5· The equilibrium constant, Kp, for the gas phase isomerization of borneol (C10

Question

5· The equilibrium constant, Kp, for the gas phase isomerization of borneol (C10HnOH) to isoborneol at 503 K is 0.106. 21.50 g of borneol is placed in a container of volume 6.25 liters, heated to 503 K, and allowed to come to equilibrium. Calculate the mass of each substance at equilibrium. Note that the reaction is borneol isoborneol. Both borneol and isoborneol have the same molecular formula and hence the same molecular weight, but they arrange the atoms differently so they are different compounds

Explanation / Answer

Kp = 0.106

Kp = IB/B

mol of B= mass/MW = 21.50/154.25 = 0.13938

V = 6.25 L

T = 503 K

PV = nRT

P-B = nRT/V = (0.13938/6.25 )(0.082*503) = 0.9198

Kp = IB/B

initially

IB = 0

B = 0.9198

in equilbirium

IB = 0 + x

B =  0.9198 - x

Kp = IB/B

0.106 = X/(0.9198 - x)

1/0.106 *x = 0.9198 - x

(1/0.106+1)x = 0.9198

x = 0.9198 /(1/0.106+1) = 0.0881

IB = 0 + x = 0.0881

B =  0.9198 - 0.0881= 0.8317

these are Partial pressures so

nIB = PV/(RT) = (0.0881)(6.25)/(0.082*503) = 0.0133

nB = PV/(RT) = (0.8317)(6.25)/(0.082*503) = 0.1260

change to mass --> mol*MW =

nIB = 0.0133 *154.25 = 2.051525 g

nB = 0.1260*154.25 = 19.4355 g

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.