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faltersainse [42]
3 years ago
5

The elementary gas-phase reaction takes place isobarically and isothermally in a PFR where 63.2% conversion is achieved. The fee

d is pure A. It is proposed to put a CSTR of equal volume upstream of the PFR. Based on the entering molar flow rate to A to the first reactor, what will be the intermediate from the CSTR, X 1 , and exit conversion from the PFR, X 2 , based on the feed to first reactor?

Chemistry
1 answer:
a_sh-v [17]3 years ago
6 0

Answer:

Explanation:

check below for explanation in the attached files.

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
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Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

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-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

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Answer:

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Hope this helps! :) Plz mark as brainliest if u can!

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C. the rock may have diferrent textures 

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