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Calculation Cstr 40L Appendix
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2) F0 = 0.15 + 0.15 = 0.3 L/min
Information
Volume of sample, Vs = 50 mL
Concentration of NaOH in the feed vessel, CNaOH,f = 0.1 mol/L
Volume of HCL for quenching, VHCl,s = 12 mL
Concentration of HCl in standard solution, CHCls = 0.25 mol/L
Volume of titrated NaOH, V1= 28 mL
Concentration of NaOH used for titration, CNaOHs = 0.1 mol/L
i- Concentration of NaOH that entering the reactor, CNaOH0.
CNaOHo = ½ CNaOHf
= ½ (0.1)
= 0.05 mol/L
ii- Volume of unreacted quenching HCl,V2
V2 = (CNaOHs / CHCls) x V1
= (0.1/0.25) x 28
= 11.2 mL
iii- Volume of HCl reacted with NaOHin sample, V3
V3 = VHCls – V2
= 12 – 11.2
= 0.8
iv- Moles of HCl reacted with NaOH in sample, n1
n1 = (CHCls x V3) / 1000
= 0.25 x 0.8/1000
= 0.0002 mol
v- Moles of unreacted NaOH in sample, n2
n2 = n1
= 0.0002 mol
vi- Concentration of unreacted NaOH in the reactor, CNaOH
CNaOH = n2/Vs x 1000
= 0.0002/50 x 1000
= 0.004 mol/L
vii- Conversion of NaOH in the reactor, X
X = (1- CNaOH / CNaOHo) x 100%
= (1 – 0.004/0.05) x 100%
= 92.0 %
viii- Residence time, τ
τ = VCSTR / Fo
= 10 / 0.30
= 33.33 min
ix- Reaction rate constant, k
k = ( CAo – CA) / τCA2
= ( 0.05 – 0.004) / (33.33 x 0.0042)
= 86.26 M-1 min -1
x- Rate of reaction, -rA
-rA = kCA2
= 86.26 x 0.0042
= 1.380 x 10 -3 mol/L.min
3) F0 = 0.20 + 0.20 = 0.4 L/min
Information
Volume of sample, Vs = 50 mL
Concentration of NaOH in the feed vessel, CNaOH,f = 0.1 mol/L
Volume of HCL for quenching, VHCl,s = 12 mL
Concentration of HCl in standard solution, CHCls = 0.25 mol/L
Volume of titrated NaOH, V1= 28.2 mL
Concentration of NaOH used for titration, CNaOHs = 0.1 mol/L
i- Concentration of NaOH that entering the reactor, CNaOH0.
CNaOHo = ½ CNaOHf
= ½ (0.1)
= 0.05 mol/L
ii- Volume of unreacted quenching HCl,V2
V2 = (CNaOHs / CHCls) x V1
= (0.1/0.25) x 28.2
= 11.28 mL
iii- Volume of HCl reacted with NaOHin sample, V3
V3 = VHCls – V2
= 12 – 11.28
= 0.72
iv- Moles of HCl reacted with NaOH in sample, n1
n1 = (CHCls x V3) / 1000
= 0.25 x 0.72/1000
= 0.00018 mol
v- Moles of unreacted NaOH in sample, n2
n2 = n1
= 0.00018 mol
vi- Concentration of unreacted NaOH in the reactor, CNaOH
CNaOH = n2/Vs x 1000
= 0.00018/50 x 1000
= 0.0036 mol/L
vii- Conversion of NaOH in the reactor, X
X = (1- CNaOH / CNaOHo) x 100%
= (1 – 0.0036/0.05) x 100%
= 92.8 %
viii- Residence time, τ
τ = VCSTR / Fo
= 10 / 0.40
= 25.0 min
ix- Reaction rate constant, k
k = ( CAo – CA) / τCA2
= ( 0.05 – 0.0036) / (25 x 0.00362)
= 143.21 M-1 min -1
x- Rate of reaction, -rA
-rA = kCA2
= 143.21 x 0.00362
= 1.856 x 10 -3 mol/L.min
4) F0 = 0.25 + 0.25 = 0.5 L/min
Information
Volume of sample, Vs = 50 mL
Concentration of NaOH in the feed vessel, CNaOH,f = 0.1 mol/L
Volume of HCL for quenching, VHCl,s = 12 mL
Concentration of HCl in standard solution, CHCls = 0.25 mol/L
Volume of titrated NaOH, V1= 28 mL
Concentration of NaOH used for titration, CNaOHs = 0.1 mol/L
i- Concentration of NaOH that entering the reactor, CNaOH0.
CNaOHo = ½ CNaOHf
= ½ (0.1)
= 0.05 mol/L
ii- Volume of unreacted quenching HCl,V2
V2 = (CNaOHs / CHCls) x V1
= (0.1/0.25) x 28.0
= 11.2 mL
iii- Volume of HCl reacted with NaOHin sample, V3
V3 = VHCls – V2
= 12 – 11.2
= 0.8
iv- Moles of HCl reacted with NaOH in sample, n1
n1 = (CHCls x V3) / 1000
= 0.25 x 0.8/1000
= 0.0002 mol
v- Moles of unreacted NaOH in sample, n2
n2 = n1
= 0.0002 mol
vi- Concentration of unreacted NaOH in the reactor, CNaOH
CNaOH = n2/Vs x 1000
= 0.0002/50 x 1000
= 0.004 mol/L
vii- Conversion of NaOH in the reactor, X
X = (1- CNaOH / CNaOHo) x 100%
= (1 – 0.004/0.05) x 100%
= 92.0 %
viii- Residence time, τ
τ = VCSTR / Fo
= 10 / 0.50
= 20.0 min
ix- Reaction rate constant, k
k = ( CAo – CA) / τCA2
= ( 0.05 – 0.004) / (20 x 0.0042)
= 143.75 M-1 min -1
x- Rate of reaction, -rA
-rA = kCA2
= 143.75 x 0.0042
= 2.30 x 10 -3 mol/L.min
5) F0 = 0.30 + 0.30 = 0.6 L/min
Information
Volume of sample, Vs = 50 mL
Concentration of NaOH in the feed vessel, CNaOH,f = 0.1 mol/L
Volume of HCL for quenching, VHCl,s = 12 mL
Concentration of HCl in standard solution, CHCls = 0.25 mol/L
Volume of titrated NaOH, V1= 28 mL
Concentration of NaOH used for titration, CNaOHs = 0.1 mol/L
i- Concentration of NaOH that entering the reactor, CNaOH0.
CNaOHo = ½ CNaOHf
= ½ (0.1)
= 0.05 mol/L
ii- Volume of unreacted quenching HCl,V2
V2 = (CNaOHs / CHCls) x V1
= (0.1/0.25) x 28.0
= 11.2 mL
iii- Volume of HCl reacted with NaOHin sample, V3
V3 = VHCls – V2
= 12 – 11.2
= 0.8
iv- Moles of HCl reacted with NaOH in sample, n1
n1 = (CHCls x V3) / 1000
= 0.25 x 0.8/1000
= 0.0002 mol
v- Moles of unreacted NaOH in sample, n2
n2 = n1
= 0.0002 mol
vi- Concentration of unreacted NaOH in the reactor, CNaOH
CNaOH = n2/Vs x 1000
= 0.0002/50 x 1000
= 0.004 mol/L
vii- Conversion of NaOH in the reactor, X
X = (1- CNaOH / CNaOHo) x 100%
= (1 – 0.004/0.05) x 100%
= 92.0%
viii- Residence time, τ
τ = VCSTR / Fo
= 10 / 0.60
= 16.67 min
ix- Reaction rate constant, k
k = ( CAo – CA) / τCA2
= ( 0.05 – 0.004) / (16.67 x 0.0042)
= 172.466 M-1 min -1
x- Rate of reaction, -rA
-rA = kCA2
= 172.466 x 0.0042
= 2.760 x 10 -3 mol/L.min
