Tmax of parallel reactions/Steady state reactions


> 
> Below are four reactions, to this experiment inhibitor is added so that k$\_3$ = $0$ (but the other reactions are unaffected) calculate [D] when the reaction is over. The initial concentration of A is $0.016$M.
> 
> 
> A → B k$\_1$= $0.110$ s$^{-1}$
> 
> 
> A → C k$\_2$= $0.025$ s$^{-1}$
> 
> 
> B → C k$\_3$= $75.2$ s$^{-1}$
> 
> 
> C → D k$\_4$= $17.2$ s$^{-1}$
> 
> 
> 


I am having trouble understanding how to solve this problem. I know the first step is to calculate the time $t$ however I do not know how to apply what I know from other problems to this one.


I know that t$\_{max}$ will be when dD/dt = $0$, and if I remove reaction $3$, then I get that dD/dt = k$\_2$[A]. This is where I get confused.


Should I now replace [A] with the integrated equation of A from the reactions, meaning:


-dA/dt = [A](k$\_1$ + k$\_2$) => ln[A]-ln[A]$\_0$ = -(k$\_1$ + k$\_2$)t => [A]=[A]$\_0$e$^{-(k\_1 + k\_2)t}$


But then if I substitute that into dD/dt = k$\_2$[A] =$0$ then I can't solve for $t$


How would one solve this?


Thank you!


**EDIT**


Have I derived the wrong expression for dD/dt?


I get that: dB/dt = $k\_1$[A], dC/dt=k$\_2$[A]-k$\_4$[C], dD/dt= k$\_4$[C]


However, here I have to set both dB/dt and dC/dt equal to zero. So that means that [A] has to be zero and that would in turn lead to [C] also having to be zero. So how should the expression be derived?


