Find the differential equation for $z$ the concentration of $Z$ in the chemistry equation $$6Z+B\rightleftharpoons^{k_1}_{k_{-1}} 2Z+A$$
My idea:
Let $[Z]=z,\ [A]=a,\ [B]=b$. Then, $$\frac{dz}{dt}=-4k_1z^6b+4k_{-1}z^2a$$ Now, my question is, if the term ''$4k_{-1}z^2a$'' is well written (the one in the left direction)? How to completely finish the exercise?
Yes.
By conservation of matter, $$z-z_0=4(a_0-a)=4(b-b_0)$$ hence $z'(t)=f(z(t))$ with $$f(x)=-k_1x^6(x-z_0+4b_0)+k_{-1}x^2(4a_0+z_0-x)$$ is the desired autonomous differential equation describing the evolution of $z=[Z]$.
To go further, note that $f(x)=k_1x^2g(x)$ with $$g(x)=r(4a_0+z_0)-rx+(z_0-4b_0)x^4-x^5\qquad r=\frac{k_{-1}}{k_1}$$ For each given set of values of the parameters $(k_1,k_{-1},a_0,b_0,z_0)$, one can compute the sign of $g(x)$ depending on $x$, deduce the phase diagram of the system, and finally the behaviour of $z(t)$.
For example, if $$k_1=k_{-1}\qquad a_0=z_0=1\qquad b_0=0$$ the graph of the function $g$ is:
hence $z(t)\to z^*$ where $g(z^*)=0$, namely, $$z^*\approx1.56791$$
Edit: Assuming that $z_0>0$ (since, if $z_0=0$, there is nothing to prove) and setting $$r=\varrho z_0^4\qquad a_0=\alpha z_0\qquad b_0=\beta z_0$$ one gets $g(z_0u)=z_0^5h(u)$ with $$h(u)=\varrho (1+4\alpha-u)+(1-4\beta)u^4-u^5$$ which shows that $z^*=z_0u^*$ where $u^*$ solves $h(u^*)=0$ hence $u^*$ depends on the ratios $k_{-1}/(k_1z_0^4)$, $a_0/z_0$ and $b_0/z_0$ only.