Question

The gas-phase decomposition of acetic acid at 1189 k proceeds by way oftwo parallel reactions:(1) ch3coohch4+co2k1=3.74 s1(2) ch3coohch2co+h2ok2=4.65 s1what is the maximum percentage yield of the ketene ch2co obtainable atthis temperature?

Answer 1

Answer:

Let

$CH_{3} COOH----->CH_{4} +CO_{2} (R)  K_{1} =3.74s^{-1}$

$CH_{3} COOH------>H_{2} C=C=0+H_{2} (s)  K_{2} =4.65S^{-1}$

$p_{R} =\frac{dp_{R} }{dt}$=$k_{1} p_{A} -----(1)$

$p_{s} =\frac{dp_{s} }{dt} =k_{2} p_{A} ----(2)$

divide 1 by 2 we get

$\frac{dp_{R} }{dp_{s} } =\frac{k_{1} }{k_{2} }$

by integrating we get

$\int\limits^e_{eR_{0} } \,de_{R}$

=$\frac{k_{1} }{k_{2} } \int\limits^e_{e_{s0} } _de_{s}$

by solving we get

and the yield at main and side pdt in the reaction is

$Y_{R} =\frac{e_{R} }{C_{AD} } =\frac{K_{1} }{K_{1}+K_{2}  }$    t---->α

$Y_{s} =\frac{e_{s} }{C_{Ao} }$=$\frac{k_{2} }{k_{1}+k_{2}  } =\frac{4.65}{8.39} =0.55$

maximum percentage yield at the ketene is 55%