When a student plotted ln [vapor pressure of a gas] vs. inverse Kelvin temperature, she obtained a straight line with a slope equal to -1.0500 x 10^4 K.
According to the Clausius Clapeyron equation, Hvap is
kJ/mol. ?
The slope is -Hvap/R, so multiply the slope by R (the gas constant, in appropriate units) to get the heat pf vaporization, Hvap.
R = 8.317 J/mole K in case you have forgotten.
is the answer supposed to be 87297?
That depends upon the units you are using. Using the R value I gave you, the answer would be in Joules/mole. Always check and list dimensions
Divide by 1000 for kJ/mole
I get 87.33 kJ/mole. Maybe you used a slightly different R, with more significiant figures.
To determine Hvap (enthalpy of vaporization) using the Clausius Clapeyron equation with the given information, we can start by understanding the equation itself:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively,
ΔHvap is the enthalpy of vaporization,
R is the ideal gas constant, and
T1 and T2 are the corresponding inverse Kelvin temperatures.
From the given information, we have the slope of the straight line, which is equal to -1.0500 x 10^4 K. We can equate this slope to -ΔHvap/R and solve for ΔHvap.
-1.0500 x 10^4 K = -ΔHvap/R
Since R is a constant, we can rearrange the equation to solve for ΔHvap:
ΔHvap = (-1.0500 x 10^4 K) * R
Now, we need to determine the value of R. The value of the ideal gas constant (R) is typically expressed in J/mol·K. However, we want to find the value of Hvap in kJ/mol. To convert R from J/mol·K to kJ/mol·K, we divide it by 1000:
R (kJ/mol·K) = R (J/mol·K) / 1000
The value of R in J/mol·K is approximately 8.314 J/mol·K. Dividing this by 1000 gives us 0.008314 kJ/mol·K.
Substituting the value of R into our equation:
ΔHvap = (-1.0500 x 10^4 K) * (0.008314 kJ/mol·K)
Calculating this expression yields the value of ΔHvap in kJ/mol.