Question

in this lab, you will determine the vapour pressure (p) of water at various temperatures and use this data to find its enthalpy of vaporization (δvaph).
more generally, for an unknown liquid: if a plot of ln p vs. 1/t gives a slope of -3105 k, what is its δvaph in kj/mol? do not worry about how realistic the value is. assume r = 8.314 j mol-1 k-1.
answer:

Explanation:

Step1: Recall Clausius - Clapeyron equation

The Clausius - Clapeyron equation in linear form is $\ln P=-\frac{\Delta_{vap}H}{R}\cdot\frac{1}{T}+C$, where the slope of the plot of $\ln P$ vs. $\frac{1}{T}$ is $m =-\frac{\Delta_{vap}H}{R}$.

Step2: Solve for $\Delta_{vap}H$

We know that the slope $m=- 3105\ K$ and $R = 8.314\ J\ mol^{-1}\ K^{-1}$. From $m=-\frac{\Delta_{vap}H}{R}$, we can re - arrange the formula to get $\Delta_{vap}H=-m\times R$.

Substitute the values of $m$ and $R$ into the formula:
$\Delta_{vap}H=-(-3105\ K)\times8.314\ J\ mol^{-1}\ K^{-1}$
$\Delta_{vap}H = 3105\times8.314\ J\ mol^{-1}$
Calculate $3105\times8.314$:
$3105\times8.314=(3000 + 105)\times8.314=3000\times8.314+105\times8.314=24942+872.97 = 25814.97\ J\ mol^{-1}$

Convert the units from $J\ mol^{-1}$ to $kJ\ mol^{-1}$ by dividing by 1000:
$\Delta_{vap}H=\frac{25814.97\ J\ mol^{-1}}{1000}=25.81\ kJ\ mol^{-1}$ (approximate value, more accurately $3105\times8.314 = 3105\times8+3105\times0.314=24840+975.07 = 25815.07\ J\ mol^{-1}\approx25.8\ kJ\ mol^{-1}$)

Answer:

$\approx25.8\ kJ/mol$ (or more precisely, using $3105\times8.314 = 25815.07\ J/mol=25.82\ kJ/mol$)