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Chemical Reaction Engineering
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The following gas phase reactions are carried out isothermally in a CSTR.
A → 2R; r1 = k1pA;
k1 = 20 mole/(sec.m3.bar)
A → 3S; r2 = k2pA;
k2 = 40 mole/(sec.m3.bar)
What is the maximum possible value of FR (mole/sec)?

Answer & Solution
Correct Answer: Option C
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1 Comment
Abrar Nasser
Abrar Nasser 5 months ago
r =r1+r2
r=K1+K2
rR=2r1
rA=20+40pA
rR=2×20pA
rR/rA=60pA/40PA
=60/40
=3/2