Answer:
Explanation:
Given that:
The flow rate Q = 0.3 m³/s
Volume (V) = 200 m³
Initial concentration
= 2.00 ms/l
reaction rate K = 5.09 hr⁻¹
Recall that:







where;







Thus; the concentration of species in the reactant = 102.98 mg/l
b). If the plug flow reactor has the same efficiency as CSTR, Then:
![t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=t%20_%7BPFR%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B5.09%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7B200%7D%7B102.96%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D0.196%20%5CBig%20%5B%20In%20%28%201.942%29%20%5CBig%20%5D)





The volume of the PFR is ≅ 140 m³
Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature
= 
= rate constant at temperature
=
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)

The activation energy Ea for this reaction is 22689.8 J/mol
Neutralization reaction is the reaction between acid and base.
There the neutralization reaction is:
HBr + LiOH ----------> LiBr + H2O.
Hope this helps!
I’m so sorry i have to do this 5L 24
heterogeneous, because it does not have a uniform texture
hope That helps