Problem 113 - Chemistry The Central Science In Si Units

Question

At \(35^{\circ} \mathrm{C}\) the vapor pressure of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is 47.9 \(\mathrm{kPa}\), and that of carbon disulfide, \(\mathrm{CS}_{2}\), is \(66.7 \mathrm{kPa}\). A solution composed of an equal number of moles of acetone and carbon disulfide has a vapor pressure of \(86.7 \mathrm{kPa}\) at \(35^{\circ} \mathrm{C} .(\mathbf{a})\) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Based on the behavior of the solution, predict whether the mixing of acetone and carbon disulfide is an exothermic \(\left(\Delta H_{\text {soln }}<0\right)\) or endothermic \(\left(\Delta H_{\text {soln }}>0\right)\) process.

Step-by-Step Solution

Answer

(a) 57.3 kPa is the ideal vapor pressure. (b) The mixing is endothermic, \( \Delta H_{\text{soln}} > 0 \).

Step 1: Define Ideal Behavior

For a solution exhibiting ideal behavior, the vapor pressure can be found using Raoult's Law: \[ P_{ ext{solution}} = X_{ ext{acetone}} \cdot P_{ ext{acetone}}^0 + X_{ ext{CS}_2} \cdot P_{ ext{CS}_2}^0 \]where \( X \) is the mole fraction and \( P^0 \) is the pure component's vapor pressure.

Step 2: Calculate Mole Fractions

Since we have an equal number of moles of acetone and carbon disulfide, the mole fractions for each component are both \( X_{ ext{acetone}} = X_{ ext{CS}_2} = 0.5 \).

Step 3: Calculate Ideal Vapor Pressure

Substituting the mole fractions and vapor pressures into Raoult's Law gives:\[P_{ ext{solution}} = 0.5 \cdot 47.9 \, \text{kPa} + 0.5 \cdot 66.7 \, \text{kPa} \]\[P_{ ext{solution}} = 23.95 \, \text{kPa} + 33.35 \, \text{kPa} = 57.3 \, \text{kPa}\]Thus, the ideal vapor pressure is 57.3 kPa.

Step 4: Compare with Actual Vapor Pressure

The actual vapor pressure of the solution is 86.7 kPa, which is higher than the ideal vapor pressure of 57.3 kPa. This indicates a deviation from ideal behavior.

Step 5: Determine Enthalpy Change

Since the actual vapor pressure is greater than the ideal, this suggests the solution exhibits positive deviation from ideality. Positive deviation is typically associated with an endothermic mixing process, where \( \Delta H_{\text{soln}} > 0 \).

Key Concepts

Vapor Pressure

Vapor pressure is a measure of how much a liquid's particles tend to escape into the gas phase. It's the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. For instance, in your exercise, acetone and carbon disulfide have their own vapor pressures defined at 35°C: 47.9 kPa and 66.7 kPa, respectively.
Understanding vapor pressure is crucial because it helps predict how a liquid will behave when mixed with another. A high vapor pressure indicates that the substance evaporates easily, while a low vapor pressure suggests it's less volatile. When you combine two liquids, their individual vapor pressures contribute to the overall vapor pressure of the mixture. This forms the basis of Raoult's Law, blending in their mole fractions as next steps show, to determine the behavior of solutions under ideal conditions.

Ideal Behavior

When we talk about ideal behavior in solutions, we're discussing how substances act when they completely follow theoretical predictions, specifically Raoult's Law.
According to Raoult’s Law, the vapor pressure of an ideal solution equals the sum of the individual vapor pressures of the components, each weighted by its mole fraction. Mathematically, it's expressed as:

• \( P_{\text{solution}} = X_{\text{acetone}} \cdot P_{\text{acetone}}^0 + X_{\text{CS}_2} \cdot P_{\text{CS}_2}^0 \)

The exercise asks us to calculate the vapor pressure assuming ideal behavior. By computing the weighted contributions of each component's vapor pressure, we determine the theoretical vapor pressure of a perfect mixture.
Ideal behavior assumes no interaction beyond individual contributions, meaning the molecules don't attract or repel each other. This setting is extremely instrumental in understanding deviations as well, where real solutions show differences because of molecular interactions not accounted for in the ideal scenario.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It conveys the number of moles of a component relative to the total moles in the mixture. It's a dimensionless quantity, calculated by dividing the moles of one component by the total number of moles in the solution.
In the exercise, the solution comprises equal moles of acetone and carbon disulfide, which leads us to straightforwardly define their mole fractions as both 0.5. This simplicity eases the calculations when applying Raoult's Law. The mole fraction essentially relays the proportionate presence and contribution of each substance to overall properties, such as vapor pressure.
Understanding mole fractions is key because it directly relates the pure component properties to the mixed behavior of solutions. The deviation in expected outcomes often hints at complex interactions and paves the way to explore further thermal effects, like exothermic or endothermic processes, as seen in your original problem set.