Lomefloxacin hydrochloride is a quinolone antibacterial drug that can be used to treat acute and chronic infectious diseases caused by various Gram-positive and Gram-negative bacteria. Now, an adult patient is given lomefloxacin hydrochloride by intravenous infusion at a rate of $\rm 0.200~g/h$, and the infusion time is $\rm T = 1~h$.

After lomefloxacin hydrochloride enters the body, it quickly diffuses to various parts to reach equilibrium. Therefore, the entire organism can be regarded as a single "compartment". Let $V_d$ be the ratio of the amount of drug in the body to the blood drug concentration at this time, then $V_d$ is equivalent to the volume of the "compartment" (apparent volume of distribution). The metabolism of lomefloxacin hydrochloride in the human body can be regarded as a first-order reaction (rate constant is $k$). The relationship between the concentration of lomefloxacin hydrochloride in plasma $\rm C (\mu g/mL)$ and time is shown in the following table.

| 时间 (h) | 0.00 | 0.50 | 1.00 | 4.00 | 7.00 | 11.00 | 15.00 | +∞ |
|---|---|---|---|---|---|---|---|---|
| $\rm C (\mu g/mL)$ | 0 | 2.54 | 4.49 | 2.43 | 1.49 | 0.79 | 0.45 | 0 |

The following statements are correct (if the calculated result has an error greater than 3%, the statement is considered incorrect):
1. Using linear fitting, the rate constant of drug metabolism can be calculated as $\rm k=0.163~ \rm h^{-1}$
2. The apparent volume of distribution is $\rm V_d=45~ \rm L$ (to maintain the uniqueness of the results, if you want to substitute the data in the table for calculation, uniformly choose $\rm t=1.00~h$)
3. If intravenous infusion is continued at the same rate, after a period of time, the blood drug concentration in the patient's body can reach a relatively stable level, which is called the steady-state blood drug concentration $\rm C_{ss}$. Under the conditions of the problem, the drug's $\rm C_{ss}=30.06~ \mu \rm g/mL$
4. If the intravenous infusion rate is increased, the time $\rm t$ for the drug concentration to reach 90% of the steady-state blood drug concentration will be shorter
5. Lomefloxacin hydrochloride is more evenly distributed in human body fluids
盐酸洛美沙星是一种喹诺酮类抗菌药，可用于治疗各种革兰氏阳性菌和阴性菌引起的急、慢性感染性疾病。现对某成年病人采取静脉滴注的方式，按照$\rm 0.200~g/h$的给药速率滴注盐酸洛美沙星，滴注时间$\rm T = 1~h$。

盐酸洛美沙星在进入体内后，迅速扩散至各部位达到平衡，因此可将整个机体看做是单一"房室"。记此时体内药量与血药浓度的比值为$V_d$，则$V_d$相当于该"房室"的容积（表观分布容积）。盐酸洛美沙星在人体内的代谢可视为一级反应（速率常数为$k$）。血浆中盐酸洛美沙星的浓度$C (\mu \rm g/mL)$随时间的变化关系如下表所示。

| 时间 (h) | 0.00 | 0.50 | 1.00 | 4.00 | 7.00 | 11.00 | 15.00 | +∞ |
|---|---|---|---|---|---|---|---|---|
| $\rm C (\mu g/mL)$ | 0 | 2.54 | 4.49 | 2.43 | 1.49 | 0.79 | 0.45 | 0 |

下列说法正确的有（如计算得到的结果误差大于3%则认为该说法错误）：
1. 利用线性拟合，可以算得该药物代谢的速率常数$\rm k=0.163~ \rm h^{-1}$
2. 表观分布容积$\rm V_d=45~ \rm L$（为保持结果的唯一性，计算时若要代入表中数据，统一选择$\rm t=1.00~h$的）
3. 若按相同的速率持续保持静脉滴注给药，经一段时间后，患者体内的血药浓度可达到相对稳定的水平，称此时的血药浓度为稳态血药浓度$\rm C_{ss}$。题设条件下药物的$\rm C_{ss}=30.06~ \mu \rm g/mL$
4. 若增加静脉滴注给药速率，则药物浓度达到稳态血药浓度的90%的时间$\rm t$变小
5. 盐酸洛美沙星较均匀地分布在人体体液中
All other options are incorrect
1,2
2,3,4
3,4,5
1,4,5
2,5
2,3,4,5
1,3,5
1,2,4,5
2,3
1,4
1,3
3,5
3
5
1
其他选项均不正确
1,2
2,3,4
3,4,5
1,4,5
2,5
2,3,4,5
1,3,5
1,2,4,5
2,3
1,4
1,3
3,5
3
5
1
**Statement 1**:

For a first-order reaction, $\rm \ln C = \ln C_0 - kt$; therefore, plotting $\rm \ln C$-$t$, the slope is $\rm k$.

Calculate the logarithm of each $\rm C$, $\rm \ln C$:

| Time (h) | 1.00 | 4.00 | 7.00 | 11.00 | 15.00 |
|---|---|---|---|---|---|
| $\rm C (\mu g/mL)$ | 4.49 | 2.43 | 1.49 | 0.79 | 0.45 |
| $\rm \ln C$ | 1.50185 | 0.88789 | 0.3988 | -0.23572 | -0.79851 |

Linear regression yields $\rm k=0.163~h^{-1}$, statement 1 is correct.

<Checkpoint>[$\rm k=0.163~h^{-1}$](2)</Checkpoint>

**Statement 2**:

For the intravenous infusion administration phase, the drug amount in the patient's body changes with time according to:

$\rm dC/dt = r - kC$

$\rm -\frac{1}{k} \ln(-kC + r) = t + c$

When $\rm t=0$, $\rm C=0$; therefore, $\rm c = -\frac{1}{k}\ln(r)$

So: $\rm t = \frac{1}{k}\ln(r) - \frac{1}{k}\ln(-kC + r)$

Simplified: $\rm \exp(-kt) = 1 - \frac{kC}{r}$

<Checkpoint>[$\rm \exp(-kt) = 1 - \frac{kC}{r}$](2)</Checkpoint>

$\rm \exp(-0.163~h^{-1} \cdot t) = 1 - \frac{0.163~h^{-1} \cdot C}{r}$

Substituting $\rm t=1.00~h$, $\rm C=4.49~\mu g/mL$, we get: $\rm r=4.9~\mu g/(mL·h)$

The apparent volume of distribution is: $\rm V_d = 0.200~g/h \div 4.9~\mu g/(mL·h) = 41~L$, statement 2 is incorrect.

<Checkpoint>[The apparent volume of distribution is: $\rm V_d = 41~L$](2)</Checkpoint>

**Statement 3**:

$\rm C_{ss} = r/k = 4.9~\mu g/(mL\cdot h) \div 0.163~h^{-1} = 30.06~\mu g/mL$, statement 3 is correct.

<Checkpoint>[$\rm C_{ss} = 30.06~\mu g/mL$](1)</Checkpoint>

**Statement 4**:

From $\rm \exp(-kt) = 1 - \frac{kC}{r}$ and $\rm C_{ss} = r/k$, we know $\rm C = C_{ss}[1-exp(-kt)]$, so the time $\rm t$ for the drug concentration to reach 90% of the steady-state blood drug concentration is only related to $\rm k$. Increasing the intravenous infusion rate does not change $\rm t$, statement 4 is incorrect.

<Checkpoint>[The time $\rm t$ for the drug concentration to reach 90% of the steady-state blood drug concentration is only related to $\rm k$. Increasing the intravenous infusion rate does not change $\rm t$](1)</Checkpoint>

**Statement 5**: The apparent volume of distribution is $\rm V_d = 41~L$, which is comparable to the total body fluid volume, so lomefloxacin hydrochloride is relatively evenly distributed in the human body, statement 5 is correct.

<Checkpoint>[Lomefloxacin hydrochloride is relatively evenly distributed in the human body](1)</Checkpoint>

Statements 1, 3, and 5 are correct, choose H
**说法1**：

对于一级反应，$\rm \ln C = \ln C_0 - kt$；因此，作$\rm \ln C$-$t$图，其斜率即为$\rm k$。 
 
算出每个$\rm C$对应的对数值$\rm \ln C$：  

| 时间 (h) | 1.00 | 4.00 | 7.00 | 11.00 | 15.00 |  
|---|---|---|---|---|---|  
| $\rm C (\mu g/mL)$ | 4.49 | 2.43 | 1.49 | 0.79 | 0.45 |  
| $\rm \ln C$ | 1.50185 | 0.88789 | 0.3988 | -0.23572 | -0.79851 |  

线性回归可得$\rm k=0.163~h^{-1}$，说法1正确

<Checkpoint>[$\rm k=0.163~h^{-1}$](2)</Checkpoint>

**说法2**：

对于静脉滴注给药阶段，患者体内药量随时间的变化关系满足：

$\rm dC/dt = r - kC$

$\rm -\frac{1}{k} \ln(-kC + r) = t + c$

当$\rm t=0$时，$\rm C=0$；因此，$\rm c = -\frac{1}{k}\ln(r)$

所以：$\rm t = \frac{1}{k}\ln(r) - \frac{1}{k}\ln(-kC + r)$

化简得：$\rm \exp(-kt) = 1 - \frac{kC}{r}$

<Checkpoint>[$\rm \exp(-kt) = 1 - \frac{kC}{r}$](2)</Checkpoint>

$\rm \exp(-0.163~h^{-1} \cdot t) = 1 - \frac{0.163~h^{-1} \cdot C}{r}$

将$\rm t=1.00~h$，$\rm C=4.49~\mu g/mL$代入，求得：$\rm r=4.9~\mu g/(mL·h)$

表观分布容积为：$\rm V_d = 0.200~g/h \div 4.9~\mu g/(mL·h) = 41~L$，说法2错误

<Checkpoint>[表观分布容积为：$\rm V_d = 41~L$](2)</Checkpoint>

**说法3**：

$\rm C_{ss} = r/k = 4.9~\mu g/(mL\cdot h) \div 0.163~h^{-1} = 30.06~\mu g/mL$，说法3正确

<Checkpoint>[$\rm C_{ss} = 30.06~\mu g/mL$](1)</Checkpoint>

**说法4**：

由$\rm \exp(-kt) = 1 - \frac{kC}{r}$、$\rm C_{ss} = r/k$可知$\rm C = C_{ss}[1-exp(-kt)]$，故药物浓度达到稳态血药浓度的90%的时间$\rm t$仅与$\rm k$有关，增加静脉滴注给药速率时$\rm t$不变，说法4错误

<Checkpoint>[药物浓度达到稳态血药浓度的90%的时间$\rm t$仅与$\rm k$有关，增加静脉滴注给药速率时$\rm t$不变](1)</Checkpoint>

**说法5**：表观分布容积为$\rm V_d = 41~L$，这与人体总体液量相当，故盐酸洛美沙星较均匀地分布在人体中，说法5正确

<Checkpoint>[盐酸洛美沙星较均匀地分布在人体中](1)</Checkpoint>

说法1、3、5正确，选H