<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.
1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.
2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.
3) Halogenation of Cyclohexene (
Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.
4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>
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 same number of each element present before the reaction takes place must also be present on the product side of the equation. Coefficients are placed in front of a chemical formula to show the number of moles of that substances that are necessary for the reaction to occur.
Explanation:
<u>Answer:</u> The partial pressure of helium gas is 0.80 atm
<u>Explanation:</u>
Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.
To calculate the partial pressure of helium gas, we use the law given by Dalton, which is:

We are given:
Total pressure of the cylinder,
= 1.00 atm
Vapor pressure of oxygen gas,
= 0.20 atm
Putting values in above equation, we get:

Hence, the partial pressure of helium gas is 0.80 atm