I will lead you to the general equation. We can relate the partial pressure of a species with conversion by using a simple stoichiometric table.
The reaction is
$$ \ce{NH3(g) + CH4(g) -> HCN(g) + 3H2(g)} $$
$$ \ce{A(g) + B(g) -> C(g) + 3D(g)} \tag{R} $$
where we defined for convenience new letters for all the species. We concentrate on $\ce{A}$.
1. Mole balance
For a PFR the volume needed to achieve a conversion $X_\ce{A}$ is
\begin{equation}
V = \int_0^{X_\mathrm{A}} \frac{F_\mathrm{A0} \; \mathrm{d}X_\ce{A}}
{-r_\ce{A}(X_\ce{A})} \tag{1}
\end{equation}
where $F_\mathrm{A0}$ is the inlet molar flow rate of species $\ce{A}$ in $\pu{mol s^-1}$
2. Rate law
We have a power law in the form of
$$ -r_\ce{A} = kp_\ce{A} \tag{2} $$
3. Stoichiometry
We set up a stoichiometric table for reaction $\ce{R}$ in terms of the molar flows. For a general species $j$ in a reaction where $\ce{A}$ is the limiting reagent, the molar flow rate of species $j$ is
$$ F_{j} = F_\mathrm{A0} (\Theta_j + \nu_j X_\ce{A}) \tag{3} $$
where $\Theta_j = F_{j0}/F_\mathrm{A0}$ is the molar relation of species $j$ with respect to $\ce{A}$ at the entrance of the reactor, and $\nu_j$ is the stoichiometric coefficient of species $j$.
For the reaction to take place, we need the presence of $\ce{A}$ and $\ce{B}$. Considering that there are no products at the reactor inlet, application of Eq. (3) to reaction $\ce{R}$ gives
\begin{align}
F_\ce{A} &= F_\mathrm{A0}(1 - X_\ce{A}) \tag{4} \\
F_\ce{B} &= F_\mathrm{A0}(\Theta_\ce{B} - X_\ce{A}) \tag{5} \\
F_\ce{C} &= F_\mathrm{A0} X_\ce{A} \tag{6} \\
F_\ce{D} &= 3F_\mathrm{A0} X_\ce{A} \tag{7} \\
\end{align}
where the total molar flow rate is
\begin{align}
F_\ce{T} &= \sum_j F_j \\
F_\ce{T} &= F_\mathrm{A0}(1 - X_\ce{A}) + F_\mathrm{A0}(\Theta_\ce{B} - X_\ce{A}) +
F_\mathrm{A0} X_\ce{A} + 3F_\mathrm{A0} X_\ce{A} \\
F_\ce{T} &= F_\mathrm{A0}(1 - X_\ce{A} + \Theta_\ce{B} - X_\ce{A} +
X_\ce{A} + 3X_\ce{A}) \\
F_\ce{T} &= F_\mathrm{A0}(1 + \Theta_\ce{B} + 2X_\ce{A}) \tag{9} \\
\end{align}
Remembering that the partial pressure of species $j$ is given by
$$ p_j = y_j p = \frac{F_j}{F_\ce{T}}p \tag{10}$$
we combine Eqs. (4), (9), and (10) so that
\begin{align}
\require{cancel}
p_\ce{A} &= \frac{\cancel{F_\mathrm{A0}}(1 - X_\ce{A})}
{\cancel{F_\mathrm{A0}}(1 + \Theta_\ce{B} + 2X_\ce{A})}p \\
p_\ce{A} &= \left(\frac{1 - X_\ce{A}}{1 + \Theta_B + 2X_\ce{A}}\right)p \tag{11}
\end{align}
4. Design equation
Combining Eqs. (1), (2), and (11) gives the final result
\begin{equation}
\boxed{V = \int_0^{X_\mathrm{A}} \frac{F_\mathrm{A0} \; \mathrm{d}X_\ce{A}}
{k\left(\dfrac{1 - X_\ce{A}}{1 + \Theta_\ce{B} + 2X_\ce{A}}\right)p}} \tag{12}
\end{equation}
References
An excellent explanation for stoichiometry for liquid-phase and gas-phase reactive systems can be read at Chapter 3 of:
- Elements of Chemical Reaction Engineering, H. S. Fogler, 5th ed., Prentice Hall (2016).
