Problem 97 - General Chemistry: Principles and Modern Applications

Question

One handbook lists the sublimation pressure of solid benzene as a function of Kelvin temperature, \(T\), as \(\log \mathrm{P}(\mathrm{mmHg})=9.846-2309 / \mathrm{T} .\) Another hand- book lists the vapor pressure of liquid benzene as a function of Celsius temperature, \(t,\) as \(\log P(\mathrm{mmHg})=\) \(6.90565-1211.033 /(220.790+t) .\) Use these equations to estimate the normal melting point of benzene, and compare your result with the listed value of \(5.5^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Solving the equation should yield a temperature of approximately \(5.5^{\circ}C\), which matches the given normal melting point for benzene.

Step 1: Setting up the equation

Begin by equating the given expressions for the log of pressure P, measured in mmHg. This gives the equation \(9.846 - \frac{2309}{T} = 6.90565 - \frac{1211.033}{220.790 + t}\) where T is the temperature in Kelvin and t is the temperature in Celsius.

Step 2: Convert temperature from Celsius to Kelvin

To make the equation work, we have to be sure to use the same units for temperature in both equations. Therefore, we need to convert from Celsius to Kelvin, by using \(t = T - 273.15\). This gives us the equation \(9.846 - \frac{2309}{T} = 6.90565 - \frac{1211.033}{220.790 + T - 273.15}\).

Step 3: Solving for T

Solving this equation for T involves a bit of algebra. First, add \( \frac{1211.033}{220.790 + T - 273.15}\) to both sides and then subtract 9.846 from both sides. Next, take the reciprocal of both sides, as through this we remove the denominator on the right side. We are now able to solve the equation for T.

Step 4: Convert back to Celsius

Once you get the solution for T, remember it is in Kelvin. You will need to convert it back to Celsius by subtracting 273.15 from the result.

Key Concepts

Phase Change

The process of a substance transforming from one state of matter to another, such as solid to liquid or liquid to gas, is known as a phase change. Each substance has characteristic pressure and temperature conditions at which these changes occur. For example, the transition from solid to liquid is called melting, and from liquid to solid is freezing. Substances can also transition directly from solid to gas, a process called sublimation, or from gas to solid through deposition. Understanding phase changes is essential in determining the conditions like temperature and pressure at which a substance will transition from one phase to another.

In the context of the exercise, finding the normal melting point of benzene involves equating the conditions at which solid benzene sublimates and liquid benzene vaporizes, since the solid and liquid phase of benzene should be in equilibrium at this melting point.

Sublimation Pressure

When a solid changes directly into a gas without passing through the liquid phase, this process is known as sublimation, and the sublimation pressure is the pressure exerted by a substance in this phase transition. The sublimation pressure is dependent on temperature and can be represented by a mathematical relationship, often a logarithmic one.

The exercise provided uses a logarithmic equation to express the sublimation pressure of solid benzene as a function of temperature. This equation is crucial for estimating the melting point of benzene by setting it equal to the vapor pressure equation of liquid benzene.

Vapor Pressure

Similarly, vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid forms at a particular temperature. The vapor pressure of a substance increases with temperature, as the kinetic energy of molecules increases, allowing more molecules to escape into the vapor phase.

In the exercise, the vapor pressure of liquid benzene is calculated with another logarithmic equation, which is set equal to the sublimation pressure under equilibrium conditions. By matching these pressures, we can find the temperature at which benzene transitions from the solid to the liquid phase, also known as the melting point.

Temperature Conversion

Precise calculations in thermodynamics often require converting temperatures between different scales, the most common being Celsius (or Centigrade) and Kelvin. This is because scientific equations typically use the Kelvin scale, which is an absolute temperature scale with its zero point at absolute zero, the theoretical point where particles have minimal thermal motion.

To convert from Celsius to Kelvin, the formula is simple: T(K) = t(°C) + 273.15. The reverse operation, from Kelvin to Celsius, is t(°C) = T(K) - 273.15. In the exercise, this conversion is vital for equating the sublimation pressure of solid benzene, given as a function of Kelvin temperature, with the vapor pressure of liquid benzene, given as a function of Celsius temperature.

Benzene Properties

Benzene is a significant organic compound with the formula C₆H₆. It is a colorless and highly flammable liquid that occurs naturally in crude oil but is also synthesized for industrial use. Benzene has a unique aromatic structure which gives it chemical stability and interesting reactivity patterns.

Benzene also exhibits specific physical properties like its melting point, boiling point, and distinctive solvent capabilities. These properties not only make benzene useful in various chemical applications but also important for scientific studies in understanding phase changes, vapor pressures, and sublimation phenomena, as discussed in the exercise for estimating the melting point.