3) (a) Calculate the rate constant for the oxidation of Fe2+ (below) in a water

Question

3)

(a) Calculate the rate constant for the oxidation of Fe2+ (below) in a water sample taht is open to the atmosphere given that half of the Fe2+ is consumed after 23 seconds at 25 degrees C. Assume first order kinetics with respect to Fe2+/

Fe2+(aq) + O2(aq) <--> Fe3+ (aq) + O2^(-) (aq)

(b) Calculate the activation energy for the oxidation of Fe2+ if half of the Fe2+ is consumed after 15 seconds.

(c) For the same water used in part a above that is not open to the atmosphere, half of the Fe2+ is consumed after 45 seconds at 25 degrees C. Can you explain the difference?

Explanation / Answer

3. a.

We know from first order kinetics

k = 1/t ln [A]0/ [A]

ere k = rate constant

t = time oxidized of Fe2+= 23 s

[A]0 = x (suppose)

[A] = x/2

Hence

      k=1/23 ln 2x/x

     => k = 0.693 /26

=> k = 0.03013 s-1

3. b.

We know from ARrhenius equation

k = Ae-(Ea/RT)

Here,

          k = 0.693/ 15 = 0.0462 s-1 = rate constant

          A =pre-exponential factor (frequecny factor)

          Ea = activation energy for the oxidation of Fe2+

          R = gas constant = 8.314 JK-1mol-1

          T = absoulte temperature

By taking log in both sides

log k = logA - Ea/2.303RT

=> Ea = 1/RT [log A/k]

By knowing T and A we can calculate Ea

3.c.

For 23 s

k (23) = 0.03013 s-1

for 45 s

k (45) = 0.693 / 45 = 0.0154 s-1

Because, rate constant depends on concentration and time for oxidation (cosumed) not on pressure.

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