 - x)

29.95 (0.39 - x) = x (0.43 + x)

11.681 - 29.95x = x² + 0.43x --------> At this point, we can rearrange the equation to a 2nd grade equation:

x² + 0.43x + 29.95x - 11.681 = 0

x² + 30.38x - 11.681 = 0

x = -30.38 (30.38² + 4x1x11.681)^1/2 / 2x1

x = -30.38 31.14 / 2

x₁ = 0.38

x₂ = -34.76

The value we are going to take would be 0.38 so:

pCl5 = 0.39 - 0.38 = 0.01 atm

pCl₃ = 0.43 + 0.38 = 0.81 atm

pCl₂ = 0.38 atm

This wuld be the correct values, and it's just one way to solve this problem.

The other way is doing in a longer way by calculating Kc,applying this expression: Kc = Kp (RT)ⁿ then with the ideal gas equation PV = nRT or P = cRT calculate c to all the gases, and doing later the same procedure as before to calculate x, and then converting again into pressure. You can choose the way you like the most.

queen_honey_blossom answered 11 months ago

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