text:
\section*{Supplementary Information}

\subsection{Feshbach resonance}

The scattering length near the Feshbach resonance $\widetilde{a
(B)=a_{\mathrm{bg}}[1-\Delta B/(B-B_{0})]$ has been determined in Ref.
\cite{Rui2017s} by fitting the binding energies of the Feshbach molecules using
the universal model. For the Feshbach resonance between A and B, we obtain
$a_{\mathrm{bg}}=-455$ $a_{0}$, $B_{0}=138.71$ G and $\Delta B=-34.60$ G. For
the Feshbach resonance between A and C, we have $a_{\mathrm{bg}}=126$ $a_{0}$,
$B_{0}=130.637$ G and $\Delta B=4.0$ G. The binding energies of the Feshbach
molecules are given by$\ E_{\mathrm{ij}}^{\mathrm{{b}}}=\hbar^{2}/2\mu
_{d}a_{\mathrm{ij}}{}^{2}$, where $a_{\mathrm{ij}}=\widetilde{a}_{\mathrm{ij
}(B)-\bar{a},$ with $\bar{a}=51$ $a_{0}$ being the mean scattering length. The
scattering lengths and binding energies are shown in Fig. \ref{figs1}. Here
$a_{\mathrm{AB}}$ and $a_{\mathrm{AC}}$ are the input parameters in the STM
equations. From the binding energy curves, we obtain $E_{\mathrm{AB
}^{\mathrm{{b}}}=E_{\mathrm{AC}}^{\mathrm{{b}}}$ at $B=130.24(1)$ G, which is
about 10 mG smaller than $B_{tn}=130.249(2)$ G which was determined from the direct
measurement. Therefore, in the numerical calculations, the magnetic field is
shifted by 10 mG to compare with the experiments.

\begin{figure}[ptb]\centering
\includegraphics[width=8cm]{fig_s1}
\renewcommand\thefigure{S1}
\caption{\textbf{Overlapping Feshbach resonance.} \textbf{a}.  The scattering lengths of the collision channels $\rm A+B$ (red) and $\rm A+C$ (blue). Solid lines represent the scattering lengths from the universal model. Dashed lines represent the results from the coupled-channel calculations. \textbf{b}. The binding energies of the weakly bound AB and AC Feshbach molecules obtained from the universal model, which predict the two curves to intersect at 130.24 G. The grey shaded region denotes the range of magnetic fields in which the current experiment is performed. }\label{figs1
\end{figure}

\subsection{Raman photoassociation and molecular bound-bound transition}

In this work, we employ Raman photoassociation to form AC Feshbach molecules~\cite{Fu2014s}. We use two blue-detuned Raman light fields to couple the $|9/2,-1/2\rangle$
and C states with a single-photon detuning of $\Delta\approx2\pi\times251$
GHz, with respect to the D2 line transition of the $^{40}$K atom, as shown in
Fig. \ref{figs2}. The two Raman beams have a frequency difference of
approximately $2\pi\times37$ MHz at the magnetic fields of around 130 G. Thus,
they can be simply created by using two acousto-optic modulators with the light
field from a single laser. The width of each Raman beam is about 0.8 mm at the
position of the atoms, and the power is 30 mW. The Raman Rabi frequency for
the $|9/2,-1/2\rangle\rightarrow\mathrm{{C}}$ transition is about $2\pi
\times90$ kHz.

To associate the AC molecules from the $\mathrm{A+|9/2,-1/2\rangle}$ mixture,
we use a 300-$\mu$s Gaussian pulse with a full width at half maximum (FWHM)
duration of about 150 $\mu$s. Using a Gaussian pulse, the high Fourier side
lobes can be efficiently suppressed, and thus the transfer of free atoms from
the $|9/2,-1/2\rangle$ to the C state can be neglected. The Raman light fields
have no influence on the active stabilization loop of the magnetic field, and
thus it does not cause a shift of the magnetic field.

\begin{figure}[ptb]\centering
\includegraphics[width=6cm]{fig_s2} \renewcommand\thefigure{S2}
\caption{\textbf{Raman photoassociation.} The atom mixture is initially prepared in the $\rm A + |9/2,-1/2\rangle$ state. The AC molecule is formed by the Raman light. $E_b$ denotes the binding energy of the AC molecules. $E_0$ denotes the free atomic transition frequency between the two spin states of K atoms. }\label{figs2
\end{figure}

After creating the AC molecules, we determine the molecular bound-bound
transition frequency $\mathrm{\nu_{AC\rightarrow AB}}$ by measuring the
molecule radio-frequency (rf) spectrum. This is achieved by first transferring
the AC molecule to the AB molecule via a rf pulse, and then the AB molecules
are detected by dissociating them into the $\mathrm{A+|9/2,-7/2\rangle}$ state for
imaging. Besides, we also measure the free atomic transition frequency
$\mathrm{\nu_{C\rightarrow B}}$ by measuring the rf spectrum between the C and
B states. From these measurements, the reaction
energy change $\Delta E=\nu_{\mathrm{AC\rightarrow AB}}-\nu
_{\mathrm{C\rightarrow B}}$ can be accurately determined, and the results are
shown in Fig. 1 in the main text. After determining the bound-bound
transition frequency, the Rabi oscillation between the AB and AC molecule
states is measured, and thus the duration of
the $\pi$ pulse is determined.

\subsection{Remove the remaining A atoms and the AC molecule products}

\begin{figure}[ptb]\centering
\includegraphics[width=8cm]{fig_s3} \renewcommand\thefigure{S3}
\caption{\textbf{Removal of the remaining A atoms.} \textbf{a}. The energy levels of the two-photon Raman transition used for transferring the Na atoms from the A to the $|2,2\rangle$ state. After the Raman transfer, the Na atoms in the $|2,2\rangle$ state are resonantly blasted out of the optical trap by a light pulse  resonant with the cycling transition. \textbf{b}. The time sequence of the preparation process including the removal of the remaining A atoms.}\label{figs3
\end{figure}

To demonstrate that the discrepancy between the time constants $\tau
_{\mathrm{{AB}}}$ and $\tau_{\mathrm{{B}}}$ for the endothermic reaction and
nearly thermoneutral reaction is caused by the reverse reaction, we remove the
AC molecule products continuously during the reaction process by dissociating
them into the $\mathrm{A+|9/2,-1/2\rangle}$ state. However, the remaining A
atoms and the residual $\mathrm{|9/2,-1/2\rangle}$ atoms due to the
imperfection of the $\pi$ pulse may cause a competition between the
dissociation and association process. Therefore, we first remove the A atoms
after the AB molecules have been prepared. This is achieved as follows. We
first transfer Na atoms from the A to the $|2,2\rangle$ state using a blue-detuned
Raman $\pi$ pulse with a single-photon detuning of $\Delta\approx2\pi
\times215$ GHz with respect to the D2 line of Na atom, and then blast them out
of the optical trap by applying a light pulse resonantly coupling the
$|2,2\rangle\rightarrow$ $|3^{\prime},3^{\prime}\rangle$ cycling transition,
as shown in Fig. \ref{figs3}a. The Raman fields for Na atoms have a width of
about 1.1 mm at the position of the atoms, and each beam has a power of about
50 mW, which yields a Raman Rabi frequency of $2\pi\times46$ kHz for the
$\mathrm{A\rightarrow|2,2\rangle}$ transition. The Raman-Blast sequences are
applied twice to completely remove the remaining A atoms. The time sequence
including  the removal of the A atoms is shown in Fig. \ref{figs3}b. The Raman-Blast
process also causes small loss of the AB molecules. Typically, we still keep
$90\%$ of the AB molecules after removing the remaining A atoms.

\begin{figure*}[ptb]\centering
\includegraphics[width=12cm]{fig_s4} \renewcommand\thefigure{S4}
\caption{\textbf{Lifetime of the molecules under different conditions}. The measurements are taken after removing the remaining A atoms at 130.26 G. \textbf{a} and \textbf{b}. The lifetime of the AC molecules without and with the Raman removing fields, respectively. The two-photon frequency is tuned to be about 220 kHz above the bound-free transition frequency. The AC molecules are detected by first transferring them to the AB state and then dissociating them into the $\rm A+|9/2,-7/2\rangle$ state. \textbf{c} and \textbf{d}. The lifetime of the AB molecules without and with the Raman removing fields, respectively. Error bars represent $\pm1$ s.d.}\label{figs4
\end{figure*}

Then we can remove the AC molecule products during the reaction process by
continuously dissociating them into the $\mathrm{A+|9/2,-1/2\rangle}$ state.
This is achieved by using the Raman light coupling the
$\mathrm{|9/2,-1/2\rangle}$ and the C states, which is also used to associate the AC
molecules. For the dissociation, the two-photon detuning is tuned to be
$220-320$ kHz above the bound-free transition, and the powers of both beams are
reduced to obtain half the peak intensity as was used in the association process. This
Raman dissociation light reduces the lifetime of the AC molecules to  around
$80~\mu$s, as is shown in Fig. \ref{figs4}b, and thus the AC molecules can be
quickly eliminated. We checked that the Raman removing pulse has negligible
effects on the C atoms. The Raman removing pulse also affects the lifetime of
the AB molecules. However, the AB molecule still has a lifetime of about 2.7
ms under the Raman removing fields, as is shown in Fig. \ref{figs4}d, which is long
enough for the study of the reaction dynamics.

We also compare the time evolutions when only removing the remaining A atoms
and the time evolutions when both the remaining A atoms and the AC molecules are removed. After removing the A
atoms, the dominant atoms in the mixture are the C atoms. The AC molecule has
a long lifetime of about 8 ms when coexisting with the C atoms, as shown in
Fig. \ref{figs4}a. In this case, the effects of the reverse reaction may also
be observed on a long time scale. As shown in Fig.\ref{figs5}, when only the A
atoms are removed, we observe a long-term decay for the B atom products.
However, this long-term decay disappears when the AC products are continuously
removed during the reaction process. Therefore, we attribute this long-term decay
to the influence of the reverse reaction.

Besides, we also measure the influence of the Raman removing light on the
molecular and atomic transition frequencies. Our measurements show that the atomic
transition frequency $\nu_{\mathrm{C\rightarrow B}}$ is shifted by
about $+0.3$ kHz, whereas the molecular bound-bound transition frequency
$\nu_{\mathrm{AC\rightarrow AB}}$ is shifted by about $-3.4$ kHz. The energy
change $\Delta E=\nu_{\mathrm{AC\rightarrow AB}}-\nu_{\mathrm{C\rightarrow B
}$ under the Raman removing fields is plotted in Fig. \ref{figs6}. The
polynomial fitting shows the magnetic field of zero energy change is shifted
to 130.258(2) G under the Raman removing fields.

\subsection{Reaction dynamics}

\begin{figure}[ptbh]\centering
\includegraphics[width=8cm]{fig_s5} \renewcommand\thefigure{S5}
\caption{\textbf{Time evolution after removing the remaining A atoms.} The red (blue) data points represent the time evolution without (with) applying the Raman removing light at 130.26 G. \textbf{a}. The time evolution of the AB reactants. The time constants are measured to be $\tau_{\rm{AB}}=0.319(21)$ ms (red) and $\tau_{\rm{AB}}=0.262(28)$ ms (blue), respectively. \textbf{b}, The time evolutions of the B products, which are fitted with the model, $N_{\rm B}(t)=N^{0}_{\rm B}(e^{-t/\tau_2}-e^{-t/\tau_1})$. The fitting gives $\tau_1=0.184(24)$ ms and $\tau_2=21(4)$ ms (red), and $\tau_1=0.220(18)$ ms and $\tau_2=+\infty$ (blue).  \textbf{(Inset)}. The short term evolution of the B products.  The simple exponential fitting gives a time constant of $\tau_{\rm{B}}=0.166(22)$ ms (red), and $\tau_{\rm{B}}=0.220(18)$ ms (blue). Error bars represent $\pm1$ s.d. }\label{figs5
\end{figure}

\begin{figure}[ptbh]\centering
\includegraphics[width=8cm]{fig_s6} \renewcommand\thefigure{S6}
\caption{\textbf{Influence on the energy change of the Raman removing fields.} The red points represent the measured energy change under the Raman removing fields. The blue points represent the energy change in the main text. The solid lines are polynomial fitting to the measured values, which shows the zero-energy-change magnetic field is shifted by 9 mG under the Raman fields. The grey dashed line is the energy change calculated from the universal binding energy curves. Error bars inside the circles represent $\pm1$ s.d.}\label{figs6
\end{figure}

Taking into account the reverse reaction, the time evolution of the AB
reactant and B product may be described by
\begin{align}
\dot{N}_{\mathrm{AB}}  &  =-\gamma_{\mathrm{AB}}N_{\mathrm{AB}}+k_{re
N_{\mathrm{AC}}N_{\mathrm{B}},\\
\dot{N}_{\mathrm{B}}  &  =\beta_{r}\bar{n}_{\mathrm{C}}N_{\mathrm{AB}
-k_{re}N_{\mathrm{AC}}N_{\mathrm{B}},
\end{align}
where the first and second term of the right side account for the
forward and reverse reaction respectively. The time evolution of the AC
molecule may be written in a similar form as the AB molecule. Since the time
evolution of the AC molecule is difficult to detect, its equation is not
shown. Here the overall loss rate of $N_{\mathrm{AB}}$ is $\gamma
_{\mathrm{AB}}=\beta_{\mathrm{A}}\bar{n}_{\mathrm{A}}+\beta_{\mathrm{C}
\bar{n}_{\mathrm{C}}+\beta_{r}\bar{n}_{\mathrm{C}}$, with $\bar{n
_{\mathrm{A}}$ and $\bar{n}_{\mathrm{C}}$ the mean densities of the A and C
atoms, respectively. The first term accounts for the loss due to inelastic
collisions with the remaining A atoms, with $\beta_{\mathrm{A}}$ the loss
rate coefficient. When the A atoms are removed, this term can be neglected. The other two
terms describe the losses due to collisions with the atom reactant C, which
include the desired reactive collision with a reaction rate coefficient of
$\beta_{r}$, and the losses due to reactions in other channels and vibrational
relaxations with an overall loss rate of $\beta_{\mathrm{C}}$. The collisions
between the AB molecules are neglected since the Feshbach molecules are
fermionic molecules. We have assumed $\bar{n}_{\mathrm{A}}$ and $\bar
{n}_{\mathrm{C}}$ are constants since the number of molecules is about one
order of magnitude smaller than that of the atoms. The reverse reaction is
caused by the reactive collisions between the AC molecule and B atom products
with $k_{re}$ the reverse reaction rate.

In our experiment, the time evolutions of $N_{\mathrm{AB}}(t)$ and
$N_{\mathrm{B}}(t)$ are measured. To extract the reaction rate coefficient, we us

\begin{equation}
\beta_{r}=\frac{\dot{N}_{\mathrm{B}}(0)}{\bar{n}_{\mathrm{C}}N_{\mathrm{AB
}(0)}. \label{rate2
\end{equation}
This is reasonable as we start from the $\mathrm{AB+C}$ mixture and thus at
$t=0$ the number of products $N_{\mathrm{AC}}$ and $N_{\mathrm{B}}$ can be
neglected. The exact form of $N_{\mathrm{AB}}(t)$ and $N_{\mathrm{B}}(t)$ are
difficult to know. Therefore, we assume they may be approximated by
exponentials and use exponentials to fit the data points. The reaction rate
coefficients are extracted with $N_{\mathrm{AB}}(0)$ and $\dot{N}_{\mathrm{B
}(0)$ obtained from the fitting curves. Note that when the reverse reaction is
suppressed or can be neglected, the analytic solutions of $N_{\mathrm{AB}}(t)$
and $N_{\mathrm{B}}(t)$ are exponentials, and thus reaction rate coefficients
obtained from Eq. (\ref{rate2}) are exact.

The initial molecule number has to be corrected by taking into account the
finite rf dissociation rate \cite{Rui2017s}, since the reaction is fast. The
finite $\pi$-pulse transfer efficiency of the K atoms from the B to
$|9/2,-7/2\rangle$ states also needs to be included, which is about 90\% with
the Na atoms, and 95\% without the Na atoms. Finally, all uncertainties of the
parameters are included to give the uncertainty in the reaction rate coefficients.

\subsection{STM equations}

The scattering amplitudes satisfy the STM equation
\begin{equation}
\begin{split}
t_{ii}(k,p,E)  &  =\frac{m_{\mathrm{{A}}}}{\pi\mu_{d}}\sqrt{\frac
{a_{\mathrm{{AC}}}}{a_{\mathrm{{AB}}}}}\int^{\Lambda}dq\frac{q}{2k
K(k,q,E) \\
              & \times D_{\mathrm{{AC}}}(q,E) t_{fi}(q,p,E)\\
t_{fi}(k,p,E)  &  =\frac{2\pi\hbar^{4}}{\mu_{d}^{2}\sqrt{a_{\mathrm{{AB}
}a_{\mathrm{{AC}}}}}\frac{m_{\mathrm{{A}}}}{2pk}K(k,p,E)\\
               &+\frac{m_{\mathrm{{A
}}}{\pi\mu_{d}}\sqrt{\frac{a_{\mathrm{{AB}}}}{a_{\mathrm{{AC}}}}}\int
^{\Lambda}dq\frac{q}{2k}K(k,q,E)\\
               & \times D_{\mathrm{{AB}}}(q,E)t_{ii}(q,p,E)
\end{split}
\end{equation}
wher
\begin{align*}
K(k,p,E)  &  =\log\frac{2\mu_{d}E-p^{2}-k^{2}+2\mu_{d}pk/m_{\mathrm{{A}}
}{2\mu_{d}E-p^{2}-k^{2}-2\mu_{d}pk/m_{\mathrm{{A}}}}\\
D_{\mathrm{{jl}}}(q,E)  &  =[-\frac{\hbar}{a_{\mathrm{{jl}}}}+\sqrt{-2\mu
_{d}(E-\frac{q^{2}}{2\mu_{ad}})}]^{-1}.
\end{align*}

We denote the input channel $\mathrm{{AB+C}}$ by channel $i$ and the reaction
channel $\mathrm{{AC+B}}$ by channel $f$, and thus $t_{ii}(k,p,E)$ is the
elastic scattering amplitude, or in other words, the back-reflection amplitude
of scattering into the input channel, and $t_{fi}(k,p,E)$ is the reactive
scattering amplitude, with $p$ and $k$ being the relative momentum of the
incoming and outgoing atom-dimer pairs and $E$ the total energy, $\mu
_{d}=m_{\mathrm{{A}}}m_{\mathrm{{B}}}/(m_{\mathrm{{A}}}+m_{\mathrm{{B}}})$ is
the reduced mass, and $\mu_{ad}=m_{\mathrm{{A}}}(m_{\mathrm{{A}
}+m_{\mathrm{{B}}})/(2m_{\mathrm{{A}}}+m_{\mathrm{{B}}})$ is the reduced of
mass of atom and dimer. In the integral equations, the only input parameters
are the scattering lengths $a_{\mathrm{{AB}}}$ and $a_{\mathrm{{AC}}}$, which
have been given in Sec. A in the Supplementary Information, and the unknown
three-body parameter $\Lambda$, which will be determined by comparing the
reaction rates. Since the magnetic field window in the experiment is narrow,
we assume $\Lambda$ is a constant.

We numerically solve the STM integral equations to calculate the on-shell
reaction amplitudes $t_{fi}(k_{f},p_{i},E)$ with $E=\hbar^{2}k_{f}^{2
/2\mu_{d}-E_{\mathrm{AC}}^{\mathrm{{b}}}=\hbar^{2}p_{i}^{2}/2\mu
_{d}-E_{\mathrm{AB}}^{\mathrm{{b}}}$ using the method introduced in Ref.
\cite{Braaten2009s,Helfrich2010s}. The incident relative momentum is calculated
for $p_{\min}\leq p_{i}\leq p_{\max}$, where $p_{\min}=0$ for the exothermic
reactions and $p_{\min}=\sqrt{2\mu_{d}|\Delta E|}$ for the endothermic
reactions, and $p_{\max}=$ $\sqrt{2\mu_{d}E_{\mathrm{AB}}^{\mathrm{{b}}}}$,
i.e., we only consider the collisional kinetic energy smaller than the binding
energy of the AB molecule. This is a good approximation since in our
experiment we have $E_{\mathrm{AB}}^{\mathrm{{b}}}>4k_{\mathrm{B}}T$. The
total reaction cross section is given by $\sigma_{r}(p_{i})=(8\pi^{3
/h^{4})\mu_{ad}^{2}(k_{f}/p_{i})|t_{fi}(k_{f},p_{i},E)|^{2}.$ The reaction
rate coefficient is thus calculated by $\beta_{r}=\int v\sigma_{r}(v)f(v)dv$
where we have $v=p/\mu_{ad}$ and  $f(v)=4\pi v^{2}[\mu_{ad}/(2\pi
k_{\mathrm{{B}}}T)]^{3/2}\exp[-(\mu_{ad}v^{2})/(2k_{\mathrm{{B}}}T)]$ is the
Maxwell-Boltzmann distribution. In the calculation, the temperature is assumed
to be $700$ nK, and the magnetic field has been shifted $10$ mG so that the
overlapping magnetic field obtained from the universal binding energy curves
is consistent with the direct measurement.

The calculated rate coefficients are compared with the experiment for the
magnetic fields between $129.44$ G to $130.38$ G, where the data points
between $129.44$ G and $129.91$ G are obtained from our previous work. The
reaction rate at $130.03$ G is also measured in our previous work
\cite{Rui2017s}. However, due to the suppression of the bound-free
Franck-Condon coefficients, the number of the AB molecule that can be formed
from the A+C mixture is only about 5000. In comparison, using the indirect
method in this work, $1.5\times10^{4}$ AB molecules can be formed. Therefore,
at 130.03 G, the reaction rate measured in this work has much better
signal-to-noise ratio and thus is used in the fitting. For the magnetic fields
larger than 130.03 G, the reaction rates can only be measured using the
current method.

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