The vapor pressure of liquid chloroform, CHCl3, is 400.0 torr at 24.1°C and 100.0 torr at -6.3°C. What is ΔHvap (in the unit of kJ) of chloroform?
To calculate the enthalpy of vaporization (ΔHvap) of chloroform, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the vapor pressures of chloroform at temperatures T1 and T2 respectively,
ΔHvap is the enthalpy of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)),
T1 and T2 are the temperatures in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin:
T1 = 24.1°C + 273.15 = 297.25 K
T2 = -6.3°C + 273.15 = 266.85 K
Next, we substitute the values into the equation:
ln(100.0 torr / 400.0 torr) = (-ΔHvap/8.314 J/(mol·K)) * (1/266.85 K - 1/297.25 K)
Now, we solve for ΔHvap:
ln(0.25) = (-ΔHvap/8.314 J/(mol·K)) * (-0.003762 K^-1)
Rearrange the equation:
ΔHvap/8.314 J/(mol·K) = ln(0.25) / 0.003762 K^-1
Now, solve for ΔHvap:
ΔHvap = (ln(0.25) / 0.003762 K^-1) * 8.314 J/(mol·K)
Calculating this expression, we get:
ΔHvap = -30.4 kJ
So, the enthalpy of vaporization of chloroform is approximately -30.4 kJ.
To find the enthalpy of vaporization (ΔHvap) of chloroform (CHCl3), we can make use of the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
where P1 and P2 are the initial and final vapor pressures, T1 and T2 are the initial and final temperatures, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant (8.314 J/(mol·K)).
Firstly, we need to convert the temperatures to Kelvin:
T1 = 24.1°C + 273.15 = 297.25 K
T2 = -6.3°C + 273.15 = 266.85 K
Next, we can substitute the given values into the equation. We will use the natural logarithm (ln) in place of log base e:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
ln(100.0 torr / 400.0 torr) = -ΔHvap/(8.314 J/(mol·K)) * (1/266.85 K - 1/297.25 K)
Now we can solve for ΔHvap:
ln(0.25) = -ΔHvap/(8.314 J/(mol·K)) * (0.00375 K^-1)
Using the natural logarithm rules, we can isolate ΔHvap:
ΔHvap = -(8.314 J/(mol·K)) * ln(0.25) / (0.00375 K^-1)
Calculating this expression, we find:
ΔHvap ≈ 24.0 kJ
Therefore, the enthalpy of vaporization (ΔHvap) of chloroform is approximately 24.0 kJ.
ln(pressure)=ΔHvap/RT + C
so you have two conditions:
ln(400)= ΔHvap/(R*297)+C
ln(100)=ΔHvap/(R*266.7) + C
Hvap=(R*297)(ln400-C)
Hvap=(R*266.7)(ln100-C)
two equations, two unknowns. set the two equations equal, solve for C first, then go back into either equation, and solve for Hvap.
R is (8.314 J/mole K)
most solve this on graph paper, plotting ln(Pressure) vs 1/Temp, then measuring the slope, it will be Hvap, on the graph, C does not come into play.