Given two independent liquid-phase reactions
$$
\begin{array} {c} {{\mathrm{A \!+\! 2 B \longrightarrow P }}} \\ {{\mathrm{2 P \!+\! B \longrightarrow S}}} \\ \end{array}
$$
If the initial concentrations of the reactions are $C_{\mathrm{A 0}} \,, \, C_{\mathrm{B 0}} \,, \, C_{\mathrm{P 0}} \,, \, C_{\mathrm{S 0}}$, where A is the key component and its conversion rate is $x_{\mathrm{A}}$, and the yield of the desired product $\mathbf{P}$ is $\varphi$, assuming that changes in material density during the reaction can be neglected, what are the concentrations of reactant $\mathrm{B}$ and products $\mathrm{P}$ and $\mathrm{S}$?

Options:
A. $$
c_{B} \!=\! c_{B 0} \!-\! c_{A 0} x_{A} \!-\! \frac{1} {3} c_{A 0} ( \varphi \!-\! x_{A})
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{B 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! c_{\mathrm{{A 0}}} \! \left( \varphi \!-\! x_{\mathrm{{A}}}\right)
$$
B. $$
c_{B} \!=\! c_{B 0} \!-\! 2 c_{A 0} x_{A} \!+\! \frac{1} {4} c_{A 0} ( \varphi \!-\! x_{A})
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\frac{1} {3} \varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
C. $$
c_{B} \!=\! c_{B 0} \!-\! 2 c_{A 0} \varphi \!-\! c_{A 0} ( x_{A} \!+\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! {\frac{3} {4}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
D. $$
c_{B} \!=\! c_{8 0} \!-\! c_{A 0} x_{A} \!-\! \frac{1} {2} c_{A 0} ( x_{A} \!+\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!-\! {\frac{1} {2}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!+\! \varphi\right)
$$
E. $$
c_{B} \!=\! c_{B 0} \!-\! c_{A 0} x_{A} \!-\! \frac{1} {2} c_{P 0} ( x_{A} \!-\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{B 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! {\frac{1} {2}} c_{\mathrm{{A 0}}} \! \left( \varphi \!-\! x_{\mathrm{{A}}}\right)
$$
F. $$
c_{B} \!=\! c_{8 0} \!-\! 2 c_{A 0} x_{A} \!-\! \frac{1} {2} c_{A 0} ( x_{A} \!-\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! {\frac{1} {2}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
G. $$
c_{B} \!=\! c_{B 0} \!-\! 2 c_{A 0} \varphi \!-\! \frac{1} {3} c_{A 0} ( x_{A} \!-\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\frac{1} {2} \varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
H. $$
c_{B} \!=\! c_{B 0} \!-\! 3 c_{A 0} x_{A} \!-\! \frac{1} {2} c_{A 0} ( x_{A} \!+\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+ x_{\mathrm{A}} c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! {\frac{1} {4}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
I. $$
c_{B} \!=\! c_{B 0} \!-\! 2 c_{A 0} x_{A} \!-\! c_{A 0} ( x_{A} \!-\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\frac{3} {2} \varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!+\! {\frac{1} {3}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!+\! \varphi\right)
$$
J. $$
c_{B} \!=\! c_{B 0} \!-\! 2 c_{A 0} x_{A} \!+\! \frac{1} {2} c_{A 0} ( x_{A} \!-\! \varphi)
$$
$$
c_{\mathrm{P}}=c_{\mathrm{P 0}}+\varphi c_{\mathrm{A 0}}
$$
$$
c_{\mathrm{{S}}} \!=\! c_{\mathrm{{S 0}}} \!-\! {\frac{1} {2}} c_{\mathrm{{A 0}}} \! \left( x_{\mathrm{{A}}} \!-\! \varphi\right)
$$
Answer: F