Water contains 7.0 mg/L of soluble ion (Fe2 ) that is to be oxidized by aeration to a concentration of 0.25 mg/L. The pH of the
water is 6.0 and temperature is 12oC. Assume the dissolved oxygen in the water is in equilibrium with the surrounding atmosphere. Laboratory results indicate the pseudo first-order rate constant for oxygenation of Fe2 is 0.175/min. Assuming steady-state operations and a flow rate of 40,000 m3 /d, calculate the minimum detention time and reactor volume necessary for oxidation of Fe2 to Fe3 . Perform the calculations for both a CMFR and PFR. (You should be able to work this out from information provided in Chapters 3 and 4.)
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Answer:
The correct option is <em>b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)</em>
Explanation:
The relation between Kc and Kp is given by the following equation:
where R is the gas constant (0,082 L.atm/K.mol), T is the temperature (in K) and dn is the change in moles.
The change in moles (dn) is calculated as:
dn = moles of products - moles reactants
If dn=0, RT= 1 ⇒ Kc=Kp
We calculate dn for each reaction from the estequiometrial coefficients of products and reactants as follows:
a) 4 NH₃(g) + 3 O₂(g) ⇌ 2 N₂(g) + 6 H₂O(g)
dn= (2+6) - (4+3) = 1 ⇒ Kc ≠ Kp
b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)
dn = (1+1) - (1+1)= 0 ⇒ Kc = Kp
c) 2 N₂(g) + O₂(g) ⇌ 2 N₂O(g)
dn= 2 - (2+1) = -1 ⇒ Kc ≠ Kp
d) 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
dn = 2 - (2+1) = -1 ⇒ Kc ≠ Kp
The reaction in which Kc=Kp is b), because reactants and products have the same number of moles.