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I am trying to draw the following reaction schemes using the chemfig package

enter image description here

enter image description here

To draw the triangular scheme I used the commands

\setchemfig{scheme debug=false}
\schemestart
\chemfig{A \+ B}
\arrow(bb--cc){<=>[$\k_{1}$][$\k_{-1}$]}
\chemfig{C}
\arrow(@cc--P)[-135]P %
\arrow(@bb--@P)[-45] %[$\k_{0}$]
\schemestop

It sort of does the job but I can not write the rate constant $k_{0}$ and set P at the vertex of the triangle.

For the second scheme I used the code


\setchemfig{scheme debug=false}
\schemestart
\chemfig{A_1}
\arrow(aa--bb){[$k_{1}$][$k_{-1}$]}
\chemfig{A_2}
\arrow{->[*{0}$k_{22}$][]}[-90]\chemfig{P_2}
\arrow(@aa--pp)[-90]\chemfig{P_1}
\schemestop

As in the first scheme, I am unable to draw the rate constant $k_{21}$ besides the righthandside downarrow.

1 Answer 1

3

something like that?

\documentclass{article}
\usepackage{chemfig}

\begin{document}
\setchemfig{scheme debug=false}
\schemestart
\chemfig{A \+ B}
\arrow(bb--cc){<=>[$k_{1}$][$k_{-1}$]}[,2]
\chemfig{C}
\arrow(@cc--P){->[*{0}\quad $k_2$]}[240,2]P %
\arrow(@bb--@P){->[][*{0} $k_0$]}[300,2] %[$\k_{0}$]
\schemestop

\vspace{1cm}

\setchemfig{scheme debug=false}
\schemestart
\chemfig{A_1}
\arrow(aa--bb){<=>[$k_{1}$][$k_{-1}$]}[0,2]
\chemfig{A_2}
\arrow{->[*{0}$k_{22}$][]}[-90,2]\chemfig{P_2}
\arrow(@aa--pp){->[*{0}$k_{21}$]}[-90,2]\chemfig{P_1}
\schemestop

\end{document}

enter image description here

1
  • Works beautifully, thanks.
    – PAEP
    Commented Jul 30, 2022 at 15:51

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