Question

# 1) A solution made by dissolving 0.658 moles of non electrolyte solution in 855 grams of...

1) A solution made by dissolving 0.658 moles of non electrolyte solution in 855 grams of benzene. Calculate the freezing point of the solution.
2) write the equilibrium constant expression for the reaction: A(s)+ 3B(l)⇌2C(aq)+D (aq) in terms of [A], [B], [C] and [D] as needed.
3) The equilibrium constant for the chemical equation: N2(g)+3H2(g)⇌2NH3(g) is Kp=0.0451 at 235 degrees Celsius. Calculate the value of the Kc for the reaction at 235 degrees Celsius.

1)

Colligative properties:

dTf = -Kf*m

m = molality = mol per kg solvent

kg solvent = 855 g = 0.855 kg

mol solute = 0.658

mol = mass/MW = 0.658/(0.855) = 0.76959 molal

for benzene

Tf = 5.5ºC

Kf = –5.12

then

dTf = -5.12*0.76959 = -3.94

Tmix = Tf - dTf = 5.5-3.94 = 1.56 ºC

2)

for an equilibrium

Kc = [products]^p / [reactans]^r

Kc = (C^2)(D)/(A)(B^3)

since, for liquid and solids, activity = 1; then

Kc = (C^2)(D)/(A)(B^3) turns to

Kc = [C]^2[D]

3)

Kp = 0.0451

Kc = ?

the relationship:

Kp = Kc*(RT)^dn

dn = nf- ni = 2-4 = -2

then

Kp = Kc*(RT)^-2

Kc= Kp*(RT)^2

T = 235+273 = 508

Kc = (0.0451)(0.082*508)^2 = 78.2585

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