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aleksklad [387]
3 years ago
10

Water with a flowrate of 1 m3/min and benzene concentration of 150 mg/L flows through a 10 meter long, 0.1 m diameter steady sta

te and conservative PFR, after which it is fed into a CMFR with a volume of 100 m3. A second water stream with a flowrate of 0.5 m3/min and benzene concentration of 150 mg/L is also fed into the CMFR. The CMFR is steady state and conservative. Calculate the effluent concentration of the CMFR.
Chemistry
1 answer:
padilas [110]3 years ago
6 0

Answer:

the effluent concentration of the CMFR is 150 mg/L

Explanation:

Given that;

flowrate of first water Q₁ = 1 m³/min

benzene concentration C₁ = 150 mg/L

flowrate of second water Q₂ = 0.5 m3/min

benzene concentration C₂ =  150 mg/L

To Calculate the effluent concentration of the CMFR, we use the expression;

C_{effluent} = (Q₁C₁ + Q₂C₂) / Q₁ + Q₂

we substitute

C_{effluent} = (1 × 150 + 0.5 × 150) / 1 + 0.5

C_{effluent} = 225 / 1.5

C_{effluent} = 150 mg/L

Therefore, the effluent concentration of the CMFR is 150 mg/L

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Answer:

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Explanation:

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2. For the reaction, 2SO2(g) + O2(g) 2SO3(g), at 450.0 K the equilibrium con¬stant, Kc, has a value of 4.62. A system was charge
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Answer:

Q >> Kc

We have more products than reactans. To reach the equilibrium, the balance will shift to the left.

Explanation:

Step 1: Data given

Temperature = 450.0 K

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When Kc > Q, we have more reactants than products. To reach the equilibrium, the balance will shift to the right

When Kc < Q, we have more products than reactans. To reach the equilibrium, the balance will shift to the left.

When Kc = Q,the equiation isatequilibrium

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Step 2: The balanced equation

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Step 3: Calculate the Q

Q = [SO3]² / [O2][SO2]²

Q = 0.254²/ (0.500 * 0.00855²)

Q = 1765

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