Figure 1 shows the elementary reactions of ethylene oxide in series take place in a continuous constant-flow stirred tank reactor (CSTR) under isothermal conditions.(Refer Figure 1)The rate constant, k2=k1 + k3 and with initial CA=CA0 and CB0= CC0= CD0 =0. Given v0= 1000L·h-1, k1= 1 min-1, k2= 2 min-1 and k3=1 min-1 and CA0= 1 mol·h-1. Determine the expression of concentration of each components as a function of timei.e. CA(ꞇ), CB(ꞇ), CC(ꞇ) and CD(ꞇ). ꞇ (CH3CHO)* (B) –CHĄ(C)k3HCOCH2CO2H (D)CH20CH2 (A)Figure 1 Reaction of ethylene oxides in series CONFIDENTIALAPPENDIX 1(1)CE/FEB 2022/CHE502ARRHENIUS LAWk = Ae-E/RTr2k2Ег11In= In -k1r1-R [T,T2.IDEAL GAS CONSTANTdm3 · atmm3 . atmcalR = 0.082= 0.082= 1.987= 8.314mol · Kkmol · Kmol · Kтol - KENERGY BALANCEQ - W, – FA0 )O Cpi(T – T.) – FaoX[AH; + AG,(T – TR)] = 0i=1ADIABTIC TEMPERATUREx[-AHrx(tr)] + E 0;CpiTioΣ 0,cρTο + ΧΔCTT =E 0;Cpi + XAC,EQUILIBRIUM CONVERSIONE 0;Ĉp{(T – T.)-AHRX(TR) + 4Cp (T – TR)X EB =NUMERICAL METHODSTrapezoidal rulehf (X) dX =2zF(X.) + f(X,)]ХоSimpson's one-third rulehf(X) dX =[f (X.) + 4f (X1) + f(X2)]3ХоSimpson's three-eighths rulex33F(X) ax =A[S(X,) + 3/(X,) + 3/(X2) + f(X3)Ih[f (X,) + 3f(X1) + 3f(X2) + f(X3)]8.Xo© Hak Cipta Universiti Teknologi MARACONFIDENTIALCONFIDENTIALAPPENDIX 1(2)EH/JUN 2015/CHE584/594Five-point quadrature formulaX4h"rx) ax =tr(x,) + 4f(X,) + 2f(X,) + 4f(X;) + f(X,)]|[f (X,) + 4f (X,)+2f(X2) + 4f(X3) + f(X4)]3Integralsdx1= ln1 - x1- xdx11(1 – x)21– X21- x1dxJo (1 – x)21 - x*-I _ z(x =dx1= - In(1 + ɛx)1+ EX(1+ e)²x(1 –x)•* (1+ ex)² dx= 2ɛ(1+ ɛ) In(1 – x) + e²x +(1 – x)2Differentiationd[cf (x)] = cf'(x)dxd[f(x)g(x)] = f(x)g'(x) + f'(x)g(x)dxf (x)g'(x) + f'(x)g(x)[S(x)] _ g(x)f'(x) – f(x)g'(x)dx g(x)-[g(x)]²d(x")dx— пхп-1© Hak Cipta Universiti Teknologi MARACONFIDENTIAL

Question

Figure 1 shows the elementary reactions of ethylene oxide in series take place in a continuous constant-flow stirred tank reactor (CSTR) under isothermal conditions.(Refer Figure 1)The rate constant, k2=k1 + k3 and with initial CA=CA0 and CB0= CC0= CD0 =0. Given v0= 1000L·h-1, k1= 1 min-1, k2= 2 min-1 and k3=1 min-1 and CA0= 1 mol·h-1. Determine the expression of concentration of each components as a function of timei.e. CA(ꞇ), CB(ꞇ), CC(ꞇ) and CD(ꞇ). ꞇ (CH3CHO)* (B) –CHĄ(C)k3HCOCH2CO2H (D)CH20CH2 (A)Figure 1 Reaction of ethylene oxides in series CONFIDENTIALAPPENDIX 1(1)CE/FEB 2022/CHE502ARRHENIUS LAWk = Ae-E/RTr2k2Ег11In= In -k1r1-R [T,T2.IDEAL GAS CONSTANTdm3 · atmm3 . atmcalR = 0.082= 0.082= 1.987= 8.314mol · Kkmol · Kmol · Kтol - KENERGY BALANCEQ - W, – FA0 )O Cpi(T – T.) – FaoX[AH; + AG,(T – TR)] = 0i=1ADIABTIC TEMPERATUREx[-AHrx(tr)] + E 0;CpiTioΣ 0,cρTο + ΧΔCTT =E 0;Cpi + XAC,EQUILIBRIUM CONVERSIONE 0;Ĉp{(T – T.)-AHRX(TR) + 4Cp (T – TR)X EB =NUMERICAL METHODSTrapezoidal rulehf (X) dX =2zF(X.) + f(X,)]ХоSimpson's one-third rulehf(X) dX =[f (X.) + 4f (X1) + f(X2)]3ХоSimpson's three-eighths rulex33F(X) ax =A[S(X,) + 3/(X,) + 3/(X2) + f(X3)Ih[f (X,) + 3f(X1) + 3f(X2) + f(X3)]8.Xo© Hak Cipta Universiti Teknologi MARACONFIDENTIALCONFIDENTIALAPPENDIX 1(2)EH/JUN 2015/CHE584/594Five-point quadrature formulaX4h&#34;rx) ax =tr(x,) + 4f(X,) + 2f(X,) + 4f(X;) + f(X,)]|[f (X,) + 4f (X,)+2f(X2) + 4f(X3) + f(X4)]3Integralsdx1= ln1 - x1- xdx11(1 – x)21– X21- x1dxJo (1 – x)21 - x*-I _ z(x =dx1= - In(1 + ɛx)1+ EX(1+ e)²x(1 –x)•* (1+ ex)² dx= 2ɛ(1+ ɛ) In(1 – x) + e²x +(1 – x)2Differentiationd[cf (x)] = cf'(x)dxd[f(x)g(x)] = f(x)g'(x) + f'(x)g(x)dxf (x)g'(x) + f'(x)g(x)[S(x)] _ g(x)f'(x) – f(x)g'(x)dx g(x)-[g(x)]²d(x&#34;)dx— пхп-1© Hak Cipta Universiti Teknologi MARACONFIDENTIAL

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