8. The following reaction is a reduction reaction of magnetite (Fe3O4). Fe304 (s

Question

8. The following reaction is a reduction reaction of magnetite (Fe3O4). Fe304 (s) 2C(s, graphite) 2CO2(g) 3Fe(s) a) Calculate standard Gibbs free energy of this reaction. b) Calculate equilibrium constant (K) of this reaction. c) If you start this reaction with 2.0gof Fe3O4, l.0g of graphite and 0.010 g of Co2, calculate equilibr concentration of CO2 and mass of Fe3O4 at the equilibrium. 9. Calculate pH of saturated lead (II) fluoride solution in water. 10. Thorium-230 (2 Th) undergoes alpha decay gradually. If you have 2.0 g of Thorium-230, what will i remaining thorium-230 amount after year? Also, what will be the amount of Radium-226 after lyea life of thorium-230 is 17.23 days. 11. The following shows reaction mechanism of NO2 reaction with Co. Step 1 NO2 NO2 NO3 NO slow Step 2 NO CO NO2 CO2 fast a) Write the rate law of this reaction. b If 500 torr of NO2 and 300 torr of CO were reacted, what will be the concentration of NO2 and CO minutes? Rate constant of this reaction is 1.90 x 10" /torr-s.

Explanation / Answer

reduction of magnetite is

Fe3O4(s) + 2C(s, graphite) --> 2CO2(g) + 3Fe(s)

a) Standard Gibbs free energy :

Delta G of formation of Fe3O4 = 1015.46 KJ / mole

Delta G of formation of CO2 = 394.38 KJ / mole

Delta G of formation of C and Fe = 0

So Delta G of reduction = 2X Delta G of formation of CO2 - Delta G of formation of Fe3O4

Delta G of reduction = 2X(394.38) - (-1015.46) = 226.7 KJ / mole

b) Equilibrium constant is related to equlibrium constant as

Delta G = -RT lnK

R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1

T = let it be 298 K

so Delta G = - 8.314 X 298 X ln K

lnK= - 226.7 X 1000 / 8.314 X 298 = -91.50

So K = 1.82 X 10^-40

9)The lead flouride will dissociate as

PbF2(S) --> Pb2+(aq) + 2F-(aq)

Let solubility of PbF2 = s

so Ksp = sX(2s)2

s = (ksp/4)1/3

The Ksp of PbF2 = 3.6 X 10^-8

solubility = (3.6 x 10-8 /4)1/3

= 2.08 X 10^-2 M

therefore concentration of F- = 2s = 2*0.00208 = 0.00416 M

Now let us calculate the pH.

we know that HF is a weak acid so F- will be a strong base.

           F-(aq) +     H2O(l)   ---> HF(aq) + OH-(aq)
Initial 0.00416                             0         0

Change                 -x                                   +x          +x

equilibrium       0.00416 -x                          x            x

Kb = [OH-][HF]/[F-] = 1.48 X 10-11

(1.48*10-11) = x2 / (0.00416 -x)

We can ignore x in denominator

0.00616 X 10^-11 = x2

x = 2.48X10-7 M

Concentration of OH- = 2.48*10^-7 M

pOH = -log(2.48*10^(-7))

     = 6.605

so pH = 14- pOH

pH = 14-6.605 = 7.395

10 ) Half life of Thorium = 17.23 days

so we can calculate the amount left after 1 year (365 days)

(1/2)n = (1/2)(365/(17.23) = (0.5)21.18 = 4.21 X10^-7 = 4.21 X 10^-5 %

remaining amount from 2 gram of thorium will be = 4.21 X 10^-5 X 2 / 100 = 2.105 X 10^-7 grams will be left

11) The rate law depends upon the slow step

so Rate = K [NO2]2

Rate constant = [NO3][NO] / [NO2]NO2]

b) quesiton image is incomplete.

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