Question

question 7 of 10 > points possible: 1.00 not yet answered the slope of your linear graph of ln p vs. 1/t is found to be -4800 k. what is the experimental value for the heat of vaporization (δhvap) (in j/mol)? (the universal gas constant, r, is 8.31 j/mol·k). select one: -577 j/mol 4800 j/mol 39,900 j/mol 577 j/mol

Explanation:

Step1: Recall Clausius - Clapeyron equation

The Clausius - Clapeyron equation for a linear plot of $\ln P$ vs. $\frac{1}{T}$ is $\ln P=-\frac{\Delta H_{vap}}{R}\times\frac{1}{T}+C$, where the slope $m =-\frac{\Delta H_{vap}}{R}$.

Step2: Solve for $\Delta H_{vap}$

We are given that $m=- 4800\ K$ and $R = 8.31\ J/(mol\cdot K)$. Rearranging the slope formula for $\Delta H_{vap}$ gives $\Delta H_{vap}=-m\times R$.
Substitute the values: $\Delta H_{vap}=-(-4800\ K)\times8.31\ J/(mol\cdot K)$.
$\Delta H_{vap}=4800\times8.31\ J/mol = 39900\ J/mol$ (rounded to three - significant figures).

Answer:

39,900 J/mol