The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.
<h3>How to calculate the effluent flow?</h3>
It should be noted that the total inflow will be equal to the total outflow. Therefore,
0.2 + 0.048 = 0.05 + We
Collect like terms
Qe = 0.2 + 0.048 - 0.05
Qe = 0.198m³/sec
The concentration will be:
= (360 × 1000)/0.05
= 7200mg/L.
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