Solid naphthalene (C_10H_8, MW = 128) sublimes, so that it may be used to fumiga

Question

Solid naphthalene (C_10H_8, MW = 128) sublimes, so that it may be used to fumigate enclosed spaces with its vapors: C_10H_8(s) harr C_10H_8(g), Kc = 4.29 times 10^-6 (at 298 K) Suppose that initially 1.00 g of the solid is placed into a ten-liter container at 25 degree C. Approximately what mass of the C_10H_8 did not sublime once equilibrium is established? Solid naphthalene (C_10H_8, MW = 128) sublimes, so that it may be used to fumigate enclosed spaces with its vapors: C_10H_8(s) harr C_10H_8(g), Kc = 4.29 x 10^-6 (at 298 K) Suppose that initially 1.00 g of the solid is placed into an evacuated ten-liter container at 25 degree C. At 25 degree C, approximately what equilibrium pressure will result? (in atm)

Explanation / Answer

C10H8(s) <----> C10H8(g) Kc = 4.29*10^-6

No of mol of napthalene = 1/128= 0.0078125 mol

Concentration of napthalene = 0.0078125/10 = 7.8*10^-4 M

kC = [c10h8] = 4.29*10^-6

kp = (4.29*10^-6 )*(0.0821*298)^1

   = 1.05*10^-4

kp = pC10H8 = 1.05*10^-4

EQUILIBRIUM PRESSURE = 1.05*10^-4 atm

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