At temperature T and pressure $p^\circ$, reactants are charged into a long tube. On the left side, 3 mol of A and 1 mol of B are added, while on the right side, 3 mol of D and 1 mol of He are introduced. The gases on both sides are separated by a freely movable partition (initially positioned at 0 cm, impermeable to gases), and the right side also has a freely movable piston (initially positioned at 4 cm). The bottom of the tube (the far left end) is marked at -4 cm. The following reactions are known to occur:

$\mathrm{2A(g) + B(g) = C(g); K_1^\circ= 9.87×10^4}$

$\mathrm{A(g) + 2C(g) = D(g); K_2 ^\circ(1 < K_2^\circ< 100)}$

When the reactions reach equilibrium, the piston does not move. Determine the position of the partition at equilibrium and the amount of substance of B on the left side, selecting the closest option.
在温度为T，压强为$p^\circ$下向一长管中投料，左侧加入3 mol A和1 mol B，右侧加入3 mol D和1 mol He。将两侧气体隔开的是一可自由移动的挡板（初始位于0 cm处，气体不可穿过），右侧还有一可自由移动的活塞（初始位于4 cm处），管底（管最左侧）标记为-4 cm。已知存在以下反应：

$\mathrm{2A(g) + B(g) = C(g); K_1^\circ= 9.87×10^4}$

$\mathrm{A(g) + 2C(g) = D(g); K_2 ^\circ(1 < K_2^\circ< 100)}$

当反应达到平衡后，活塞未发生移动。试求出达到平衡后挡板所在的位置以及左侧B的物质的量，选择最接近的选项。
-4 cm; $\mathrm{1.2×10^{-3}}$
-3.7 cm; $\mathrm{1.5×10^{-4}}$
-3.3 cm；；$\mathrm{5.0×10^{-3}}$
-2.9 cm; $\mathrm{8.0×10^{-4}}$
-2.7 cm; $\mathrm{1.4×10^{-5}}$
-2.1 cm; $\mathrm{2.1×10^{-4}}$
-1.8 cm; $\mathrm{2.1×10^{-5}}$
-1.2 cm; $\mathrm{8.6×10^{-3}}$
-0.8 cm; $\mathrm{1.8×10^{-4}}$
-0.4 cm; $\mathrm{6.6×10^{-5}}$
0.5 cm; $\mathrm{4.2×10^{-5}}$
1 cm; $\mathrm{1.7×10^{-2}}$
1.4 cm; $\mathrm{2.9×10^{-6}}$
1.8 cm; $\mathrm{3.5×10^{-4}}$
2.4 cm; $\mathrm{9.8×10^{-6}}$
2.7 cm; $\mathrm{9.0×10^{-3}}$
3.2 cm; $\mathrm{5.1×10^{-5}}$
3.5 cm; $\mathrm{1.8×10^{-6}}$
4.0 cm; $\mathrm{6.9×10^{-2}}$
-4 cm；$\mathrm{1.2×10^{-3}}$
-3.7 cm；$\mathrm{1.5×10^{-4}}$
-3.3 cm；；$\mathrm{5.0×10^{-3}}$
-2.9 cm；$\mathrm{8.0×10^{-4}}$
-2.7 cm；$\mathrm{1.4×10^{-5}}$
-2.1 cm；$\mathrm{2.1×10^{-4}}$
-1.8 cm；$\mathrm{2.1×10^{-5}}$
-1.2 cm；$\mathrm{8.6×10^{-3}}$
-0.8 cm；$\mathrm{1.8×10^{-4}}$
-0.4 cm；$\mathrm{6.6×10^{-5}}$
0.5 cm；$\mathrm{4.2×10^{-5}}$
1 cm；$\mathrm{1.7×10^{-2}}$
1.4 cm；$\mathrm{2.9×10^{-6}}$
1.8 cm；$\mathrm{3.5×10^{-4}}$
2.4 cm；$\mathrm{9.8×10^{-6}}$
2.7 cm；$\mathrm{9.0×10^{-3}}$
3.2 cm；$\mathrm{5.1×10^{-5}}$
3.5 cm；$\mathrm{1.8×10^{-6}}$
4.0 cm；$\mathrm{6.9×10^{-2}}$
Since $K_1^\circ$ is much greater than $K_2^\circ$, it can be assumed that the first reaction proceeds to completion, and n(B) = 0.

<Checkpoint>[$n(B) = 0$](1)</Checkpoint>

Let the amount of n(D) at equilibrium on the left side be x mol, the amount of n(A) at equilibrium on the right side be y mol, and the position of the partition at equilibrium be a cm.

On the left side: n(A) = (1-x) mol, n(B) = 0, n(C) = (1-2x) mol, n(D) = x mol;  
On the right side: n(A) = y mol, n(B) = 0, n(C) = 2y mol, n(D) = (3-y) mol.  

$\mathrm{2 - 2x + 4 + 2y = 8, y = 1 + x}$

<Checkpoint>[$\mathrm{y = 1 + x}$](1)</Checkpoint>

$K_2^\circ = \frac{x(2-2x)^2}{(1-x)(1-2x)^2} = \frac{(3-y)(4+2y)^2}{4y^3}$.  
Substituting y = 1 + x, we solve for x = 0.332, and $K_2^\circ$ = 7.84.

<Checkpoint>[$x = 0.332$](1)</Checkpoint>

<Checkpoint>[$K_2^\circ = 7.84$](1)</Checkpoint>

$K_1^\circ = \frac{(1-x)(1-2x)^2}{(1-x)^2n(B)_1} = \frac{2y(4+2y)^2}{y^2n(B)_2}$.  
Solving these, we obtain $\mathrm{n(B)_1 = 1.36×10^{-5} (mol), n(B)_2 = 6.76×10^{-4} (mol)}$.

<Checkpoint>[$\mathrm{n(B)_1 = 1.36×10^{-5} (mol), n(B)_2 = 6.76×10^{-4} (mol)}$](2)</Checkpoint>

Both $\mathrm{n(B)_1}$ and $\mathrm{n(B)_2}$ are very small, indicating that the approximation is reasonable.

$\frac{2-2x}{4} = \frac{a+4}{4}, a = -2.66$

<Checkpoint>[$a = -2.66$](1)</Checkpoint>
由于$K_1^\circ$远大于$K_2^\circ$，因此可认为第一个反应完全进行，n(B) = 0。

<Checkpoint>[$n(B) = 0$](1)</Checkpoint>

令左侧平衡后n(D) = x mol，右侧平衡后n(A) = y mol，挡板平衡后位于a cm处。

左侧n(A) = (1-x) mol，n(B) = 0，n(C) = (1-2x) mol，n(D) = x mol；
右侧n(A) = y mol，n(B) = 0，n(C) = 2y mol，n(D) = (3-y) mol

$\mathrm{2 - 2x + 4 + 2y = 8, y = 1 + x}$

<Checkpoint>[$\mathrm{y = 1 + x}$](1)</Checkpoint>

$K_2^\circ = \frac{x(2-2x)^2}{(1-x)(1-2x)^2} =  \frac{(3-y)(4+2y)^2}{4y^3}$，
代入y = 1 + x，解得x = 0.332，$K_2^\circ$ = 7.84

<Checkpoint>[$x = 0.332$](1)</Checkpoint>

<Checkpoint>[$K_2^\circ = 7.84$](1)</Checkpoint>

$K_1^\circ = \frac{(1-x)(1-2x)^2}{(1-x)^2n(B)_1} =  \frac{2y(4+2y)^2}{y^2n(B)_2}$，
解得$\mathrm{n(B)_1 = 1.36×10^{-5} (mol)，n(B)_2 = 6.76×10^{-4} (mol)}$

<Checkpoint>[$\mathrm{n(B)_1 = 1.36×10^{-5} (mol)，n(B)_2 = 6.76×10^{-4} (mol)}$](2)</Checkpoint>

$\mathrm{n(B)_1和n(B)_2}$均很小，说明近似合理。

$\frac{2-2x}{4} = \frac{a+4}{4}，a = -2.66$

<Checkpoint>[$a = -2.66$](1)</Checkpoint>