Answer :
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