Pregunta: Problema 1Consider the irreversible, solid-catalyzed gas-phase reactionH3C-CH=CH2+H2O2longrightarrowH2C-CH-CH3+H2OA+BlongrightarrowC+Dis carried out isothermally with no pressure drop. The rate law for this reaction is-rA1'=k1KBPAPB1+KBPB+KCPCmolkgcat*swhere ,k1=135molkg catalyst satmKB=0.0296atm-1KC=0.5333atm-1PC= Partial pressure of propylene oxide,

Problema $1$

Consider the irreversible, solid$-$catalyzed gas$-$phase reaction

$H_{3}C-CH=C{H}_2+H_2{O}_{2}longrightarrow{H}_{2}C-CH-C{H}_3+H_{2}O$

$A+$BlongrightarrowC$+D$

is carried out isothermally with no pressure drop. The rate law for this reaction is

$-r_{A1}^{\text{'}}=\frac{k_1{K}_B{P}_A{P}_B}{1+K_B{P}_B+K_C{P}_C}\frac{\text{m}\text{o}\text{l}}{\text{k}\text{g}\text{c}\text{a}\text{t}*s}$

where $,k_1=135mo\frac{l}{k}g$ catalyst $\frac{s}{a}tm$

$K_B=0\mathrm{.}0296at{m}^{-1}$

$K_C=0\mathrm{.}5333at{m}^{-1}$

$P_C=$ Partial pressure of propylene oxide, atm

$P_B=$ Partial pressure of hydrogen peroxide, atm

$P_A=$ Partial pressure of propylene, atm

The temperature for this gas$-$phase reaction is $45°C,$ there is no pressure drop, the initial partial pressures of each reactant are $10$atm, and the volumetric flow rate is $50d\frac{m^3}{s}.$

Cu$á$l ser$í$a el peso de catalizador si se buscase un $80\%$ de conversi$ó$n

Qu$é$ conversi$ó$n podr$í$a alcanzarse si se usase el mismo peso de catalizador encontrado en la parte #$1$ pero la alimentaci$ó$n consistiera en $P_{A0}=12,5$ atm y $P_{B0}=7,5$ atm. $¿$Es ventajoso este cambio?

Para las condiciones originales, pero ahora considerando que s$í$ hay ca$í$da de presi$ó$n y que el factor de ca$í$da es $a=0,0006k{g}^{-1}$ determine la conversi$ó$n que se alcanzar$í$a si se usase la misma cantidad de catalizador determinada en el punto #$1$