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.
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
Tf = 5.5ºC
Kf = –5.12
dTf = -5.12*0.76959 = -3.94
Tmix = Tf - dTf = 5.5-3.94 = 1.56 ºC
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]
Kp = 0.0451
Kc = ?
Kp = Kc*(RT)^dn
dn = nf- ni = 2-4 = -2
Kp = Kc*(RT)^-2
T = 235+273 = 508
Kc = (0.0451)(0.082*508)^2 = 78.2585
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