Consider the placing 3.00 g of solid naphthalene in a 5.00 L container at 298 K....

Consider the placing 3.00 g of solid naphthalene in a 5.00 L container at 298 K. After some time equilibrium is established between solid and gaseous naphthalene (KC = 4.29 x 10‐6 )

What is the pressure of gaseous naphthalene at equilibrium?

Expert Solution

moles of naphtalene = 3.00 / 128.17 = 0.0234

P V = n R T

P x 5 = 0.0234 x 0.0821 x 298

P = 0.1145 atm

C10H8 (s) <-------------> C10H8 (g)

0.1145 0

0.1145 - x x

Kp = Kc

Kp = x / 0.1145 - x

4.29 x 10^-6 = x / 0.1145 - x

x = 4.913 x 10^-7

pressure of gaseous naphthalene at equilibrium = 4.91 x 10^-7 atm

queen_honey_blossom answered 2 years ago

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