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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of water and the normal boiling point of water the molar entropy change accompanying fussion is 22.0 and that accompanying evaporation

Sagot :

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is [tex]-22Jmole^-^1k^-^1[/tex]

(b) The normal boiling point of water (J·K−1·mol−1) is [tex]-109Jmole^-^1K^-^1[/tex]

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      [tex](\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p[/tex] = [tex]-5_m(\beta )+5_m(\alpha )[/tex] =  [tex]-\frac{\Delta H}{T}[/tex]

 [tex]\alpha ,\beta[/tex] → phases

ΔH → enthalpy of transition

T → temperature transition

 [tex](\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p[/tex] =[tex]= -\frac{\Delta_fH}{T_f}[/tex]

            = [tex]\frac{-6.008kJ/mole}{273.15K}[/tex] ( [tex]\Delta_fH[/tex] is the enthalpy of fusion of water)

           = [tex]-22Jmole^-^1k^-^1[/tex]

(b) [tex](\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}[/tex]

                                  = [tex]\frac{40.656kJ/mole}{373.15K}[/tex] ([tex]\Delta_v_a_p_o_u_rH[/tex] is the enthalpy of vaporization)

                               = [tex]-109Jmole^-^1K^-^1[/tex]

(c) [tex]\Delta\mu =\Delta\mu(l)-\Delta\mu(s)[/tex] =[tex]-S_m\DeltaT[/tex]

[tex][\mu(l-5[/tex]°[tex]C)-\mu(l,0[/tex]°[tex]C)][/tex] =  [tex][\mu(s-5[/tex]°[tex]C)-\mu(s,0[/tex]°[tex]C)][/tex][tex]=-S_m[/tex]ΔT

[tex]\mu(l,-5[/tex]°[tex]C)-\mu(s,-5[/tex]°[tex]C)=-Sm\DeltaT [\mu(l,0[/tex]

[tex]\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)[/tex]

     = 109J/mole

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