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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

Respuesta :

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.

[tex]\text{Moles of A}=\frac{\text{Mass of A}}{\text{Molar mass of A}}=\frac{20g}{100g/mol}=0.2mole[/tex]

[tex]\text{Moles of B}=\frac{\text{Mass of B}}{\text{Molar mass of B}}=\frac{80g}{50g/mol}=1.6mole[/tex]

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.

[tex]\text{Enthalpy of mixing of A and B}=(\text{Moles of A}\times \text{Enthalpy of pure A})+(\text{Moles of B}\times \text{Enthalpy of pure B})[/tex]

[tex]\text{Enthalpy of mixing of A and B}=(0.2mol\times 10kJ/mol)+(1.6mol\times 6kJ/mol)[/tex]

[tex]\text{Enthalpy of mixing of A and B}=11.6kJ[/tex]

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 = [tex]\frac{9.8kJ}{1.8mol}=5.4kJ/mol[/tex]

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

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