Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

For the binary ideal solution having 1 mole of \(\mathrm{A}\) and 2 moles of B at \(298 \mathrm{~K}\) which one is correct? (a) \(\Delta \mathrm{S}_{\text {mix }}=-\mathrm{R} \sum \mathrm{x}_{1} \ln \mathrm{x}_{\mathrm{i}}\) (b) \(\Delta \mathrm{H}_{\text {mix }}=0\) (c) \(\Delta \mathrm{G}_{\text {mix }}=\mathrm{RT} \sum \mathrm{x}_{\mathrm{i}} \ln \mathrm{x}_{\mathrm{i}}\) (d) \(\Delta \mathrm{H}_{\operatorname{mix}}=(+)\) ve

Step-by-Step Solution

Answer

Options (a) and (b) are correct for the given ideal binary solution.

Step 1: Calculate Mole Fractions

To determine if the given options are correct, we need the mole fractions of the components in the solution. The mole fraction of a component A is calculated using the formula:\( x_A = \frac{n_A}{n_A + n_B} \), where \( n_A \) and \( n_B \) are the number of moles of A and B respectively. For component A: \( x_A = \frac{1}{1 + 2} = \frac{1}{3} \). For component B: \( x_B = \frac{2}{1 + 2} = \frac{2}{3} \). Therefore, mole fractions are \( x_A = \frac{1}{3} \) and \( x_B = \frac{2}{3} \).

Step 2: Assess Entropy of Mixing \( (\Delta S_{mix}) \)

\( \Delta S_{mix} \) is given by the formula:\[ \Delta S_{mix} = -R \sum x_i \ln x_i \]\( \Delta S_{mix} = -R (x_A \ln x_A + x_B \ln x_B) \)Substitute the mole fractions:\( \Delta S_{mix} = -R \left( \frac{1}{3} \ln \frac{1}{3} + \frac{2}{3} \ln \frac{2}{3} \right) \)Therefore, the formula is correctly stated. The option (a) is correct.

Step 3: Assess Enthalpy of Mixing \( (\Delta H_{mix}) \)

For ideal solutions, there is no enthalpy change upon mixing. This means \( \Delta H_{mix} = 0 \).Hence, the option (b) is correct.

Step 4: Assess Gibbs Free Energy of Mixing \( (\Delta G_{mix}) \)

\( \Delta G_{mix} \) is given by the formula:\[ \Delta G_{mix} = RT \sum x_i \ln x_i \]\( \Delta G_{mix} = RT (x_A \ln x_A + x_B \ln x_B) \)Therefore, option (c) is incorrect since it should be \( \Delta G_{mix} = RT \sum x_i \ln x_i \).

Step 5: Determine the Correctness of \( (\Delta H_{mix} = (+) \text{ve}) \)

As mentioned earlier, for an ideal solution, the enthalpy of mixing, \( \Delta H_{mix} \), is zero, not positive. Therefore, option (d) is incorrect.

Key Concepts

Mole Fraction Calculation

Mole fractions are a way to express the concentration of a component in a mixture. It is a dimensionless number indicating the ratio of the number of moles of a component to the total number of moles in the mixture. To calculate the mole fraction, you use the formula:
\[ x_i = \frac{n_i}{n_{total}} \]
where:

• \( n_i \) is the number of moles of component \( i \).
• \( n_{total} \) is the total number of moles in the solution.

In our case, we have 1 mole of \( A \) and 2 moles of \( B \), giving a total of 3 moles. Therefore:

• For component \( A \), \( x_A = \frac{1}{3} \).
• For component \( B \), \( x_B = \frac{2}{3} \).

These values help in further calculations like entropy and Gibbs free energy calculations.

Entropy of Mixing

Entropy of mixing, \( \Delta S_{mix} \), quantifies the increase in randomness or disorder when two substances are mixed. For an ideal solution, it can be calculated using the formula:
\[ \Delta S_{mix} = -R \sum x_i \ln x_i \]
Here, \( R \) is the universal gas constant (8.314 J/(mol K)). This formula accounts for the contribution of each component's concentration in terms of its mole fraction.

• Plug in \( x_A = \frac{1}{3} \) and \( x_B = \frac{2}{3} \) into the equation.
• Calculate \( \Delta S_{mix} = -R \left( \frac{1}{3} \ln \frac{1}{3} + \frac{2}{3} \ln \frac{2}{3} \right) \).

This calculation tells us the degree to which disorder increases, signifying that option (a) from the exercise is indeed correct.

Enthalpy Change

In an ideal solution, the enthalpy change, \( \Delta H_{mix} \), upon mixing is zero. This is because the intermolecular interactions in ideal solutions are similar to those in the pure components. Thus, no energy is absorbed or released upon mixing.

• The principle behind ideal solutions is that the attraction between unlike molecules is equal to the attraction between like molecules.
• This results in no net change in enthalpy, or \( \Delta H_{mix} = 0 \).

This means option (b) in the exercise correctly represents this concept. Additionally, this clarification invalidates option (d), which incorrectly suggests a positive enthalpy.

Gibbs Free Energy Calculation

The Gibbs free energy of mixing, \( \Delta G_{mix} \), for an ideal solution, is expressed as:
\[ \Delta G_{mix} = RT \sum x_i \ln x_i \]
This expression calculates the spontaneity and thermal feasibility of the mixing process. Here, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

• This equation shows that mixing decreases free energy, indicating a spontaneous process.
• It is important to note that the correct form of the free energy of mixing involves subtracting \( RT \sum x_i \ln x_i \) from zero, not adding, making the process favorable.

Although the given option (c) had the correct qualitative statement, remember the actual calculation ensures spontaneous mixing in terms of decreased Gibbs free energy.