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Chapter 4: Stoichiometry
Writing -rA Solely as a Function of X.
Write the rate law for the elementary liquid phase reaction
solely in terms of conversion. The feed to the batch reactor is equal molar A and B with CA0 = 2 mol/dm3 and kA= .01 (dm3/mol)41/s.
(a) −rA=kACACB23
(b) −rA=kACACB
(c) −rA=kACA3CB2
Hint 2: What is the concentration of A?
(a) CA=CA0(1−x)
(b) CA=CA0(2−x)
(c) CA=CA0(1−3x)
Hint 3: What is the concentration of B?
(a) CB=CA0(1−23x)
(b) CB=CA0(23−23x)
(c) CB=CA0(1−2x)
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What is the rate law?
The reaction is elementary so what happens to the coefficients from the reaction?
Hint 2
What is the concentration of A?
Liquid phase, v = vo (no volume change)
Back to Problem
Hint 3
What is the concentration of B?
What is ΘB?
Equal molar
Species A is the limiting reactant because the feed is equal molar in A and B, and two moles of B consumes 3 moles of A.
Solution
Write the rate law for the elementary liquid phase reaction
solely in terms of conversion. The feed to the batch reactor is equal molar A and B with CA0 = 2 mol/dm3 and kA= .01 (dm3/mol)41/s.
1) Rate Law: -rA=kC3AC2B
2) Stoichiometry:
Species A
Liquid phase, v = vo (no volume change)
Species B
What is ΘB?
Species A is the limiting reactant because the feed is equal molar in A and B, and two moles of B consumes 3 moles of A.
We now have -rA=f(X) and can size reactors or determine batch reaction times.