The homogeneous, reversible exothermic, liquid-phase reaction A ↔ R is being carried out in a reactor system consisting of two i
deal CSTRs in series. Both reactors operate at 150°C. The molar flow rate of A entering the first CSTR is 55,000 mol/h, the concentration of A in this stream is 6.5 M, and the concentration of R is zero. The fractional conversion of A in the outlet stream from the second CSTR is 0.75. The reaction is first order in both directions. The rate constant for the forward reaction at 150°C is 1.28 h-1 and the equilibrium constant based on concentration at 150°C is 10.0. If the volume of the second CSTR is 10,000 L, what is the required volume of the first CSTR?
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Answer:
Explanation:
Given parameters:
Molarity of KOH = 0.26M
Volume of H₂SO₄ = 19.76mL
Molarity of H₂SO₄ = 0.20M
Unknown:
Volume of KOH = ?
Solution:
This is a neutralization reaction in which an acid reacts with a base to produce salt and water:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
We solve from the known to the unknown in the reaction.
The known is the acid and from there we can find the number of moles of the acid to be completely neutralized:
Number of moles of acid = molarity x volume
Number of moles of acid = 19.76 x 0.20 = 3.95mol
From the balanced reaction equation:
1 mole of acid reacts with 2 moles of the bases KOH
3.95mole of acid would react with 3.95moles x 2 of the base
Number of moles of reacting base = 7.90moles
To find the volume of base;
Volume of base =
Volume = = 30.40mL
Learn more:
Neutralization brainly.com/question/6447527
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