Clausius–Clapeyron equation: shape of phase diagrams makes no sense
This is a summary of the equations to use to calculate phase transitions.

The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion.

The Clausius-Clapeyron equation describes solid-vapour and liquid-vapour changes because the final volume is far greater than the initial one, and is $\displaystyle \frac{dp}{dT}=p\frac{\Delta H}{RT^2}$ where $\Delta H$ the enthalpy change at the liquid–vapour or sublimation transition. Integrating this last equation from pressure $p_1 \to p_2$ and temperature $T_1 \to T_2$ gives $\displaystyle \ln\left(\frac{p_2}{p_1} \right) = -\frac{\Delta_{vap}H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right) $.

The change in the volume during fusion is $\displaystyle \Delta_{fus}V = m\left(\frac{1}{d_l}-\frac{1}{d_s} \right)$ where $m$ is the molar mass and $d_l$ and $d_s$  the densities of the liquid and solid. The pressure variation for the solid to liquid (melting or fusion) change is

$$\displaystyle p_2=p_1+\frac{\Delta_{fus}H}{\Delta_{fus}V}\ln\left(\frac{T_2}{T_1} \right)$$

and for evaporation and sublimation

$$\displaystyle p_2=p_1\exp\left( -\frac{\Delta_{vap}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1} \right) \right)$$

with the appropriate $\Delta H$. This is $\Delta_{vap}H$ for evaporation and $\Delta_{fus}H + \Delta_{vap}H$ for sublimation. Sublimation is treated as two steps merged into one; melting and instantaneous evaporation.  

The $p$ vs $T$ plot for benzene is shown in the figure. Notice how this is different to how these are generally shown. This is sometimes due to the fact that log pressure is sometimes plotted but this is not always shown on the figure. Notice also how the solid liquid line is effectively vertical.

 

The data used is

R = 8.314  #( J/mol/K)
dens_sol = 981.0  #( kg/m^3)
dens_liq = 879.0  #( kg/m^3)
mol_mass = 78.0/1000.0 #( kg/mol)
DH_vap   = 30.8*1000   #( J/mol)
DH_fus   = 10.6*1000   #( J/mol)
p3 = 36.0/760*101325 #( triple point pressure Pa)
T3 = 5.5 + 273.16    #( triple point temperature K)
DV_fus = mol_mass*(1/dens_liq-1/dens_sol)                          # delta volume fusion

Python/numpy  functions for the pressure are:  
p_liq_vap= lambda T: p3*np.exp( (DH_vap/R)*(1/T3-1/T))             
p_sol_vap= lambda T: p3*np.exp( ((DH_fus + DH_vap)/R ) * (1/T3-1/T) ) 
p_sol_liq= lambda T: p3 + DH_fus/DV_fus*(np.log(T) - np.log(T3))