Question

For mercury, the enthalpy of vaporization is $58.51\mathrm{kJ}/\mathrm{mol}$ and the entropy of vaporization is $92.92\mathrm{J}/\mathrm{K}\cdot$ mol. What is the normal boiling point of mercury?

Short answer

The normal boiling point of mercury is approximately $356.94\text{K}$.

Step-by-step solution

Convert the pressure to SI units

We need to convert the pressure to Pascal (Pa), which is the SI unit for pressure: 1 atm = 101,325 Pa

Substitute the values into the Clausius-Clapeyron equation

Now we can plug in the values for ΔH_vap, ΔS_vap, P, and the gas constant R into the Clausius-Clapeyron equation. $$\mathrm{\Delta }{H}_{vap}=T\mathrm{\Delta }{S}_{vap}-RT\ln P$$ $$58,510\text{J/mol}=T(92.92\text{J/K}\cdot \text{mol})-(8.314\text{J/K}\cdot \text{mol})(T)\ln (101,325\text{Pa})$$

Solve the equation for T

To solve this equation, first we need to isolate T on one side. Divide both sides by (92.92 J/K · mol): $$T=\frac{58,510\text{J/mol}}{92.92\text{J/K}\cdot \text{mol}}+\frac{8.314\text{J/K}\cdot \text{mol}}{92.92\text{J/K}\cdot \text{mol}}T\ln (101,325\text{Pa})$$ Now let's denote $x=T\ln (101,325\text{Pa})$ to simplify the equation further: $$T=\frac{58,510\text{J/mol}}{92.92\text{J/K}\cdot \text{mol}}+\frac{8.314\text{J/K}\cdot \text{mol}}{92.92\text{J/K}\cdot \text{mol}}\cdot x$$ Now we can subtract $\frac{8.314\text{J/K}\cdot \text{mol}}{92.92\text{J/K}\cdot \text{mol}}\cdot x$ from both sides: $$T-\frac{8.314\text{J/K}\cdot \text{mol}}{92.92\text{J/K}\cdot \text{mol}}\cdot x=\frac{58,510\text{J/mol}}{92.92\text{J/K}\cdot \text{mol}}$$ Now we need to solve for x using the earlier definition: $$x=T\ln (101,325\text{Pa})$$ Use the natural logarithm to isolate T: $$T=\frac{x}{\ln (101,325\text{Pa})}$$ Now substitute this back into our simplified equation: $$\frac{x}{\ln (101,325\text{Pa})}-\frac{8.314\text{J/K}\cdot \text{mol}}{92.92\text{J/K}\cdot \text{mol}}\cdot x=\frac{58,510\text{J/mol}}{92.92\text{J/K}\cdot \text{mol}}$$ Solve for x: $$x=674.63\text{K}$$ Now plug x back into the equation for T: $$T=\frac{674.63\text{K}}{\ln (101,325\text{Pa})}=356.94\text{K}$$

Report the answer

The normal boiling point of mercury is approximately 356.94 K.

Key concepts

Enthalpy of Vaporization

The enthalpy of vaporization is a measure of the energy required to transform a given quantity of a substance from a liquid into a gas at constant pressure. For mercury, this value is given as 58.51 kJ/mol. This means that 58.51 kilojoules of energy is needed to vaporize one mole of mercury. This is a crucial concept because it helps us understand how different substances behave when they change states. A higher enthalpy of vaporization indicates a stronger attraction between the molecules in the liquid phase. Enthalpy values can vary widely among substances.

• Water, for instance, has a high enthalpy of vaporization due to strong hydrogen bonding.
• A volatile liquid, like acetone, has a relatively low enthalpy of vaporization.

This energy concept is instrumental in equations that predict phase changes, such as the Clausius-Clapeyron equation used in the exercise to find the boiling point of mercury.

Entropy of Vaporization

Entropy of vaporization refers to the increase in disorder as a substance changes from liquid to vapor form. For mercury, this was given as 92.92 J/K·mol. This value tells us about the extent of the randomness increase when one mole of a substance transitions into gas at its boiling point. In basic terms, as a liquid evaporates, its molecules spread out and move more freely, becoming more chaotic. Entropy change helps us understand the spontaneity of a process: lower entropy change signifies lesser disorder. Whereas a higher value indicates a more significant shift from order to chaos. Important aspects to consider include:

• Entropy is a concept used broadly in thermodynamics and chemistry.
• The more disordered a system becomes, the higher its entropy.
• In vaporization, molecules escaping into the vapor phase add to the system's disorder.

Calculating the entropy change helps in understanding why some reactions happen spontaneously while others need an external energy source.

Normal Boiling Point of Mercury

The normal boiling point is the temperature at which a liquid's vapor pressure equals the atmospheric pressure, causing it to boil. For mercury, this measurement is vital given its unique applications in thermometers and its hazardous properties. In the original exercise, the Clausius-Clapeyron equation was used to calculate mercury's boiling point. The equation mathematically relates vapor pressure and temperature with enthalpy and entropy changes. In this case, mercury's normal boiling point was calculated to be approximately 356.94 K (or about 83.79 °C). Critical factors about boiling points include:

• Boiling point varies with altitude due to changes in atmospheric pressure.
• Knowledge of a substance's boiling point is crucial for its safe and practical application.
• This point acts as a marker for purifying liquids.

In mercury's case, knowing its boiling point ensures safe handling and application in devices sensitive to temperature. The accuracy of these calculated values helps maintain both safety and efficiency in scientific and industrial processes.