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To determine the enthalpy of vaporization, [tex]\(\Delta H_{\text{vap}}\)[/tex], of HF from the given vapor pressure data, we follow these steps:
### Step 1: Organize the Data
First, we list the given data:
- Temperatures, [tex]\(T\)[/tex], in Kelvin: [tex]\(250 \, \text{K}\)[/tex], [tex]\(260 \, \text{K}\)[/tex], [tex]\(270 \, \text{K}\)[/tex], [tex]\(280 \, \text{K}\)[/tex]
- Vapor pressures, [tex]\(P\)[/tex], in mmHg: [tex]\(129.63 \, \text{mmHg}\)[/tex], [tex]\(206.65 \, \text{mmHg}\)[/tex], [tex]\(318.22 \, \text{mmHg}\)[/tex], [tex]\(475.16 \, \text{mmHg}\)[/tex]
### Step 2: Convert Pressure to Natural Logarithm
We need to convert the pressure data to their natural logarithm values, [tex]\(\ln P\)[/tex]:
[tex]\[ \begin{aligned} \ln(129.63) & \approx 4.86468424 \\ \ln(206.65) & \approx 5.33102654 \\ \ln(318.22) & \approx 5.76274297 \\ \ln(475.16) & \approx 6.16365159 \end{aligned} \][/tex]
### Step 3: Convert Temperature to Inverse Temperature
Next, we convert the temperature data to their inverse values, [tex]\(\frac{1}{T}\)[/tex], in [tex]\(\text{K}^{-1}): \[ \begin{aligned} \frac{1}{250} & = 0.004 \\ \frac{1}{260} & \approx 0.00384615 \\ \frac{1}{270} & \approx 0.00370370 \\ \frac{1}{280} & \approx 0.00357143 \end{aligned} \] ### Step 4: Perform Linear Regression To determine the slope of the line in the \(\ln P\)[/tex] vs. [tex]\(\frac{1}{T}\)[/tex] plot, we perform a linear regression. The linear regression provides us with the slope of the best-fit line. The result for the slope obtained from the regression is:
[tex]\[ \text{slope} \approx -3030.8993687210695 \][/tex]
### Step 5: Calculate the Enthalpy of Vaporization
The Clausius-Clapeyron equation in its linear form is given by:
[tex]\[ \ln P = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T}\right) + C \][/tex]
where [tex]\(R\)[/tex] is the universal gas constant ([tex]\(R = 8.314 \, \text{J/(mol·K)}\)[/tex]). From the slope [tex]\(-\frac{\Delta H_{\text{vap}}}{R}\)[/tex], we can solve for [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[ \Delta H_{\text{vap}} = -\text{slope} \times R \][/tex]
Given that:
[tex]\[ \text{slope} \approx -3030.8993687210695 \][/tex]
[tex]\[ R = 8.314 \, \text{J/(mol·K)} \][/tex]
Converting the enthalpy from J/mol to kJ/mol by dividing by 1000:
[tex]\[ \Delta H_{\text{vap}} = -(-3030.8993687210695) \times 8.314 / 1000 \][/tex]
Simplifying this expression yields:
[tex]\[ \Delta H_{\text{vap}} \approx 25.19889735154697 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of vaporization, [tex]\(\Delta H_{\text{vap}}\)[/tex], of HF is approximately:
[tex]\[ \boxed{25.19889735154697 \, \text{kJ/mol}} \][/tex]
### Step 1: Organize the Data
First, we list the given data:
- Temperatures, [tex]\(T\)[/tex], in Kelvin: [tex]\(250 \, \text{K}\)[/tex], [tex]\(260 \, \text{K}\)[/tex], [tex]\(270 \, \text{K}\)[/tex], [tex]\(280 \, \text{K}\)[/tex]
- Vapor pressures, [tex]\(P\)[/tex], in mmHg: [tex]\(129.63 \, \text{mmHg}\)[/tex], [tex]\(206.65 \, \text{mmHg}\)[/tex], [tex]\(318.22 \, \text{mmHg}\)[/tex], [tex]\(475.16 \, \text{mmHg}\)[/tex]
### Step 2: Convert Pressure to Natural Logarithm
We need to convert the pressure data to their natural logarithm values, [tex]\(\ln P\)[/tex]:
[tex]\[ \begin{aligned} \ln(129.63) & \approx 4.86468424 \\ \ln(206.65) & \approx 5.33102654 \\ \ln(318.22) & \approx 5.76274297 \\ \ln(475.16) & \approx 6.16365159 \end{aligned} \][/tex]
### Step 3: Convert Temperature to Inverse Temperature
Next, we convert the temperature data to their inverse values, [tex]\(\frac{1}{T}\)[/tex], in [tex]\(\text{K}^{-1}): \[ \begin{aligned} \frac{1}{250} & = 0.004 \\ \frac{1}{260} & \approx 0.00384615 \\ \frac{1}{270} & \approx 0.00370370 \\ \frac{1}{280} & \approx 0.00357143 \end{aligned} \] ### Step 4: Perform Linear Regression To determine the slope of the line in the \(\ln P\)[/tex] vs. [tex]\(\frac{1}{T}\)[/tex] plot, we perform a linear regression. The linear regression provides us with the slope of the best-fit line. The result for the slope obtained from the regression is:
[tex]\[ \text{slope} \approx -3030.8993687210695 \][/tex]
### Step 5: Calculate the Enthalpy of Vaporization
The Clausius-Clapeyron equation in its linear form is given by:
[tex]\[ \ln P = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T}\right) + C \][/tex]
where [tex]\(R\)[/tex] is the universal gas constant ([tex]\(R = 8.314 \, \text{J/(mol·K)}\)[/tex]). From the slope [tex]\(-\frac{\Delta H_{\text{vap}}}{R}\)[/tex], we can solve for [tex]\(\Delta H_{\text{vap}}\)[/tex]:
[tex]\[ \Delta H_{\text{vap}} = -\text{slope} \times R \][/tex]
Given that:
[tex]\[ \text{slope} \approx -3030.8993687210695 \][/tex]
[tex]\[ R = 8.314 \, \text{J/(mol·K)} \][/tex]
Converting the enthalpy from J/mol to kJ/mol by dividing by 1000:
[tex]\[ \Delta H_{\text{vap}} = -(-3030.8993687210695) \times 8.314 / 1000 \][/tex]
Simplifying this expression yields:
[tex]\[ \Delta H_{\text{vap}} \approx 25.19889735154697 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of vaporization, [tex]\(\Delta H_{\text{vap}}\)[/tex], of HF is approximately:
[tex]\[ \boxed{25.19889735154697 \, \text{kJ/mol}} \][/tex]
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