Question

1. One liter of saturated actinium hydroxide solution contains .021 g of dissolved Ac(OH)3. Calculate Ksp for Ac(OH)3

CORRECT ANS: 8.8E-16 solve and explain pls

2. For a particular reaction at STP ( 298 K) CHANGE IN H= -297kJ/MOL AND CHANGE IN S= -113.3 J/MOL*K. At which of the following temperatures would the reaction become spontaneous?

CORRECT ANS: 2450 K

Solve and show steps

3. Rate data have been determined at a particular temp for the overall reaction 2NO + 2h2-> N2 + 2H2O IN WHICH ALL REACTANTS AND PRODUCTS ARE GASES

TRIAL INITIAL [NO] INTIAL [H2] INITIAL RATE (MS^-1)

1 .1M .2M .0150

2 .1M .3M .0225

3 .2M .2M .0600

(which column do you divide by what to get the power for determining that rate law expression?

4. Molten AlCl3 is electrolyzed for 5.0 hours with a Current of .4 Amps. Metallic Al is produced at one electrode and chlorine gas (Cl2) is produced at the other. How many grams of aluminum are produced?

CORRECT ANS: 0.67 g (EXPLAIN AND SHOW WORK PLS)

Answer 1

1) we know that

moles = mass / molar mass

also

molar mass or Ac(OH)3 = 274

now

moles of Ac(OH)3 = 0.021 / 274

moles of Ac(OH)3 = 7.664 x 10-5

now

molar solubility = moles / volume (L)

so

molar solubility of Ac(OH)3 = 7.664 x 10-5 / 1 = 7.664 x 10-5

now

Ac(OH)3 --> Ac+3 + 3OH-

let the molar solubility be s

then

[Ac+3]= s

[OH-] = 3s

Ksp = [Ac+3] [OH-]^3

so

Ksp = [s] [3s]^3

Ksp = 27s4

Ksp = 27 ( 7.664 x 10-5)^4

Ksp = 9 x 10-16

2) for a reaction to be spontaneous

dG < 0

now

dG = dH - dTS

so

dH - TdS < 0

(-297 x 1000) - ( T x -113.3) < 0

-297000 + 113.3T < 0

113.3 T < 297000

T < 2621

so

the temperature should be less than 2621 K

3)

let the rate law be

rate = k [ No]^a [ H2]^b

from trail 1 and 2

0.0225 /0.015 = ( 0.3 / 0.2 )^b

b = 1

from trail 1 and trail 3

0.06 / 0.015 = ( 0.2 / 0.1 )^a

a = 2

so

the rate law is

rate = k [ N0]^2 [H2]

4)

Al+3 + 3e- --> Al

according to fardays first law of electrolysis

m = I x t x M / F x Z

given

I = 0.4

time = 5 hr = 5 x 60 x 60 s

z =3 as three electrons are transferred

so

m = 0.4 x 5 x 60 x 60 x 27 / 96485 x 3

m = 0.67

so

0.67 g of Al is produced