The heat of vaporization of bromine at its boiling point (59.6 °C) at 1.00 atm is 29.54 kJ/mol. Calculate the vapor pressure of bromine at 37.8 °C.
Use the Clausius-Clapeyron equation.
To calculate the vapor pressure of bromine at 37.8 °C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P2 is the vapor pressure at 37.8°C
P1 is the vapor pressure at 59.6°C
ΔHvap is the heat of vaporization (29.54 kJ/mol)
R is the ideal gas constant (8.314 J/mol·K)
T2 is the temperature at 37.8°C (in Kelvin: 37.8 + 273.15 K)
T1 is the temperature at 59.6°C (in Kelvin: 59.6 + 273.15 K)
First, we need to convert the heat of vaporization from kJ/mol to J/mol:
ΔHvap = 29.54 kJ/mol * 1000 J/kJ = 29,540 J/mol
Now, let's substitute the values into the equation and calculate the vapor pressure:
ln(P2/1.00 atm) = (-29,540 J/mol / 8.314 J/mol·K) * (1/(37.8 + 273.15 K) - 1/(59.6 + 273.15 K))
Simplifying the equation further:
ln(P2/1.00) = (-29,540 J/mol / 8.314 J/mol·K) * (1/311.95 K - 1/332.75 K)
ln(P2) = -3.5566 * (0.0032047 - 0.0030016)
ln(P2) = -3.5566 * 0.0002031
ln(P2) = -0.0007247
Now, we can find P2 by taking the exponential of both sides:
P2 = e^(-0.0007247)
Using a scientific calculator, we find:
P2 ≈ 0.9993 atm
Therefore, the vapor pressure of bromine at 37.8 °C is approximately 0.9993 atm.
To calculate the vapor pressure of bromine at 37.8 °C, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1),
where P1 and P2 are the vapor pressures at temperatures T1 and T2, ΔHvap is the heat of vaporization, R is the gas constant (8.314 J/mol*K), and T1 and T2 are the temperatures in Kelvin.
First, let's convert the temperatures to Kelvin:
T1 = 59.6 °C + 273.15 = 332.75 K (boiling point of bromine)
T2 = 37.8 °C + 273.15 = 311.95 K
Next, we can substitute the given values into the equation:
ln(P2/1.00 atm) = (-29.54 kJ/mol / (8.314 J/mol*K)) * (1/311.95 K - 1/332.75 K)
Now, we can solve for ln(P2/1.00 atm):
ln(P2/1.00 atm) = (-29.54 kJ/mol / (8.314 J/mol*K)) * (0.003199 - 0.002931)
Calculating the values inside parentheses:
ln(P2/1.00 atm) = (-29.54 kJ/mol / (8.314 J/mol*K)) * 0.000268
Multiplying the values in the parentheses:
ln(P2/1.00 atm) = -0.96512
To isolate P2, we need to take the exponential of both sides of the equation:
e^ln(P2/1.00 atm) = e^(-0.96512)
Simplifying,
P2/1.00 atm = 0.38003
Finally, we can solve for P2 by multiplying both sides by 1.00 atm:
P2 = 1.00 atm * 0.38003
P2 = 0.38003 atm
Therefore, the vapor pressure of bromine at 37.8 °C is approximately 0.38003 atm.