Question

Copper shot is dissolved in an aerated acid solution of constant strength and temperature of 35C for secondary recovery of copper. The shot is uniform in size and has a diameter of 0.2 cm. Laboratory studies on sheet copper using the same solution and temperature measured a linear rate constant of 1.25 x 10-2 cm/hour at 35C and an activation energy of 5.0 Kcal/mole.

Determine the following:

a) Time required (hours) for 50% dissolution of the copper.

b) Retention time required for complete dissolution

c)If the acid solution above was increased in temperature to 55C, what would be the retention time for complete dissolution?

Answer 1

the unit of the rate constant cm/hr shows that reaction is of zero order . By using the formula time required for 50% = 0.1 cm dissolution of copper is 8 hours . For complete dissolution is 16 hour . At 55^o C it is 9.7 hours .

Calculations are explained below.

fk is rate conste) rate & [AIM rate = K[AIM lonen – KTAI time Here com/her - KTAIM rate of dissolution Cm - Cm [AIM mur = m=0 zero order unit of k=canlar Ais in un reaction Ao = 0.2cm for a) K= 1.25*102 cnlar zero order Rean A = Ao - kk 50% completion A- Ao = -Kt Ao-it = kt 250 0.2-0.1 = kl 0.1 - 1.25x10 2t 0.1 Wire = b 1.250 10-2 158 hr it) q Ao ord = Remaining Reactant 501. - 0.50 diameter = 0.2cm. 50% = 0.1cm = 0.08 x 10² he = t-

6) A-AO-KA 100% completion no 0.2cm O = 0.2 - 1.25x10 2t 0.2 hour to 1.250 102 16 hour - t c) en k a Ea | T2-T ) a= 500 Kcal/molo 112 @ T, = 350 = 308 K @ Ta = 55 c = 326K R = 1.99 Callmark k, = 1.25 X 10 -2 un how , K₂ = ? In ka 1. 25x102 con 1 hour 5.08103 cal mole 13284-308 K 1 1.gg callmak | 328 K X 308k) 100 X 103 201.04 x103 2 1. 25x102 cmlhaus - € 0.5 K = 1658 1.25 X 102 cm 1 hour M₂ = 2.06 x 102 con/ hour A=0 Now A = Aukt no 20.2cm 0=0.2 - 2.06 x 102 in show Odhour at & got hour