2. The data below summarize the loss of a solute from water due to a chemical re

Question

2. The data below summarize the loss of a solute from water due to a chemical reaction in a batch reactor. Determine whether the data are best fitted by a zero-order, first-order, or second-order rate ex Concentration vs. time data from a kinetics experiment for a reactive chemical in water pression, and report the rate constant with proper units. Time (h Concentration (mg L) 100 27.0 0.5 1.0 2.0 3.0 3. The stoichiometry of the reaction of hydrogen sulfide with dissolved oxygen is given by H2S(aq) + 202(ag) SO42-+ 2H- At a constant dissolved oxygen concentration of 0.25 mM (8 mg/L), this reaction is first order with respect to H2S (at pH

Explanation / Answer

2)

let us assume, it would be zero order

a = 100 mg/l a-x = 27 mg/l after 0.5 hr

k0 = a-(a-x)/t

    = (100-27)/0.5

    = 146 mg.L-1.s-1

after 1 hr

k0 = (100-11.1)/1 = 89 mg.L-1.s-1

K0 is different so that its not zero order.

let us assume, it would be first order

k1 = 2.303/t log(a/a-x)

= (2.303/0.5)log(100/27)

= 2.62 hr-1

at 1 hr

K1 = 2.303/t log(a/a-x)

= (2.303/1)log(100/11.1)

= 2.2 hr-1

let us assume, it would be second order

k2 = 1/t(1/a-x -1/a)

   = 1/0.5((1/27)-(1/100))

k2 = 0.054

after 1 hr

= 1/1((1/11.1)-(1/100))

= 0.08

K2 = 1/2((1/6.5)-(1/100))

     = 0.072

let us assume, it would be third order

k3 = 1/(n-1)t(1/(a-x)^2 -(1/a^2))

   = (1/(2*1))((1/(27^2))-(1/100^2))

    = 0.00127

k3 = (1/(2*1))((1/(11.1^2))-(1/100^2))

    = 0.004

i have solved for 0,1,2,3 orders. But no equation had given similar values.but in 2nd order it gave some what closervaues.so that it may be 2nd order.

3.) if pH<7 Rate constant k1 = 2.75*10^(-3) h-1.

k1 = 2.303/t log(a/a-x)

2.75*10^(-3) = (2.303/24)log(10^(-5)/a-x)

a-x = concentration after one day = 24 hr = 9.36*10^(-6) M

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