Question

A 20 wt% A solution is obtained by mixing A component with B in an insulated mixer at steady state. For every mole of solution 1 kJ is removed to keep the system temperature constant. Determine the enthalpy of mixture for this solution. Molecular weight of A: 100 g/mol; Molar Enthalpy of Pure A: 10 kJ/mol Molecular weight of B: 50 g/mol; Molar Enthalpy of Pure B: 6 kJ/mol

Answer 1

The enthalpy of mixture for the solution is 11.6 kJ.

Step-by-step explanation:

To determine the enthalpy of mixture for the solution, we need to calculate the moles of A and B in the solution and then use their respective molar enthalpies to calculate the overall enthalpy change.

Let's assume we have 100 g of the solution. Since it is a 20 wt% A solution, 20 g of it is A and 80 g is B.

First, we calculate the moles of A and B:

Moles of A = mass of A / molar mass of A = 20 g / 100 g/mol = 0.2 mol

Moles of B = mass of B / molar mass of B = 80 g / 50 g/mol = 1.6 mol

Next, we calculate the overall enthalpy change:

Enthalpy of mixture = (Moles of A * Molar Enthalpy of Pure A) + (Moles of B * Molar Enthalpy of Pure B)

Enthalpy of mixture = (0.2 mol * 10 kJ/mol) + (1.6 mol * 6 kJ/mol) = 2 kJ + 9.6 kJ = 11.6 kJ

Answer 2

Answer : The enthalpy of mixture for this solution is, 5.4 kJ/mol

Explanation :

As we are given that 20 wt% of A solution. That means,

In 100 g of solution, there 20 g of A and 80 g of B.

Now we have to calculate the moles of A and B.

`\text{Moles of A}=\frac{\text{Mass of A}}{\text{Molar mass of A}}=(20g)/(100g/mol)=0.2mole`

`\text{Moles of B}=\frac{\text{Mass of B}}{\text{Molar mass of B}}=(80g)/(50g/mol)=1.6mole`

Now we have to calculate the total number of moles of solution.

Total number of moles of solution = Moles of A + Moles of B

Total number of moles of solution = 0.2 + 1.6

Total number of moles of solution = 1.8 mole

Now we have to calculate the amount of energy removed.

As, 1 mole of solution released energy = 1 kJ

So, 1.8 moles of solution released energy = 1.8 × 1 kJ = 1.8 kJ

Now we have to calculate the enthalpy of mixing of A and B.

`\text{Enthalpy of mixing of A and B}=(\text{Moles of A}* \text{Enthalpy of pure A})+(\text{Moles of B}* \text{Enthalpy of pure B})`

`\text{Enthalpy of mixing of A and B}=(0.2mol* 10kJ/mol)+(1.6mol* 6kJ/mol)`

`\text{Enthalpy of mixing of A and B}=11.6kJ`

Now we have to calculate the actual enthalpy of mixing of solution.

Actual enthalpy of mixing of solution = 11.6 - 1.8 = 9.8 kJ

Now we have to calculate the enthalpy of mixture in kJ/mol.

Enthalpy of mixture =
`(9.8kJ)/(1.8mol)=5.4kJ/mol`

Therefore, the enthalpy of mixture for this solution is, 5.4 kJ/mol