Where did the term with chemical potential go in the first law of thermodynamics? Consider a system that can be defined by 
$U$, $V$, ${N}_{i}$ $\left(i=1,2,\ldots k\right)$ and that this system changes from state A to state B via a quasi-static reaction path $C$. 
Here, $U$ represents internal energy stored in the system, $V$ represents the volume of the system, and ${N}_{i}$ $\left(i=1,2,\ldots k\right)$  is the amount of $i$-th particles in the system. 
At this time, comparing the following two, the term of the chemical potential seems to remain. 


*

*the $\delta Q$, obtained by integrating along the reaction path (See (1-9), below) and, 

*the first law of thermodynamics (See (1-11) below)


In the reactions where the number of particles does not change, the "chemical potential term" will be zero because ${dN}_i$ $\left(i=1,2,\ldots k\right)$  are zero. 
However, for example, if chemical reactions occur in the system, the number of particles ${N}_{i}$, $\left(i=1,2,\ldots k\right)\ $  will be change.
【My Questions】

  
*
  
*Where did the term of chemical potential disappear in the first law of thermodynamics?(See (1-11))†
  
*Perhaps if we replace the definition of work that takes place during this process from (1-6) to following (1-6)'. And the definition of internal energy will be change into “the sum of kinetic energy and potential energy of particles in the system”※.
  
  
  $${{W}^{\prime}}_{C}\ = \int_{C}  -PdV + {\sum}_{i=1}^{r}{{\mu}_{i}{dN}_{i}}\tag  {1-6'}$$

†.In some comments, (1-11) is hold only in closed / isolated systems. (The definitions are as described in 【Note2】.)
But, how about the reaction, like followings?


*

*CO + O2 →CO2 (coefficients are omitted.) or

*H2O(Vapor)→H2O (liquid) 


In these reactions, seems not necessary for particles to go/come-from outside. Quasi-static paths may exist at least theoretically. Perhaps that theoretical pathway is very tricky like "Down to absolute zero, react in absolute zero, then increase temperature to final state.” but, we can use the path to calculate the relative entropy between the Initials and final state. Am I right?. 
(I add this sentence at 2019/11/27(JST) and modified at 2019/11/28(JST).)

※In my elementally textbooks (written in Japanese), the definition of U is "the sum of kinetic energy of particles in the system." They ignore the potential energy. (I add this sentence at 2019/11/27(JST).)
Is there any inconvenience if you rewrite the definition like this?
【Details】
Let $\delta Q$ be the thermal energy that flows into the system during a quasi-static micro-process where $U$, $V$, and ${N}_{i}$ $\left(i=1,2,\ldots k\right)$ change slightly. Then, 
$$\delta Q\ =T\ dS\ \tag  {1-1}$$
Here, T represents the temperature of this system.
On the other hand, let $S[U,\ V,\ {N}_{i}]$  be the entropy of this system, then,
$$dS\ =\ \frac{1}{T}dU+\frac{P}{T}dV-{\sum}_{i=1}^{k}{\frac{\mu_i}{\ T_i}dN}\tag  {1-2}$$
Here, Multiply $T$ for both sides of above equation, we can get
$$TdS\ =\ dU\ +\ PdV\ -\ \sum_{i=1}^{r}{\mu_idN_i} \tag  {1-3} $$
From equations (1-1) and (1-3), we get 
$$\delta\ Q\ = \ dU+ PdV - \ \sum_{i=1}^{r}{{\mu}_{i}{dN}_{i}}\tag  {1-4}$$ 
The $\delta Q\ $ is incomplete derivative therefore, integration of this will depends on the reaction path even though the reaction path is quasi-static but, we can do line integration along a quasi-static reaction path, $C$.  We denote ${Q}_{C}$ as the result of this line integral (See (1-5)) then, the ${Q}_{C}$ will represent the thermal energy that flows into the system during a quasi-static process $C$.
$${Q}_{C} = \int_{C}{\delta Q}\tag  {1-5}$$ 
Similarly, the work, ${W}_{C}$ that the system receives in this process is: 
$${W}_{C} = {\int}_{C} - PdV\tag  {1-6}$$ 
The dU is a complete derivative therefore, the change in internal energy before and after this reaction,
$$\Delta{U}_{A,B}={U}_{B}-{U}_{A}\tag  {1-7}$$ 
satisfies
$$\Delta{U}_{A,B}\ =\ {\int}_{C}{\ \ dU}\tag  {1-8}$$
Therefore, when equation (1-4) is integrated along reaction path $C$,
$${Q}_{C}\ =\ \Delta{U}_{A,B}\ -\ \ W_C\ -\ \sum_{i=1}^{r}\int_{C}{\ \ \mu_idN_i}\ \tag  {1-9}$$ 
therefore, 
$$\ \ \Delta{U}_{A,B}\ ={Q}_{C}\ +\ \ {W}_{C}\ +\ {\sum}_{i=1}^{r}\int_{C}{\ \ {\mu}_{i}{dN}_{i}}\tag  {1-10} $$ 
On the other hand, the first law of thermodynamics is;
$$\Delta{U}_{A,B}\ = {Q}_{C}\ +\ \ {W}_{C}\tag  {1-11}$$
【Notes】 (Added on 2019/11/28 JST)
Note 1:Definitions of quasi-static process, reversible process, and reversible reaction： 
Some respondents mention the feasibility of quasi-static process that change ${N}_{i}$ in a closed system.
I assume that a chemical reaction or physical change has occurred. However, as of 2019/11/28 (JST) it is 
under discussion whether these are appropriate example in which the number of particles changes in a closed system.
In order to unify terms in our discussions, the definitions of quasi-static process and reversible process are described according to my text. 
The definition of reversible reaction is not written in my text, but it came out in my comment, so I quote the definition from Wikipedia.
Here, as a definition of a quasi-static process, my text reads as follows.

Def1-1. Definition of a quasi-static process:
  "A thermodynamic concept that refers to the process by which a system slowly changes from one state to another while maintaining a thermodynamic equilibrium."(Originally in Japanese and I translated into English)

For reference, the definition of the reversible process is as follows.

Def1-2. Definition of the reversible process:
  "Even if a substance changes from one state to another, it is possible to return to the original state again and leave no change in the external world during this time."(Originally in Japanese and I translated into English)

These positions seems the same as the Wikipedia article on Reversible process. This article describes the relationship between quasi-static and reversible processes as follows:

"In some cases, it is important to distinguish between reversible and quasi static processes. Reversible processes are always quasi static, but the converse is not always true.[1]"

Here, the reference [1] of above-mentioned quotation is:
Sears, F.W. and Salinger, G.L. (1986), Thermodynamics, Kinetic Theory, and Statistical Thermodynamics, 3rd edition (Addison-Wesley.)
Note that, there seems to be some confusion of terms; There seems to be some literature that uses reversible processes and quasi-static processes synonymously.
Another confusing term is the term reversible reaction.  According to the Wikipedia, the definition is as follows.

Def1-3. Definition of reversible reaction:
  "A reversible reaction is a reaction where the reactants form products, which react together to give the reactants back."

■Note2. Definition of isolated system and closed system:
In my textbook the definition of isolated systems is as follows.

Def2-1. Definition of isolated systems: "A system in which neither material nor energy is exchanged with the outside world." (Originally written in Japanese and I translated into English.)

A system that exchanges energy but does not exchange substances seems to be called a "closed system" to distinguish it in my text book.
According to my textbook, the definition of closed system is as follows.

Def2-2.  Definition of closed systems: " A system that exchanges energy with the outside world but does not exchange materials." (Originally written in Japanese and I translated into English.)

 A: It is always important to know the assumptions behind an equation. Rarely does an equation hold for all possible scenarios.
The form of the first law you give only considers work done by/on the system and heat that enters/leaves the system. The first law is just a statement of energy conservation, so in general all such energy changes must be taken into account. In your case you have performed correct analysis, as repeated here. 
A: Let's imagine a system X consisting of 2 mol of H atoms and and 1 mol of O atoms held at constant $T$ and $P$. We can think of two states:


*

*In state "0," all of the H is in the form of H$_2$ and all of the O is in the form of O$_2$

*In state "eq," only H$_2$, O$_2$, and H$_2$O are present, and the three are in chemical equilibrium (which will tend towards mostly H$_2$O)


If we heated the system enough, the transition from state "0" to state "eq" would proceed very rapidly by irreversible chemical reactions (combustion), and would therefore not be described by equations 1-1, 1-4, 1-9, or 1-10. To make these equations apply, we would need to make the transition from state "0" to state "eq" reversible. To make the chemical reactions reversible, we would need to ensure that the chemical energy released was converted into work rather than thermal energy. We could do this using an ideal fuel cell (which converts chemical energy into electrical work).
With this change, the system would do work on its surroundings in two ways: boundary work due to expansion/contraction and electrical work from the fuel cell. For this particular reversible process, we could then identify the last two terms in Equation 1-10 as
$$
\Delta{U}_{A,B} ={Q}_{C} + \underbrace{{W}_{C}}_{\substack{\text{Inward}\\\text{Boundary}\\\text{Work}}} + \underbrace{{\sum}_{i=1}^{r}\int_{C}{\ \ {\mu}_{i}{dN}_{i}}}_{\substack{\text{Electrical work done by}\\\text{surroundings on fuel cell}\\\text{to maintain reversible}\\\text{chemical reactions}}} \tag{1-10 applied}
$$
where we expect the final term to be negative for process (0 $\rightarrow$ eq).
As alluded to in Aaron Stevens' answer, given that you define $W_c$ as the boundary work, Equation 1-11 is not the correct form of the First Law for a closed system. The correct form is instead
$$
\Delta U_{A,B} = Q_C + \underbrace{\sum_i W_i}_{\substack{\text{Sum of}\\\textit{all}\text{ works}}}
\tag{1-11 corrected}
$$
With this correction, equations 1-10 and 1-11 become consistent: the last term of 1-11 encompasses the last two terms of 1-10.
Note that, if the chemical reactions are not carried out reversibly (and chemical energy in the fuel is converted directly into thermal energy in the combustion products, rather than being carried away as work), then the there is no electrical work output. The First Law remains consistent, however, because all of the energy that would have been carried away as work is instead stored as thermal internal energy. The energy lost from the system goes down, but the energy retained goes up by the same amount. Mathematically, the final term in (1-10) is moved to the other side and then grouped back as part of $\Delta U$.
