Question

Using the sublimation relation $\log_{10}(P)=\frac{C_1}{T}+C_2$ for zinc, with $C_1=6790$ K, $C_2=9$, compute the latent heat of sublimation using the Clausius–Clapeyron equation. (Specify answer in kJ/mol up to one digit after decimal.)
 

Hint:

Always convert log base 10 to natural log using $\ln 10 = 2.3026$.

Answer 1

Correct Answer: 129

Answer should be 129.0-131.0 or 55.0-58.0
Step 1: Use the relation between slope and latent heat.
$\ln P = \frac{C_1}{T}\ln 10 + C_2\ln 10$. Therefore, slope = $\frac{d(\ln P)}{d(1/T)} = -C_1 \ln 10$.

Step 2: Clausius–Clapeyron equation.
$\frac{d(\ln P)}{d(1/T)} = -\frac{L}{R}$. Thus, $L = C_1 R \ln 10$.

Step 3: Substitute numerical values.
$L = 6790 \times 8.314 \times 2.3026 = 1.306\times 10^5$ J/mol.

Step 4: Convert to kJ/mol.
$L = 130.6$ kJ/mol.