The normal boiling point of Rh is 3727°C. What is its vapor pressure in mm Hg at 2914°C, if its heat of vaporization is 557.0 kJ/mol?
Use the Clausius-Clapeyron equation
I don't understand that equation. Which part of what's given would fit into it? T1=3727 T2=2914 then convert to kelvin. delta h is 557.0. but what do i use for R..or is that what i am looking for?
oh R is constant.
Yes, R is 8.314
Then P1 goes with T1 and P2 goes with T2.
Delta H is 557.0 kJ/mol but you should use 557,000 in the formula;i.e., convert to J/mol.
To find the vapor pressure of Rh at a given temperature, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature, in terms of the heat of vaporization and the gas constant.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -ΔH_vap/R * (1/T2 - 1/T1)
Where:
P1 is the vapor pressure at temperature T1
P2 is the vapor pressure at temperature T2
ΔH_vap is the heat of vaporization
R is the gas constant (8.314 J/(mol*K))
T1 is the initial temperature
T2 is the final temperature
In this case, we need to find the vapor pressure of Rh at 2914°C (or 3187 K), given the normal boiling point of 3727°C (or 4000 K) and the heat of vaporization of 557.0 kJ/mol (or 557000 J/mol).
Let's plug the values into the equation:
ln(P2/P1) = -ΔH_vap/R * (1/T2 - 1/T1)
ln(P2/1 atm) = -(557000 J/mol) / (8.314 J/(mol*K)) * (1/3187 K - 1/4000 K)
Now, we can solve for the natural logarithm of the vapor pressure ratio:
ln(P2/1) = -67.00 * (4.99e-4 - 3.17e-4)
Simplifying:
ln(P2) = -67.00 * (1.82e-4)
Now, we can solve for P2:
P2 = e^(-67.00 * 1.82e-4)
Plugging this into a calculator, we find:
P2 ≈ 0.728 mm Hg
Therefore, the vapor pressure of Rh at 2914°C is approximately 0.728 mm Hg.