ModelingToolkit Simple Chemical Reaction ODE (B)

Implement the following system in ModelingToolkit and solve it with the initial condition:

\[\begin{align*} y1 &= 1\\ y2,\;...,\; y7 &= 0 \\ y8 &= 0.0057, \end{align*}\]

And parameter values:

\[\begin{align*} k1 &= 1.71\\ k2 &= 280\\ k3 &= 8.32\\ k4 &= 0.69\\ k5 &= 0.43\\ k6 &= 1.81, \end{align*}\]

solve on the interval $t\in [0, 321.8122]$, and plot the solution.

\[\begin{align*} \frac{\mathrm{d} \mathrm{y1}\left( t \right)}{\mathrm{d}t} =& 0.0007 + k5 \mathrm{y2}\left( t \right) + k3 \mathrm{y3}\left( t \right) - k1 \mathrm{y1}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y2}\left( t \right)}{\mathrm{d}t} =& - 8.75 \mathrm{y2}\left( t \right) + k1 \mathrm{y1}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y3}\left( t \right)}{\mathrm{d}t} =& - 10.03 \mathrm{y3}\left( t \right) + k5 \mathrm{y4}\left( t \right) + 0.035 \mathrm{y5}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y4}\left( t \right)}{\mathrm{d}t} =& k1 \mathrm{y3}\left( t \right) + k3 \mathrm{y2}\left( t \right) - 1.12 \mathrm{y4}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y5}\left( t \right)}{\mathrm{d}t} =& - 1.745 \mathrm{y5}\left( t \right) + k5 \mathrm{y6}\left( t \right) + k5 \mathrm{y7}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y6}\left( t \right)}{\mathrm{d}t} =& k1 \mathrm{y5}\left( t \right) + k4 \mathrm{y4}\left( t \right) + k4 \mathrm{y7}\left( t \right) - k5 \mathrm{y6}\left( t \right) - k2 \mathrm{y6}\left( t \right) \mathrm{y8}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y7}\left( t \right)}{\mathrm{d}t} =& - k6 \mathrm{y7}\left( t \right) + k2 \mathrm{y6}\left( t \right) \mathrm{y8}\left( t \right) \\ \frac{\mathrm{d} \mathrm{y8}\left( t \right)}{\mathrm{d}t} =& k6 \mathrm{y7}\left( t \right) - k2 \mathrm{y6}\left( t \right) \mathrm{y8}\left( t \right) \end{align*}\]

Example Solution Plot

An example of the what the solution should look like is shown below: