As shown in the figure, a certain amide can spontaneously form an intramolecular hydrogen bond:

<MultiModal>![This is a schematic diagram of a chemical structure, illustrating the equilibrium between two conformational isomers. On the left side of the diagram, labeled as A, the molecule exhibits a linear structure. From left to right, the structure consists of a nitrogen atom bonded to a hydrogen atom and a methyl group (Me), which is further connected to a carbonyl group (C=O). The carbonyl group is linked to a chain composed of four methylene units. The other end of the chain is attached to a second carbonyl group, which is bonded to a nitrogen atom bearing two methyl groups (Me). Between molecule A and the molecule B on the right, there is a bidirectional equilibrium arrow. The right side of the diagram shows the molecular conformation labeled as B, which adopts a cyclic folded form. In this conformation, the carbonyl oxygen atom of the gem-dimethyl carbamoyl group on the right forms an intramolecular interaction with the N-H bond of the N-methyl carbamoyl group on the left, represented by a dashed arrow pointing from the hydrogen atom to the oxygen atom. All methyl (Me) groups in this conformation are explicitly labeled.](/media/uploads/1ff7e98e-39a4-416c-b657-fc6035b3cbc2.png)</MultiModal>

The reaction is represented in SMILE notation as:
O=C(CCCCC(N(C)C)=O)NC>>O=C1CCCC/C(N(C)C)=O\[H]N1C

Next, the experimenter quantitatively determined the thermodynamic information of this reaction by measuring the apparent chemical shift values ($δ_{obs}$) of the amide at different temperatures. The measured $δ_{obs}$ values at various temperatures are shown in the table below:

| \( T/K \) | \(\delta_{obs}/ppm\) |
|-----------|-----------------------|
| 220      | 6.67                 |
| 240      | 6.50                 |
| 260      | 6.37                 |
| 280      | 6.27                 |
| 300      | 6.19                 |

It is known that when the temperature is extremely low, $\delta_{obs}$ approaches $8.40\,ppm$, and when the temperature is extremely high, $\delta_{obs}$ tends toward $5.70\,ppm$. Please calculate the standard molar enthalpy change of the reaction (in $kJ\cdot mol^{-1}$), denoted as $H$, and the standard molar entropy change of the equilibrium transition (in $J\cdot mol^{-1} \cdot K^{-1}$), denoted as $S$. The following options provide a series of $(H,S)$ values. Select the option that is closest to your calculation results.
如图所示，某酰胺$\mathbf A$能够自发形成分子内氢键生成$\mathbf B$，两者在很大温度范围内保持平衡：

<MultiModal>![这幅图表示了从**A**到**B**的可逆平衡，SMILES表达为O=C(CCCCC(N(C)C)=O)NC>>O=C1CCCC/C(N(C)C)=O\[H]N1C](/media/uploads/1ff7e98e-39a4-416c-b657-fc6035b3cbc2.png)</MultiModal>

下面，实验者通过测定不同温度下该酰胺的表观化学位移值($δ_{obs}$)来定量测定该反应的热力学信息，测得不同温度下$δ_{obs}$值如下表所示：

| \( T/K \) | \(\delta_{obs}/ppm\) |
|-----------|-----------------------|
| 220      | 6.67                 |
| 240      | 6.50                 |
| 260      | 6.37                 |
| 280      | 6.27                 |
| 300      | 6.19                 |

已知当温度特别低时，$\delta_{obs}$趋于$8.40\,ppm$，温度特别高时，$\delta_{obs}$则趋于$5.70\,ppm$。请你计算该反应的标准摩尔焓变（$kJ\cdot mol^{-1}$），记为$H$，以及该转变平衡的标准摩尔熵变（ $J\cdot mol^{-1} \cdot K^{-1}$），记为$S$。下列选项中给出了一系列$(H,S)$的数值，请你选择和你的计算结果最接近的选项。
(-5, -10)
(5,10)
(-20, -50)
(-30, -50)
(10, -40)
(20, -30)
(-5, -30)
(10,40)
None of the above options are correct
（-5,-10）
(5,10)
(-20,-50)
(-30,-50)
(10,-40)
(20,-30)
(-5,-30)
(10,40)
以上选项均不正确
Given the problem, let the mole fractions of $\mathbf{A}$ and $\mathbf{B}$ be $x_A$ and $x_B$ ($x_A + x_B = 1$), respectively. Then, $\delta_{obs}$ can be expressed as:

$\delta_{obs}=x_A\delta_{A}+x_B\delta_{B}$

<Checkpoint>[$\delta_{obs}=x_A\delta_{A}+x_B\delta_{B}$](1)</Checkpoint>

At low temperatures, hydrogen bonds are stable, and $\mathbf{B}$ is the dominant conformation. At high temperatures, hydrogen bonds break, and $\mathbf{A}$ is the dominant conformation. Thus:

$\delta_{low}=\delta_{B}=8.40\,ppm$

$\delta_{high}=\delta_{A}=5.70\,ppm$

<Checkpoint>[At low temperatures, hydrogen bonds are stable, and $\mathbf{B}$ is the dominant conformation. At high temperatures, hydrogen bonds break, and $\mathbf{A}$ is the dominant conformation.](1)</Checkpoint>

<Checkpoint>[$\delta_{low}=\delta_{B}=8.40\,ppm$](1)</Checkpoint>

<Checkpoint>[$\delta_{high}=\delta_{A}=5.70\,ppm$](1)</Checkpoint>

From this, the expression for the equilibrium constant can be derived:

$K=\frac{x_B}{x_A}=\frac{\delta_{obs}-\delta_{A}}{\delta_B-\delta_{obs}}=\frac{\delta_{obs}-5.70}{8.40-\delta_{obs}}$

<Checkpoint>[$K=\frac{x_B}{x_A}=\frac{\delta_{obs}-5.70}{8.40-\delta_{obs}}$](1)</Checkpoint>

Calculate the equilibrium constant values under various temperature conditions:

| \( T/K \) | \(K\) |
|-----------|-----------------------|
| 220      | 0.5607                |
| 240      |0.4211                 |
| 260      | 0.3300                |
| 280      | 0.2676                |
| 300      | 0.2217                |

According to the van't Hoff equation, perform linear regression on $1/T - \ln K$ and calculate the slope $k = 764.2\,(K)$ and intercept $b = -4.050$.

<Checkpoint>[Perform linear regression on $1/T - \ln K$ and calculate the slope $k = 764.2\,(K)$ and intercept $b = -4.050$.](1)</Checkpoint>

Thus, calculate:
$H = -k \cdot R = -6.35\,kJ\cdot mol^{-1}$  
$S = b \cdot R = -33.7\,J\cdot mol^{-1} \cdot K^{-1}$

<Checkpoint>[$H = -6.35\,kJ\cdot mol^{-1}$, $S = -33.7\,J\cdot mol^{-1} \cdot K^{-1}$](1)</Checkpoint>

Therefore, the closest option is G.
由题，设$\mathbf{A、B}$摩尔分数分别为$x_A，x_B (x_A+x_B=1)$，则$\delta_{obs}$可以表示为：

$\delta_{obs}=x_A\delta_{A}+x_B\delta_{B}$

<Checkpoint>[$\delta_{obs}=x_A\delta_{A}+x_B\delta_{B}$](1)</Checkpoint>

低温下氢键稳定存在，$\mathbf B$为主要构象，高温下氢键断裂，$\mathbf A$为主要构象，则有：

$\delta_{low}=\delta_{B}=8.40\,ppm$

$\delta_{high}=\delta_{A}=5.70\,ppm$

<Checkpoint>[低温下氢键稳定存在，$\mathbf B$为主要构象，高温下氢键断裂，$\mathbf A$为主要构象](1)</Checkpoint>

<Checkpoint>[$\delta_{low}=\delta_{B}=8.40\,ppm$](1)</Checkpoint>

<Checkpoint>[$\delta_{high}=\delta_{A}=5.70\,ppm$](1)</Checkpoint>


由此可推导出反应平衡常数表达式：

$K=\frac{x_B}{x_A}=\frac{\delta_{obs}-\delta_{A}}{\delta_B-\delta_{obs}}=\frac{\delta_{obs}-5.70}{8.40-\delta_{obs}}$

<Checkpoint>[$K=\frac{x_B}{x_A}=\frac{\delta_{obs}-5.70}{8.40-\delta_{obs}}$](1)</Checkpoint>

计算出各个温度条件下的平衡常数值：

| \( T/K \) | \(K\) |
|-----------|-----------------------|
| 220      | 0.5607                |
| 240      |0.4211                 |
| 260      | 0.3300                |
| 280      | 0.2676                |
| 300      | 0.2217                |

根据范德霍夫方程，对$1/T-lnK$进行线性回归计算，算得$k=764.2(K)$截距$b=-4.050$

<Checkpoint>[对$1/T-lnK$进行线性回归计算，计算出斜率$k=764.2(K)$截距$b=-4.050$](1)</Checkpoint>

由此则计算出$H=-k \cdot R=-6.35\,kJ\cdot mol^{-1}$
$S=b \cdot R=-33.7\,J\cdot mol^{-1} \cdot K^{-1}$

<Checkpoint>[$H=-6.35\,kJ\cdot mol^{-1}$， $S=-33.7\,J\cdot mol^{-1} \cdot K^{-1}$](1)</Checkpoint>

故选择最为接近的G选项