Question

Calculate the Gibbs energy, entropy, and enthalpy of mixing when 2.00 mol C6H14

(hexane) is mixed with 3.00 mol C7H16 (heptane) at 298 K; treat the solution as ideal

Answer 1

First we calculate the mole fraction of hexane and heptane $XC_6H_4=\frac{nC_6H_{14}}{nC_6H_{14}+nC_7H_{15}}=\frac{1mol} {1mol+1mol} =0.5$

$XC_7H_{16}=1-0.5 mol= 0.5 mol$

Calculating Gibbs energy using the

$G_{mix} =nR(nC_6H_{14}​InxC_4H_{14}+xC_7​H_{16​}InxC_7H_{16}) ​$

(1+1)mol×(8.3145J/k/mol)(298k)×[0.5(In0.5)+0.5(In0.5)]

=3.43kJ/mol

Now we calculate entropy

$S_{mix} =-nR(nC_6H_{14}​InxC_4H_{14}+xC_7​H_{16​}InxC_7H_{16})$

(1+1)mol×(8.3145J/k/mol)*298k×[0.5(In0.5)+0.5(In0.5)]

=3.43J/K

Then we calculate enthalphy of mixing using

$H_{mix}=G_{mix}+S_{mix}$

= -3.43 + 3.43

=0J/mol