Question

By how much is the chemical potential of benzene reduced at 25*C by a solute that is present at a mole fraction of 0.10?
Express answer in J/mol.

Answer 1

The chemical potential of any substance in a solution can be calculated as:

\(\mu_{A}=\mu_{A}^{o}+R T \ln \left(\chi_{A}\right)\)

where \(\mu_{A}\) is the chemical potential of A in the

\(\mu_{A}^{o}\) is the chemical potential of pure \(A\),

\(\mathrm{R}\) is the universal gas constant ( \(J / K . m o l)\)

\(\mathrm{T}\) is the temperature (in \(\mathrm{K}\) )

\(\chi_{A}\) is the mole fraction of \(\mathrm{A}\),

Now, given a solution of benzene at \(25^{\circ} \mathrm{C}\) with mole fraction of solute \(=0.10\)

i) Mole fraction of benzene,

We know that, total mole fraction \(=1\)

Therefore,

\(\chi_{\text {Benzene }}+\chi_{\text {Solute }}=1\)

\(\chi_{\text {Benzene }}=1-\chi\) Solute

\(\chi_{\text {Benzene }}=1-0.10\)

\(\chi_{B \text { enzene }}=0.90\)

ii) Temperature in Kelvin,

We know,

\(\mathrm{T}(\) in \(\mathrm{K})=273+\mathrm{T}\left(\mathrm{in}^{\circ} \mathrm{C}\right)\)

\(\mathrm{T}(\) in \(\mathrm{K})=273+25\)

\(\mathrm{T}(\) in \(\mathrm{K})=298 \mathrm{~K}\)

We are supposed to find by how much chemical potential is reduced,

Therefore,

\(\mu_{\text {Benzene }}=\mu_{\text {Benzene }}^{o}+R T \ln \left(\chi_{\text {Benzene }}\right)\)

\(-R T \ln \left(\chi_{\text {Benzene }}\right)=\mu_{\text {Benzene }}^{o}-\mu_{\text {Benzene }}\)

\(-R T \ln \left(\chi_{\text {Benzene }}\right)=\Delta \mu_{\text {Benzene }}\)

\(\Delta \mu_{\text {Benzene }}=-R T \ln \left(\chi_{\text {Benzene }}\right)\)

\(\Delta \mu_{\text {Benzene }}=-(8.314 J / K . \mathrm{mol})(298 K) \ln (0.90)\)

\(\Delta \mu_{\text {Benzene }}=-(8.314 J / K . m o l)(298 . K) \ln (0.90)\)

\(\Delta \mu_{\text {Benzene }}=-(8.314)(298) \ln (0.90) \mathrm{J} / \mathrm{mol}\)

\(\Delta \mu_{\text {Benzene }}=261.04 \mathrm{~J} /\)

Answer 2

chemtral poteutal is ben z 二(8-314x299). Inc.I